Book 1. Introduction to Olympiad Number Theory

\(153=9\cdot17\), so \(9\mid153\). \(154=11\cdot14\), so \(11\mid154\). \(221=13\cdot17\), so \(13\mid221\). All three statements are true.

Let \(a=5x\), \(b=5y\), where \(x,y\) are integers. Then \(7a+4b=35x+20y=5(7x+4y)\). Hence \(5\mid 7a+4b\).

\(1260=126\cdot10=(2\cdot3^2\cdot7)(2\cdot5)=2^2\cdot3^2\cdot5\cdot7\).

\(840=84\cdot10=(2^2\cdot3\cdot7)(2\cdot5)=2^3\cdot3\cdot5\cdot7\). Hence the number of divisors is \((3+1)(1+1)(1+1)(1+1)=32\).

If \(d\mid n\), then \(\frac nd\) is the paired divisor. Divisors come in pairs, except when \(d=\frac nd\), i.e. \(d^2=n\). Thus an unpaired divisor exists exactly when \(n\) is a square. Therefore the number of divisors is odd exactly for perfect squares.

Among three consecutive integers, one is divisible by \(3\), and at least one is even. Thus the product is divisible by \(3\) and by \(2\). Since \(\gcd(2,3)=1\), the product is divisible by \(6\).

Among four consecutive integers, one is divisible by \(3\). Also, there are two even numbers, one of which is divisible by \(4\). These two even numbers together give a factor \(8\). Therefore the product is divisible by \(8\cdot3=24\).

From \(a\mid b\), we have \(b=ak\), and from \(b\mid c\), we have \(c=bm\), where \(k,m\) are integers. Then \(c=(ak)m=a(km)\). Since \(km\) is an integer, \(a\mid c\).

We have \(n\equiv -3\pmod{n+3}\). Hence \(n^2+n+1\equiv 9-3+1=7\pmod{n+3}\). Therefore \(n+3\mid7\). Thus \(n+3=\pm1,\pm7\), so \(n=-2,-4,4,-10\). Substitution confirms all four values.

If \(p>3\), then \(p\not\equiv0\pmod3\). If \(p\equiv1\pmod3\), then \(p+8\equiv0\pmod3\); if \(p\equiv2\pmod3\), then \(p+4\equiv0\pmod3\). In both cases, one number is greater than \(3\) and divisible by \(3\), hence composite. Checking \(p=3\), we get \(3,7,11\), all prime. The answer is \(p=3\).

Possible products \((\alpha_i+1)\) are \(12\), \(6\cdot2\), \(4\cdot3\), and \(3\cdot2\cdot2\). They give candidates \(2^{11}\), \(2^5\cdot3=96\), \(2^3\cdot3^2=72\), and \(2^2\cdot3\cdot5=60\). The smallest is \(60\). Indeed, \(60=2^2\cdot3\cdot5\) has \((2+1)(1+1)(1+1)=12\) divisors.

The exponent of \(2\) can be \(1,2,3,4,5\): \(5\) choices. The exponent of \(3\) has \(3\) choices: \(0,1,2\). The exponent of \(5\) has \(2\) choices: \(0,1\). Hence there are \(5\cdot3\cdot2=30\) even divisors.

We have \(n(n^2-1)(n^2-4)=n(n-1)(n+1)(n-2)(n+2)\), the product of five consecutive integers \(n-2,n-1,n,n+1,n+2\). Among them one number is divisible by \(5\), one is divisible by \(3\), and there are enough factors of \(2\) to make \(8\): among five consecutive integers, either one is divisible by \(4\) and another is even, or one is divisible by \(8\). Hence the product is divisible by \(8\cdot3\cdot5=120\).

For every \(n\), the product of three consecutive integers is divisible by \(6\), so \(m=6\) works. If some \(m\) works for all \(n\), then at \(n=1\), it divides \(1\cdot2\cdot3=6\). Hence the greatest possible \(m\) is \(6\).

The product of four consecutive integers is always divisible by \(24\): one factor is divisible by \(3\), and the two even factors together give a factor \(8\). Thus \(24\) works. If \(m\) works for all \(n\), then at \(n=1\), it divides \(1\cdot2\cdot3\cdot4=24\). Therefore the greatest \(m\) is \(24\).

If \(2n+1\mid n^2+n+3\), then \(2n+1\mid 4(n^2+n+3)\). But \(4(n^2+n+3)=4n^2+4n+12=(2n+1)^2+11\). Hence \(2n+1\mid 11\). Since \(n\) is positive, \(2n+1\ge3\), so \(2n+1=11\), giving \(n=5\). Check: \(n^2+n+3=33\) and \(2n+1=11\), so it works.

Since \(p>3\) is prime, it is odd and not divisible by \(3\). The numbers \(p-1\) and \(p+1\) are two consecutive even numbers; one is divisible by \(4\), the other by \(2\), so their product is divisible by \(8\). Also, among \(p-1,p,p+1\), one is divisible by \(3\), but \(p\) is not, so \(3\mid(p-1)(p+1)\). Since \(8\) and \(3\) are coprime, \(24\mid p^2-1\).

If \(p\ne3\), then \(p^2\equiv1\pmod3\), so \(p^2+2\equiv0\pmod3\). For \(p>3\), this number is greater than \(3\), hence composite. For \(p=2\), \(p^2+2=6\), also composite. For \(p=3\), we get \(11\), prime. The answer is \(p=3\).

Let \(p^\alpha\) be the power of a prime \(p\) in \(a\), and \(p^\beta\) its power in \(b\). From \(a^2\mid b^2\), we get \(2\alpha\le2\beta\), hence \(\alpha\le\beta\). This holds for every prime \(p\), so all prime powers of \(a\) occur in \(b\). Therefore \(a\mid b\).

If \(n=p^2\), where \(p\) is prime, its divisors are \(1,p,p^2\), exactly \(3\). Conversely, let \(n=q_1^{\alpha_1}\cdots q_s^{\alpha_s}\). Then \(\tau(n)=(\alpha_1+1)\cdots(\alpha_s+1)=3\). Since \(3\) is prime, the product can have only one factor equal to \(3\). Hence \(s=1\) and \(\alpha_1=2\), so \(n=q_1^2\), the square of a prime.

Every number from \(1\) to \(2n\) is uniquely written as \(2^tq\), where \(q\) is odd. There are at most \(n\) possible odd values \(q\): \(1,3,5,\ldots,2n-1\). Among \(n+1\) chosen numbers, by the pigeonhole principle two have the same odd part \(q\): \(2^aq\) and \(2^bq\), with \(a

If \(\tau(n)=15\), then the possible structures are \(p^{14}\) or \(p^4q^2\). The first is not minimal and, under divisibility by \(12\), would be very large. In the second case, the larger exponent should be placed on the smaller prime. Since the number must be divisible by \(12=2^2\cdot3\), take \(2^4\cdot3^2=144\). It has \((4+1)(2+1)=15\) divisors and is divisible by \(12\). The alternative \(2^2\cdot3^4=324\) is larger, and adding other primes only increases the number. The answer is \(144\).

