#1 Divisibility and Prime Factorisation

\(153=9\cdot17\), 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.