Modulo \(2\), always \(n^5\equiv n\), so \(2\mid n^5-n\). Modulo \(3\), checking residues \(0,1,2\) gives \(n^5\equiv n\). Modulo \(5\), for residues \(0,\pm1,\pm2\), we also get \(n^5\equiv n\). Hence \(30=2\cdot3\cdot5\mid n^5-n\) for all \(n\). On the other hand, at \(n=2\), \(2^5-2=30\), so every universal divisor \(m\) must divide \(30\). Therefore the greatest \(m\) is \(30\).

Consider the \(k\) consecutive numbers \((k+1)!+2,(k+1)!+3,\ldots,(k+1)!+(k+1)\). For each \(j=2,3,\ldots,k+1\), the number \((k+1)!\) is divisible by \(j\). Hence \((k+1)!+j\) is also divisible by \(j\). Also \((k+1)!+j>j\), so \(j\) is a nontrivial divisor of this number. Therefore all these \(k\) consecutive numbers are composite.

\(\gcd(252,198)=\gcd(198,54)=\gcd(54,36)=\gcd(36,18)=18\). The answer is \(18\).

\(84=2^2\cdot3\cdot7\), \(126=2\cdot3^2\cdot7\). Hence \(\gcd=2\cdot3\cdot7=42\), \(\operatorname{lcm}=2^2\cdot3^2\cdot7=252\).

\(96=2^5\cdot3\), \(180=2^2\cdot3^2\cdot5\). Thus \(\gcd=2^2\cdot3=12\), \(\operatorname{lcm}=2^5\cdot3^2\cdot5=1440\). We get \(96\cdot180=17280\) and \(12\cdot1440=17280\).

If \(d\mid n\) and \(d\mid n+1\), then \(d\mid(n+1)-n=1\). Hence \(d=1\). Therefore \(\gcd(n,n+1)=1\).

Any common divisor \(d\) of \(n\) and \(2n+1\) divides \(2n\), hence divides \((2n+1)-2n=1\). Therefore \(d=1\), so the GCD is \(1\).

\(\gcd(48,180)=12\), \(\operatorname{lcm}(48,180)=720\). Since both numbers are multiplied by \(n\), we get \(\gcd(48n,180n)=12n\) and \(\operatorname{lcm}(48n,180n)=720n\).

We have \(90n=18\cdot630\), so \(n=126\). Check: \(\gcd(126,90)=18\), \(\operatorname{lcm}(126,90)=630\), so the answer is correct.

Let \(d=\gcd(n,n+2)\). Then \(d\mid(n+2)-n=2\), so \(d\in\{1,2\}\). If \(n\) is even, both numbers are even and the GCD is \(2\). If \(n\) is odd, both numbers are odd, so they do not share the divisor \(2\), and the GCD is \(1\).

If \(d\mid n\) and \(d\mid n^2+n+1\), then \(d\mid n(n+1)\). Hence \(d\) divides the difference \((n^2+n+1)-n(n+1)=1\). Thus \(d=1\).

Let \(d\mid a+b\) and \(d\mid a-b\). Then \(d\mid (a+b)+(a-b)=2a\) and \(d\mid(a+b)-(a-b)=2b\). Since \(\gcd(a,b)=1\), no odd prime can divide \(d\). Therefore \(d\mid2\).

Modulo \(n+2\), we have \(n\equiv -2\). Then \(n^2+3n+5\equiv4-6+5=3\). Therefore \(\gcd(n+2,n^2+3n+5)=\gcd(n+2,3)\). It is greater than \(1\) exactly when \(3\mid n+2\), i.e. \(n\equiv1\pmod3\).

\(36n=12\cdot180\), so \(n=60\). Check: \(\gcd(60,36)=12\), \(\operatorname{lcm}(60,36)=180\). The answer is \(60\).

Suppose a prime \(p\) divides both \(a+b\) and \(ab\). From \(p\mid ab\), we get \(p\mid a\) or \(p\mid b\). If \(p\mid a\), then from \(p\mid a+b\), we get \(p\mid b\), contradicting \(\gcd(a,b)=1\). The case \(p\mid b\) is similar. Thus there are no common prime divisors, and the GCD is \(1\).

Modulo \(n+3\), we have \(n\equiv -3\), so \(n^2+1\equiv10\). Hence \(\gcd(n^2+1,n+3)=\gcd(10,n+3)\). This GCD is greater than \(1\) if \(n+3\) is divisible by \(2\) or by \(5\). Thus \(n\) is odd or \(n\equiv2\pmod5\).

Modulo \(n-1\), we have \(n\equiv1\). Then \(n^2+n+1\equiv1+1+1=3\). Therefore the GCD is \(\gcd(n-1,3)\), so it divides \(3\). It equals \(3\) exactly when \(3\mid n-1\), i.e. \(n\equiv1\pmod3\).

For numbers of the form \(a^m-1\), \(\gcd(a^m-1,a^n-1)=a^{\gcd(m,n)}-1\). Hence the required GCD is \(2^6-1=63\).

If a prime \(p\) divides \(a^m\), then \(p\mid a\). If the same \(p\) divided \(b^n\), then \(p\mid b\), contradicting \(\gcd(a,b)=1\). Thus \(a^m\) and \(b^n\) have no common prime divisors, and their GCD is \(1\).

Let \(d=\gcd(a^2+b^2,a+b)\). Then \(a\equiv -b\pmod d\), so \(a^2+b^2\equiv2b^2\pmod d\). Hence \(d\mid2b^2\). But \(\gcd(a+b,b)=\gcd(a,b)=1\), so \(\gcd(d,b)=1\). Therefore \(d\mid2\).

Modulo \(n+6\), we have \(n\equiv -6\), so \(n^2+4\equiv36+4=40\). Thus the GCD is \(\gcd(n+6,40)\). It is greater than \(1\) if \(n+6\) is divisible by \(2\) or \(5\). Therefore \(n\) is even or \(n\equiv4\pmod5\).

Let \(d=\gcd(n^2+n+1,n^2+2n+3)\). Then \(d\mid(n^2+2n+3)-(n^2+n+1)=n+2\). Modulo \(d\), we have \(n\equiv-2\), so from \(d\mid n^2+n+1\), we get \(d\mid4-2+1=3\). Hence \(d\) is \(1\) or \(3\). Both expressions are divisible by \(3\) exactly when \(n\equiv1\pmod3\). Therefore the GCD is \(3\) for \(n\equiv1\pmod3\), and \(1\) otherwise.

Let \(m\ge n\) and \(m=qn+r\). Since \(a^n\equiv1\pmod{a^n-1}\), we have \(a^m=a^{qn+r}\equiv a^r\pmod{a^n-1}\). Therefore \(\gcd(a^m-1,a^n-1)=\gcd(a^r-1,a^n-1)\). This is exactly the Euclidean step for the exponent pair \((m,n)\). Repeating, we reach \(d=\gcd(m,n)\). The GCD is \(a^d-1\).

Let \(a=12x\), \(b=12y\), \(\gcd(x,y)=1\). Then \(\operatorname{lcm}(a,b)=12xy=720\), so \(xy=60\). Since \(x\) and \(y\) are coprime, each prime power of \(60=2^2\cdot3\cdot5\) goes entirely to either \(x\) or \(y\). The unordered pairs \((x,y)\) are \((1,60)\), \((3,20)\), \((4,15)\), \((5,12)\). Hence the pairs \((a,b)\) are \((12,720)\), \((36,240)\), \((48,180)\), \((60,144)\).

In a perfect cube, all prime exponents are multiples of \(3\). For a cube to be divisible by \(18=2\cdot3^2\), the exponent of \(2\) must be at least \(3\), and the exponent of \(3\) must also be at least \(3\). Hence every such cube is divisible by \(2^3\cdot3^3=216\). This value is attainable: take all three cubes equal to \(216=6^3\). Therefore the smallest possible GCD is \(216\).

Let \(d=\gcd(a,b)\), \(a=dx\), \(b=dy\), where \(\gcd(x,y)=1\). Then \(\operatorname{lcm}(a,b)=dxy\), and \(a+b=d(x+y)\). Since \(\gcd(x+y,xy)=1\), we get \(\gcd(a+b,\operatorname{lcm}(a,b))=d\). Therefore \(d=\gcd(154,840)=14\).

Then \(a=14x\), \(b=14y\), \(x+y=11\), and \(\operatorname{lcm}(a,b)=14xy=840\), so \(xy=60\). The numbers \(x,y\) are roots of \(t^2-11t+60=0\), giving \(\{x,y\}=\{5,6\}\). Hence \(\{a,b\}=\{70,84\}\).

\(7\cdot289=2023\), so \(2026=7\cdot289+3\). The remainder is \(3\).

If \(n=2k\), then \(n^2=4k^2\equiv0\pmod4\). If \(n=2k+1\), then \(n^2=4k^2+4k+1\equiv1\pmod4\).

The squares of residues \(0,1,2,3,4,5,6,7\) modulo \(8\) are \(0,1,4,1,0,1,4,1\). Hence only \(0,1,4\) are possible.

The last digits of powers of \(3\) are \(3,9,7,1\), with cycle length \(4\). Since \(2025\equiv1\pmod4\), the last digit is \(3\).

Squares modulo \(5\): \(0^2\equiv0\), \(1^2\equiv1\), \(2^2\equiv4\), \(3^2\equiv4\), \(4^2\equiv1\). Thus \(n\equiv1\) and \(n\equiv4\pmod5\) work.

A square modulo \(4\) is \(0\) or \(1\). Hence \(x^2+y^2\) can have residue \(0,1,2\), but not \(3\). The right side has residue \(3\). Contradiction.

Squares modulo \(8\) are \(0,1,4\). A sum of two such residues can be \(0,1,2,4,5\), but not \(7\). The right side has residue \(7\). Contradiction.

The values of \(n^2+n+1\) modulo \(7\) for \(n=0,1,2,3,4,5,6\) are \(1,3,0,6,0,3,1\). Therefore \(n\equiv2\) or \(n\equiv4\pmod7\).

Every square modulo \(4\) is \(0\) or \(1\). Residues \(2\) and \(3\) do not occur, so a square cannot have these remainders.

The cycle is \(7,9,3,1\). Since \(2026\equiv2\pmod4\), take the second digit of the cycle. The answer is \(9\).

If \(n\equiv0\pmod2\), then \(n^2+n\equiv0\). If \(n\equiv1\pmod2\), then \(n^2+n\equiv1+1\equiv0\pmod2\). Thus the expression is always even.

Each cube modulo \(9\) is \(0,\pm1\). A sum of three such residues is among \(-3,-2,-1,0,1,2,3\), i.e. modulo \(9\) among \(6,7,8,0,1,2,3\). Residues \(4\) and \(5\) do not occur.

A square modulo \(8\) is \(0,1\), or \(4\). A sum of three such residues can give \(0,1,2,3,4,5,6\), but not \(7\). Hence a sum of three squares cannot be congruent to \(7\) modulo \(8\). The number \(8t+7\) has residue \(7\). Contradiction.

The right side \(3y^2+2\equiv2\pmod3\). But a square modulo \(3\) is only \(0\) or \(1\). Thus the left side cannot have residue \(2\). Contradiction.

If a number is not divisible by \(3\), its square has residue \(1\) modulo \(3\). If at least one of \(x,y\) is not divisible by \(3\), the sum of squares has residue \(1\) or \(2\), not \(0\), except when both squares are \(0\). Therefore both \(x\) and \(y\) are divisible by \(3\).

Checking residues modulo \(13\) gives zeros only for \(n\equiv3\) and \(n\equiv9\). Indeed, \(3^2+3+1=13\), \(9^2+9+1=91=7\cdot13\). Other residues give nonzero values. The answer is \(n\equiv3\) or \(9\pmod{13}\).

For \(n=0,1,\ldots,10\), the values of \(n^2+n+1\) modulo \(11\) are \(1,3,7,2,10,9,10,2,7,3,1\). Zero does not occur, so divisibility by \(11\) is impossible.

If \(x\) is even, then \(x^4\equiv0\pmod{16}\); if \(x\) is odd, then \(x^2\equiv1\) or \(9\pmod{16}\), and \(x^4\equiv1\pmod{16}\). Thus \(x^4+y^4\) can have residue \(0,1,2\), but not \(15\). The right side has residue \(15\). Contradiction.

For residues \(0,1,2,3,4,5,6\), the cubes modulo \(7\) are \(0,1,1,6,1,6,6\). Only \(0,1,6\) are possible. Residue \(2\) does not occur, so there are no solutions.

\(3^{20}=(3^4)^5=81^5\). Modulo \(100\): \(81^2\equiv61\), \(81^4\equiv61^2\equiv21\), \(81^5\equiv21\cdot81\equiv1\). Therefore the last two digits are \(01\).

The equation implies \(3\mid x^2+y^2\). By the table of squares modulo \(3\), we get \(3\mid x\) and \(3\mid y\). Let \(x=3x_1\), \(y=3y_1\). Then \(9x_1^2+9y_1^2=3z^2\), so \(z^2=3(x_1^2+y_1^2)\), and \(3\mid z\). Thus all of \(x,y,z\) are divisible by \(3\). If a nonzero solution existed, we could divide all three numbers by \(3\) indefinitely, impossible for nonzero integers. Hence no nonzero solutions exist.

Modulo \(5\), we get \(x^2\equiv2\pmod5\). But squares modulo \(5\) are only \(0,1,4\). Residue \(2\) is impossible. Therefore there are no solutions.

The quadratic residues modulo \(7\) are \(0,1,2,4\). For the sum of two such residues to be \(0\), the only possibility is \(0+0\): the negatives of \(1,2,4\) are \(6,5,3\), which are not square residues. Thus \(x^2\equiv0\) and \(y^2\equiv0\pmod7\), so \(7\mid x\) and \(7\mid y\).

From the problem on three squares modulo \(8\), a sum of three squares cannot have residue \(7\) modulo \(8\). Therefore no number of the form \(8t+7\) can be represented as \(x^2+y^2+z^2\). There are infinitely many such positive integers: \(7,15,23,31,\ldots\). Hence infinitely many required numbers exist.

\(x\equiv1-3\equiv-2\equiv5\pmod7\).

\(3\cdot5=15\equiv1\pmod7\). Hence \(x\equiv5\pmod7\).

\(5\cdot5=25\equiv1\pmod{12}\). Therefore the inverse of \(5\) modulo \(12\) is \(5\).

Numbers congruent to \(2\) modulo \(3\) are \(2,5,8,11,\ldots\). Among them, \(11\equiv1\pmod5\). Hence \(x\equiv11\pmod{15}\).

The number \(2x\) is even, so modulo \(4\) it can have residue \(0\) or \(2\). Residue \(1\) is impossible. Also \(\gcd(2,4)=2\nmid1\), so there are no solutions.

\(\gcd(4,10)=2\), and \(2\mid6\). Divide by \(2\): \(2x\equiv3\pmod5\). The inverse of \(2\) modulo \(5\) is \(3\), so \(x\equiv9\equiv4\pmod5\). Modulo \(10\): \(x\equiv4\) or \(9\).

Divide \(6,12,18\) by \(6\): \(x\equiv2\pmod3\). Modulo \(18\), these are the classes \(2,5,8,11,14,17\).

We get \(3x\equiv2\pmod5\). The inverse of \(3\) modulo \(5\) is \(2\), so \(x\equiv4\pmod5\). Modulo \(15\): \(x\equiv4,9,14\).

Among \(3,7,11,15,19,\ldots\), which are congruent to \(3\) modulo \(4\), the number \(7\) has residue \(2\) modulo \(5\). The answer is \(x\equiv7\pmod{20}\).

The first condition gives \(x\equiv2\pmod3\). The second gives \(x\equiv0\pmod3\). These cannot hold simultaneously. No solutions exist.

Substitute: \(6k+4\equiv1\pmod9\), so \(6k\equiv6\pmod9\). Divide by \(3\): \(2k\equiv2\pmod3\), hence \(k\equiv1\pmod3\). Then \(x=6(3t+1)+4=18t+10\). The answer is \(x\equiv10\pmod{18}\).

Numbers of the form \(5k+2\) are \(2,7,12,17,\ldots\). The first one with residue \(3\) modulo \(7\) is \(17\). The answer is \(17\).

The congruence is equivalent to the integer equation \(ax-my=b\). All numbers of the form \(ax-my\) are divisible by \(d=\gcd(a,m)\), so \(d\mid b\) is necessary. Conversely, if \(d\mid b\), Bezout's identity gives \(u,v\) with \(au+mv=d\). Multiplying by \(\frac bd\), we represent \(b\) as \(aX+mY\), giving a solution to the congruence.

If \(2n+1\mid n^2+n+7\), then \(2n+1\mid4(n^2+n+7)\). But \(4(n^2+n+7)=(2n+1)^2+27\). Hence \(2n+1\mid27\). For positive \(n\), \(2n+1=3,9,27\), so \(n=1,4,13\). All three work.

From \(x\equiv4\pmod{12}\), we get \(x\equiv1\pmod3\). But the first condition requires \(x\equiv2\pmod3\). Contradiction. Therefore there are no solutions.

Let \(x=8a+5\). Modulo \(9\): \(8a+5\equiv2\), so \(-a\equiv-3\), hence \(a\equiv3\pmod9\). Thus \(x\equiv29\pmod{72}\). Now \(x=29+72t\), and modulo \(5\): \(4+2t\equiv1\), so \(2t\equiv2\), \(t\equiv1\pmod5\). We obtain \(x\equiv101\pmod{360}\).

The congruence \(4x\equiv8\pmod{12}\), after division by \(4\), gives \(x\equiv2\pmod3\). Combine with \(x\equiv3\pmod5\). Among \(3,8,13,\ldots\) modulo \(5\), the number \(8\equiv2\pmod3\). Hence \(x\equiv8\pmod{15}\). Modulo \(60\), this gives \(x\equiv8,23,38,53\).

The conditions \(n\equiv1\pmod2\), \(n\equiv2\pmod3\) give \(n\equiv5\pmod6\). Checking \(5,11,17,23,\ldots\), the first number congruent to \(3\) modulo \(5\) is \(23\). The answer is \(23\).

Since \(6\mid18\), divide by \(6\): \(2x\equiv3\pmod5\). The inverse of \(2\) modulo \(5\) is \(3\), so \(x\equiv9\equiv4\pmod5\). Modulo \(30\): \(x\equiv4,9,14,19,24,29\).

The number \(7\) has residue \(3\) modulo \(4\) and residue \(7\) modulo \(25\). Since \(4\) and \(25\) are coprime, the solution is unique modulo \(100\). The answer is \(x\equiv7\pmod{100}\).

Let \(d=\gcd(m,n)\). If a solution \(x\) exists, then \(x-r\) is divisible by \(m\), and \(x-s\) is divisible by \(n\), so both are divisible by \(d\). Then \(r-s=(x-s)-(x-r)\) is divisible by \(d\).

Conversely, suppose \(d\mid r-s\). Look for \(x=r+mk\). We need \(r+mk\equiv s\pmod n\), i.e. \(mk\equiv s-r\pmod n\). This linear congruence has a solution exactly because \(\gcd(m,n)=d\mid s-r\). Hence the original system is compatible.

If \(3n+2\mid n^2+5n+9\), then it divides \(9(n^2+5n+9)=9n^2+45n+81\). Subtract \((3n+2)^2=9n^2+12n+4\): we get \(3n+2\mid33n+77\). Now \(3(33n+77)-33(3n+2)=165\), so \(3n+2\mid165\). The positive divisors of \(165\) congruent to \(2\) modulo \(3\) are \(5\) and \(11\). Thus \(3n+2=5\) or \(11\), giving \(n=1\) or \(3\). Both values check.

The first three conditions give \(n\equiv29\pmod{30}\). Let \(n=7k\). Then \(7k\equiv29\pmod{30}\). The inverse of \(7\) modulo \(30\) is \(13\), so \(k\equiv29\cdot13\equiv17\pmod{30}\). The smallest positive \(k\) is \(17\), hence \(n=119\).

Reduce \(6x\equiv12\pmod9\) by \(3\): \(2x\equiv4\pmod3\), so \(-x\equiv1\pmod3\), hence \(x\equiv2\pmod3\). Now solve the system with coprime moduli \(3,4,5,7\): \(x\equiv2\pmod3\), \(x\equiv1\pmod4\), \(x\equiv2\pmod5\), \(x\equiv3\pmod7\).

Combine \(x\equiv1\pmod4\) and \(x\equiv2\pmod5\): this gives \(x\equiv17\pmod{20}\). The number \(17\) also satisfies \(17\equiv2\pmod3\) and \(17\equiv3\pmod7\). Since \(3,4,5,7\) are pairwise coprime, the answer is unique modulo \(420\): \(x\equiv17\pmod{420}\).

Divide by \(2\): \(2x+3y=5\). One solution is \(x=1\), \(y=1\). Hence all solutions are \(x=1+3t\), \(y=1-2t\), where \(t\in\mathbb Z\). Check: \(2(1+3t)+3(1-2t)=5\).

From \(2x=18-3y\), the number \(y\) must be even. Since \(y>0\) and \(3y<18\), we have \(y=2\) or \(y=4\). Then \(x=6\) or \(x=3\). The solutions are \((6,2)\), \((3,4)\).

The positive divisors of \(24\) are \(1,2,3,4,6,8,12,24\). For each \(x\), we have \(y=\frac{24}{x}\). Thus the pairs are \((1,24)\), \((2,12)\), \((3,8)\), \((4,6)\), \((6,4)\), \((8,3)\), \((12,2)\), \((24,1)\).

We have \((x-y)(x+y)=15\). The positive factor pairs with the same parity are \((1,15)\) and \((3,5)\). They give \((x,y)=(8,7)\) and \((4,1)\). These are all solutions.

A square modulo \(4\) is \(0\) or \(1\). Therefore a difference of two squares modulo \(4\) can have residue \(0,1\), or \(3\), but not \(2\). The left-hand side cannot equal \(2\). No solutions exist.

One solution is \(x=-3\), \(y=2\), since \(7(-3)+11\cdot2=1\). All solutions are \(x=-3+11t\), \(y=2-7t\), where \(t\in\mathbb Z\).

Modulo \(5\), \(8y\equiv67\), so \(3y\equiv2\pmod5\). Hence \(y\equiv4\pmod5\). Since \(8y<67\), we have \(y\le8\), so \(y=4\). Then \(x=\frac{67-32}{5}=7\). The answer is \((7,4)\).

We get \((x+1)(y+1)=48\). Thus \(x+1\) and \(y+1\) are divisors of \(48\), both at least \(2\). The solutions are \((1,23)\), \((2,15)\), \((3,11)\), \((5,7)\), \((7,5)\), \((11,3)\), \((15,2)\), \((23,1)\).

We have \(xy-2x-5y=0\), hence \((x-5)(y-2)=10\). The positive divisors of \(10\) are \(1,2,5,10\). Therefore the solutions are \((6,12)\), \((7,7)\), \((10,4)\), \((15,3)\).

\((x-y)(x+y)=56\). The two factors must have the same parity, so both are even. The suitable pairs are \((2,28)\) and \((4,14)\). They give \((x,y)=(15,13)\) and \((9,5)\).

From \(8x+8y=xy\), we get \(xy-8x-8y=0\), hence \((x-8)(y-8)=64\). Therefore \(x=8+d\), \(y=8+\frac{64}{d}\), where \(d\) runs over the positive divisors of \(64\).

Modulo \(3\), we get \(x^2\equiv2\pmod3\). But a square modulo \(3\) can have only residue \(0\) or \(1\). This is a contradiction, so no solutions exist.

We have \(xy+2x+3y+6=60\), so \((x+3)(y+2)=60\). Since \(x\ge1\), \(y\ge1\), we need \(x+3\ge4\), \(y+2\ge3\). Checking the divisors of \(60\), we get \((1,13)\), \((2,10)\), \((3,8)\), \((7,4)\), \((9,3)\), \((12,2)\), \((17,1)\).

\((x-y)(x+y)=68\). The only suitable positive factor pair with the same parity is \((2,34)\). The pair \((4,17)\) has different parity. Thus \(x=\frac{2+34}{2}=18\), \(y=\frac{34-2}{2}=16\). The answer is \((18,16)\).

Multiplying gives \(10x+10y=xy\). Then \(xy-10x-10y=0\), and \((x-10)(y-10)=100\). Hence \(x=10+d\), \(y=10+\frac{100}{d}\), where \(d\) is a positive divisor of \(100\). This describes all solutions.

Rewrite as \((x+1)^2-y^2=21\). Then \((x+1-y)(x+1+y)=21\). The positive factor pairs are \((1,21)\) and \((3,7)\). They give respectively \((x,y)=(10,10)\) and \((4,2)\). There are no other solutions.

We have \(xy-x-2y+2=19\), so \((x-2)(y-1)=19\). Since \(19\) is prime, the possible factor pairs are \((1,19)\) and \((19,1)\). Thus \((x,y)=(3,20)\) and \((21,2)\).

A square modulo \(8\) is \(0,1\), or \(4\). A sum of two such residues can be \(0,1,2,4,5\) modulo \(8\), but not \(6\). The right-hand side has residue \(6\), a contradiction.

We get \(xy-x-y+1=2\), so \((x-1)(y-1)=2\). Hence \((x-1,y-1)=(1,2)\) or \((2,1)\). The solutions are \((2,3)\), \((3,2)\).

From \(xy-4x-4y=5\), we get \((x-4)(y-4)=21\). The positive divisors of \(21\) are \(1,3,7,21\). This gives \((5,25)\), \((7,11)\), \((11,7)\), \((25,5)\). Negative factors do not give positive \(x,y\).

The equation is equivalent to \((x-y)^2=x+y\). If \(x=y\), then \(0=2x\), impossible for positive \(x\). Suppose \(x>y\) and let \(d=x-y\). Then \(d^2=2y+d\), so \(y=\frac{d^2-d}{2}\), \(x=\frac{d^2+d}{2}\). We need \(d\ge2\). By symmetry, all solutions are \(\left(\frac{d(d+1)}{2},\frac{d(d-1)}{2}\right)\) and \(\left(\frac{d(d-1)}{2},\frac{d(d+1)}{2}\right)\), where \(d\ge2\) is an integer.

Let \(x\le y\). Then \(xy\le x+y+2\). If \(x\ge3\), then \(xy\ge3y\), while \(x+y+2\le2y+2\), impossible for \(y\ge3\). Thus \(x=1\) or \(x=2\). For \(x=1\), \(y\mid y+3\), so \(y\mid3\), and \(y=1,3\). For \(x=2\), \(2y\mid y+4\), so \(y\le4\); checking \(y=2,3,4\) leaves only \(y=4\). By symmetry, the solutions are \((1,1)\), \((1,3)\), \((3,1)\), \((2,4)\), \((4,2)\).

Factor: \((x-y)(x+y)=2(x+y)+15\). Hence \((x+y)(x-y-2)=15\). Let \(A=x+y\), \(B=x-y-2\). Then \(AB=15\), \(A>0\), \(B>0\). Check the pairs \((15,1)\), \((5,3)\), \((3,5)\), \((1,15)\). Since \(y=\frac{A-B-2}{2}\), a positive \(y\) occurs only for \((A,B)=(15,1)\). Then \(y=6\), \(x=9\). The answer is \((9,6)\).

The equation is symmetric, so assume \(x\ge y\). Then \(x^2\le x^2+xy+y^2=x+y+30\le2x+30\). Thus \(x^2-2x-30\le0\), so \(x\le6\).

Check \(x=1,2,3,4,5,6\), viewing the equation as a quadratic in \(y\). For \(x=1\), \(y^2=30\), no solution. For \(x=2\), \(y^2+y-28=0\), no positive integer \(y\). For \(x=3\), \(y^2+2y-24=0\), so \(y=4\), contradicting \(x\ge y\). For \(x=4\), \(y^2+3y-18=0\), so \(y=3\). For \(x=5\), \(y^2+4y-10=0\), no integer \(y\). For \(x=6\), \(y^2+5y=0\), no positive \(y\). Hence under \(x\ge y\) only \((4,3)\) appears. By symmetry, \((3,4)\) also appears. The solutions are \((4,3)\), \((3,4)\).

If \(a\) is odd, then \(a=2k+1\). Then \(a^2=4k^2+4k+1\), which is odd. Hence if \(a^2\) is even, \(a\) cannot be odd, so \(a\) is even.

Since \(a^2=a\cdot a\) and the prime \(p\) divides this product, Euclid's lemma gives that \(p\) divides at least one factor. Both factors are \(a\), so \(p\mid a\).

If \(x^2=2y^2\), then \(x^2\) is even, so \(x=2u\). Then \(4u^2=2y^2\), hence \(y^2=2u^2\), and \(y\) is also even: \(y=2v\). Then \(u^2=2v^2\), a smaller positive solution. Infinite descent is impossible.

Let \(\sqrt{2}=\frac{a}{b}\), where \(\gcd(a,b)=1\). Then \(a^2=2b^2\). Hence \(a\) is even, \(a=2c\). Then \(4c^2=2b^2\), so \(b^2=2c^2\), and \(b\) is even. Thus \(a\) and \(b\) have common divisor \(2\), a contradiction.

From \(x^2=3y^2\), we get \(3\mid x^2\), so \(x=3u\). Then \(9u^2=3y^2\), hence \(y^2=3u^2\). Therefore \(3\mid y\). Dividing both \(x\) and \(y\) by \(3\) gives a smaller positive solution, impossible by descent.

Let \(\sqrt{3}=\frac{a}{b}\), \(\gcd(a,b)=1\). Then \(a^2=3b^2\), so \(3\mid a\), \(a=3c\). We get \(9c^2=3b^2\), hence \(b^2=3c^2\), and \(3\mid b\). This contradicts \(\gcd(a,b)=1\).

Let \(\sqrt{5}=\frac{a}{b}\) in lowest terms. Then \(a^2=5b^2\), so \(5\mid a\). Let \(a=5c\). Then \(25c^2=5b^2\), hence \(b^2=5c^2\), and \(5\mid b\). This contradicts lowest terms.

From \(x^2=8y^2\), \(x\) is even. Let \(x=2u\). Then \(4u^2=8y^2\), so \(u^2=2y^2\). But this equation has no positive integer solutions. Hence the original equation has none.

From \(x^2=12y^2\), \(x\) is even: \(x=2u\). Then \(4u^2=12y^2\), so \(u^2=3y^2\). But \(u^2=3y^2\) has no positive integer solutions by descent with prime \(3\). Contradiction.

Squares modulo \(3\) are only \(0\) and \(1\). For the sum of two such residues to be divisible by \(3\), the only possibility is \(0+0\). Thus \(x^2\equiv0\pmod3\) and \(y^2\equiv0\pmod3\). Therefore \(3\mid x\) and \(3\mid y\).

From \(3\mid x^2+y^2\), we get \(3\mid x\) and \(3\mid y\). Let \(x=3x_1\), \(y=3y_1\). Then \(9x_1^2+9y_1^2=3z^2\), hence \(z^2=3x_1^2+3y_1^2\), so \(3\mid z\). Dividing \(x,y,z\) by \(3\) gives a smaller positive solution. Infinite descent is impossible.

Squares modulo \(5\) are \(0,1,4\). Thus \(2y^2\) can have residues \(0,2,3\). The sum \(x^2+2y^2\) can be \(0\) modulo \(5\) only as \(0+0\). Therefore \(x^2\equiv0\) and \(y^2\equiv0\pmod5\), hence \(5\mid x\), \(5\mid y\).

From \(5\mid x^2+2y^2\), we get \(5\mid x\), \(5\mid y\). Let \(x=5x_1\), \(y=5y_1\). Then \(25x_1^2+50y_1^2=5z^2\), so \(z^2=5x_1^2+10y_1^2\), hence \(5\mid z\). Dividing all variables by \(5\) gives a smaller positive solution. Descent is impossible.

The quadratic residues modulo \(7\) are \(0,1,2,4\). No two nonzero elements of \(\{1,2,4\}\) sum to \(0\) modulo \(7\). Hence \(x^2+y^2\equiv0\pmod7\) is possible only when \(x^2\equiv y^2\equiv0\pmod7\). Therefore \(7\mid x\), \(7\mid y\).

The equation implies \(7\mid x^2+y^2\), so \(7\mid x\) and \(7\mid y\). Let \(x=7x_1\), \(y=7y_1\). Then \(49x_1^2+49y_1^2=7z^2\), hence \(z^2=7x_1^2+7y_1^2\), so \(7\mid z\). Dividing all variables by \(7\) gives a smaller solution of the same type. Infinite descent is impossible.

Let \(\sqrt{6}=\frac{a}{b}\) in lowest terms. Then \(a^2=6b^2\), so \(a\) is even: \(a=2c\). We get \(4c^2=6b^2\), so \(2c^2=3b^2\). The right-hand side is even, hence \(b^2\) is even, and \(b\) is even. This contradicts lowest terms.

From \(x^2=2y^2\), \(x\) is even, \(x=2u\). Then \(y^2=2u^2\), so \(y\) is even. Thus \(x\) and \(y\) have common divisor \(2\), contradicting coprimality.

Since \(3\mid x^2\), we have \(x=3u\). Then \(9u^2=45y^2\), hence \(u^2=5y^2\). But \(u^2=5y^2\) has no positive integer solutions: it forces both \(u\) and \(y\) to be divisible by \(5\), starting descent. Contradiction.

From \(a^2=2b^2\), both \(a\) and \(b\) are even. Then \(a=2a_1\), \(b=2b_1\), and \(\left(\frac{a_1}{b_1}\right)^2=2\). But \(b_1=\frac{b}{2}

Suppose \(3\mid x^2+y^2\). Then modulo \(3\), both squares must be \(0\), so \(3\mid x\) and \(3\mid y\). This contradicts \(\gcd(x,y)=1\). Therefore \(x^2+y^2\) is not divisible by \(3\).

Assume a solution exists and choose one with minimal \(x+y\). If \(x=y\), then \(2x^2=3x^2\), impossible. Let \(x>y\). The equation \(x^2-3xy+y^2=0\), viewed as a quadratic in \(x\), has second root \(x'=3y-x\). By Vieta, \(xx'=y^2\), so \(x'>0\), and since \(x>y\), \(x'=\frac{y^2}{x}

As a quadratic in \(x\), the equation is \(x^2-4yx+y^2=0\). Its discriminant must be a square: \(D=16y^2-4y^2=12y^2\). Thus there is an integer \(u\) such that \(u^2=12y^2\). But this equation has no positive integer solutions, as proved earlier. Contradiction.

Rewrite as \(x^2-5yx+2y^2=0\). The discriminant \(D=25y^2-8y^2=17y^2\) must be an integer square. This would mean \(u^2=17y^2\) for some positive \(u\). But \(u^2=17y^2\) is impossible by the same descent with prime \(17\). Contradiction.

Assume a solution exists and choose one with minimal sum \(x+y\). Let \(x\ge y\). If \(y=1\), then \(x^2+1=3x+2\), so \(x^2-3x-1=0\), which has no integer root.

Thus \(y\ge2\). View the equation as a quadratic in \(x\): \(x^2-3yx+(y^2-2)=0\). The second root \(x'=3y-x\) is an integer. By Vieta, \(xx'=y^2-2>0\), so \(x'>0\). If \(x=y\), then \(2y^2=3y^2+2\), impossible, hence \(x>y\). Therefore \(x'=\frac{y^2-2}{x}

We get \(2,4,3,1\), then the cycle repeats with period \(4\). Since \(17\equiv1\pmod4\), \(2^{17}\equiv2\pmod5\).

The cycle of last digits is \(3,9,7,1\). Since \(25\equiv1\pmod4\), the last digit is \(3\).

\(4^2=16\equiv2\pmod7\), \(4^3\equiv8\equiv1\pmod7\). Hence \(4^{12}=(4^3)^4\equiv1\pmod7\).

The number \(7\) is prime and \(7\nmid3\). By Fermat's little theorem, \(3^6\equiv1\pmod7\). Therefore \(7\mid3^6-1\).

\(\gcd(2,8)=2\ne1\), so Euler's theorem cannot be used. But \(2^3=8\equiv0\pmod8\), so every \(2^n\) with \(n\ge3\) is divisible by \(8\). Hence \(2^{10}\equiv0\pmod8\).

Since \(11\) is prime, \(5^{10}\equiv1\pmod{11}\). Also \(2026\equiv6\pmod{10}\), so \(5^{2026}\equiv5^6\). Now \(5^2\equiv3\), \(5^4\equiv9\), hence \(5^6\equiv9\cdot3=27\equiv5\pmod{11}\).

By Fermat, \(2^{12}\equiv1\pmod{13}\). Since \(100\equiv4\pmod{12}\), \(2^{100}\equiv2^4=16\equiv3\pmod{13}\).

\(7^2=49\equiv4\pmod9\), \(7^3\equiv4\cdot7=28\equiv1\pmod9\). Since \(50\equiv2\pmod3\), \(7^{50}\equiv7^2\equiv4\pmod9\).

\(\gcd(3,100)=1\), and \(\varphi(100)=40\). By Euler's theorem, \(3^{40}\equiv1\pmod{100}\). Thus the last two digits are \(01\).

\(11\equiv-1\pmod{12}\). The exponent \(2025\) is odd, so \(11^{2025}\equiv(-1)^{2025}\equiv-1\equiv11\pmod{12}\).

\(3^1\equiv3\), \(3^2\equiv2\), \(3^3\equiv6\), \(3^4\equiv4\), \(3^5\equiv5\), \(3^6\equiv1\pmod7\). Since \(1\) did not appear earlier, the order is \(6\).

From the previous problem, \(\operatorname{ord}_7(3)=6\). Therefore \(3^n\equiv1\pmod7\) if and only if \(6\mid n\). The answer is all positive multiples of \(6\).

Since \(2026\equiv2\pmod4\), \(2^{2026}\equiv2^2\equiv4\pmod5\) and \(3^{2026}\equiv3^2\equiv9\equiv4\pmod5\). The sum is \(4+4=8\equiv3\pmod5\).

Since \(13\) is prime and \(13\nmid5\), \(5^{12}\equiv1\pmod{13}\). Then \(5^{12k}=(5^{12})^k\equiv1^k\equiv1\pmod{13}\). Hence \(13\mid5^{12k}-1\).

\(9^1\equiv9\), \(9^2\equiv81\), \(9^3\equiv29\), \(9^4\equiv61\), \(9^5\equiv49\), \(9^6\equiv41\), \(9^7\equiv69\), \(9^8\equiv21\), \(9^9\equiv89\), \(9^{10}\equiv1\pmod{100}\). Since \(2026\equiv6\pmod{10}\), \(9^{2026}\equiv9^6\equiv41\pmod{100}\). The last two digits are \(41\).

Modulo \(4\), \(3^{2026}\equiv1\), since the exponent is even. Modulo \(7\), \(3^6\equiv1\), and \(2026\equiv4\pmod6\), so \(3^{2026}\equiv3^4=81\equiv4\pmod7\). We need a number congruent to \(1\) modulo \(4\) and to \(4\) modulo \(7\). It is \(25\). The answer is \(25\).

If \(p\mid a\), then \(a^p-a\equiv0-0\equiv0\pmod p\). If \(p\nmid a\), Fermat gives \(a^{p-1}\equiv1\pmod p\). Multiplying by \(a\), \(a^p\equiv a\pmod p\). In both cases, \(p\mid a^p-a\).

By Fermat, \(a^{p-1}\equiv1\pmod p\). But \(a\cdot a^{p-2}=a^{p-1}\). Therefore \(a\cdot a^{p-2}\equiv1\pmod p\), so \(a^{p-2}\) is the inverse.

Direct computation: \(7\cdot2=14\equiv1\pmod{13}\). Hence the inverse of \(7\) is \(2\). This agrees with Fermat: \(7^{11}\) must also be an inverse of \(7\).

\(2^5=32\equiv1\pmod{31}\). Since \(1000\) is divisible by \(5\), \(2^{1000}=(2^5)^{200}\equiv1\pmod{31}\).

For \(p=2\), \(2^2+1=5\), not divisible by \(2\). Let \(p\) be an odd prime. By Fermat in the form \(a^p\equiv a\pmod p\), \(2^p\equiv2\pmod p\). Then \(2^p+1\equiv3\pmod p\). For \(p\mid2^p+1\), we need \(p\mid3\), so \(p=3\). Check: \(2^3+1=9\), divisible by \(3\). The answer is \(p=3\).

We have \(a^2\equiv-1\pmod p\). Then \(a^4\equiv1\pmod p\), but \(a^2\not\equiv1\pmod p\), since that would give \(-1\equiv1\pmod p\), hence \(p=2\). Thus the order of \(a\) modulo \(p\) is \(4\). The order divides \(p-1\), so \(4\mid p-1\), i.e. \(p\equiv1\pmod4\).

By Fermat, \(3^p\equiv3\pmod p\) for every prime \(p\). Hence \(3^p+2\equiv5\pmod p\). If \(p\mid3^p+2\), then \(p\mid5\), so \(p=5\). Check: \(3^5+2=243+2=245\), divisible by \(5\). The answer is \(p=5\).

From the condition, \(2^{2^n}\equiv-1\pmod p\). Squaring gives \(2^{2^{n+1}}\equiv1\pmod p\). Thus the order of \(2\) modulo \(p\) divides \(2^{n+1}\). But it does not divide \(2^n\), because \(2^{2^n}\equiv-1\not\equiv1\pmod p\). Since all divisors of \(2^{n+1}\) are powers of two, the order is \(2^{n+1}\). The order of an element modulo prime \(p\) divides \(p-1\), hence \(2^{n+1}\mid p-1\). Therefore \(p\equiv1\pmod{2^{n+1}}\).

The numbers \(x\equiv1\pmod4\) are \(1,5,9,13,17,\ldots\). The first with residue \(2\) modulo \(5\) is \(17\). The answer is \(17\), and all solutions are \(x\equiv17\pmod{20}\).

The numbers are \(4,9,14,19,24,29,34,\ldots\). The first congruent to \(6\) modulo \(7\) is \(34\). Thus \(x\equiv34\pmod{35}\).

Substitute: \(8k+5\equiv7\pmod9\), so \(8k\equiv2\pmod9\). Since \(8\equiv-1\), \(-k\equiv2\), hence \(k\equiv7\pmod9\). Then \(x=8(9t+7)+5=72t+61\). The answer is \(x\equiv61\pmod{72}\).

If \(x\equiv0\pmod4\), then \(x\) is even, so \(x\equiv0\pmod2\). This contradicts \(x\equiv1\pmod2\). No solutions exist.

The conditions mean that \(x-3\) is divisible by both \(5\) and \(7\). Since \(5\) and \(7\) are coprime, \(35\mid x-3\). Hence \(x\equiv3\pmod{35}\).

From \(x\equiv2\pmod4\), check \(2,6,10,14,\ldots\). Residue \(1\) modulo \(3\) first occurs at \(10\), so \(x\equiv10\pmod{12}\). Now \(10,22,34,46,58,\ldots\); the first with residue \(3\) modulo \(5\) is \(58\). Thus \(x\equiv58\pmod{60}\).

We get \(n\equiv1\pmod2\), \(n\equiv1\pmod3\), \(n\equiv2\pmod5\). The number \(7\) works: \(8\) is divisible by \(2\), \(9\) by \(3\), \(10\) by \(5\). Since the moduli are pairwise coprime, all solutions are \(n\equiv7\pmod{30}\), and the smallest positive is \(7\).

The residues \(5\) and \(9\) are both \(1\) modulo \(4\), so the system is compatible. Check \(5,13,21,\ldots\) modulo \(8\); \(21\equiv9\pmod{12}\). The answer is \(x\equiv21\pmod{24}\).

Let \(x=6k+4\). Then \(6k+4\equiv10\pmod{15}\), so \(6k\equiv6\pmod{15}\). Divide by \(3\): \(2k\equiv2\pmod5\), hence \(k\equiv1\pmod5\). Then \(x=6(5t+1)+4=30t+10\). The answer is \(x\equiv10\pmod{30}\).

From \(x\equiv4\pmod6\), \(x\) is even. From \(x\equiv9\pmod{10}\), \(x\) is odd. Contradiction. No solutions exist.

The conditions are equivalent to \(x\equiv-1\pmod3\), \(x\equiv-1\pmod5\), \(x\equiv-1\pmod7\). Thus \(x+1\) is divisible by \(3\cdot5\cdot7=105\). The smallest positive \(x\) is \(104\). All solutions are \(x\equiv104\pmod{105}\).

The first three congruences give \(x\equiv23\pmod{30}\). Check \(23,53,83,\ldots\) modulo \(7\). We have \(53\equiv4\pmod7\). Thus the smallest solution is \(53\), and all solutions are \(x\equiv53\pmod{210}\).

Let \(d=\gcd(m,n)\). If a solution \(x\) exists, then \(x-a\) is divisible by \(m\), and \(x-b\) is divisible by \(n\), so both differences are divisible by \(d\). Hence \(a-b=(x-b)-(x-a)\) is divisible by \(d\).

Conversely, suppose \(d\mid a-b\). Look for \(x=a+mt\). We need \(a+mt\equiv b\pmod n\), i.e. \(mt\equiv b-a\pmod n\). This linear congruence is solvable because \(\gcd(m,n)=d\mid b-a\). Therefore the system has a solution.

Take \(n=60\). Then \(n+2=62\) is divisible by \(2\), \(n+3=63\) by \(3\), \(n+4=64\) by \(4\), and \(n+5=65\) by \(5\). All these numbers are larger than the corresponding divisors, so they are composite.

Consider \((k+1)!+2,(k+1)!+3,\ldots,(k+1)!+(k+1)\). For each \(r=2,3,\ldots,k+1\), the number \((k+1)!+r\) is divisible by \(r\), because \((k+1)!\) is divisible by \(r\). It is larger than \(r\), so it is composite. There are \(k\) such numbers.