A Number Theoretical Nod To 2015

We first note that $2015 = 5 \times 13 \times 31$ , and so $2015^{2} = 5^{2} \times 13^{2} \times 31^{2}$ . Now as we want the product of the consecutive integers to have at least two factors of $31$ , we can either have two different integers in the sequence that are divisible by $31$ or one that is divisible by $31^{2}$ . To have two divisible by $31$ means that there would be at least $32$ integers in the sequence, so we'll try instead to find a shorter sequence which includes one integer divisible by $31^{2} = 961$ .

The four-digit integers divisible by $961$ are $1922, 2883, 3844, 4805, 5766, 6727, 768, 8649$ and $9610$ . Now we also require that the product have at least two factors of $13$ , so we can have two integers each divisible by $13$ or one divisible by $13^{2} = 169$ . In the former case we would then require a sequence of at least $14$ integers, so let us first see if we can find an integer, divisible by $169$ , with an absolute difference from one of the above list of multiples of $961$ of less than $14$ . Calculating these multiples modulo $169$ , we find that the least absolute difference is $10$ , as $2883 \equiv 10 \pmod{169}$ . This gives us the sequence $2873$ through $2883$ , the product of which also has $3$ factors of $5$ , (two from $2875$ and one from $2880$ ). Thus the smallest value of $n$ is $2873 - 2773 + 1 = \boxed{11}$ .

The product must contain each of the prime factors 5, 13, 31 at least twice.

If $n \leq 31$ , then one of the numbers should be a multiple of $31^2$ . Likewise, if $n \leq 13$ , then one of them should be a multiple of $13^2$ . If $n \geq 10$ , we are guaranteed to have at least two factors 5, so we won't worry about that factor yet.

So I need multiples of $13^2$ and $31^2$ that are relatively close together. A good place to start is the ratio $\frac{31^2}{13^2} \approx 5.686.$ We like to approximate this ratio as a fraction $a/b$ , where $a$ and $b$ are sufficiently small; then $31^2\:b \approx 13^2\:a$ define the sequence of numbers that we look for. The requirement of four digits limits us to $b \leq 10$ and $a \leq 59$ .

The only candidates for approximating $5.686\dots$ with a fraction of denominator no greater than 10 are $5\tfrac12 = 5.500$ (poor), $5\frac 23 \approx 5.667$ ; $5\tfrac 57\approx 5.714$ ; $5\tfrac 7{10} = 5.700$ . The lower denominators will result into smaller values for $n$ .

• $5\tfrac12\ \to\ a = 11,\ b = 2$ gives $31^2\ b = 1922$ and $13^2\ a = 1859$ , which are rather far apart. We can do better!

• $5\tfrac23\ \to\ a = 17,\ b = 3$ gives $2883$ and $2873$ , so that $n = 11$ .

• $5\tfrac57\ \to\ a = 40, b = 7$ gives $6727$ and $6760$ , so that $n = 34$ .

• $5\tfrac7{10}\ \to\ a = 57, b = 10$ gives $9633$ and $9610$ , so that $n = 24$ .

Our best option, then, is the second one, with $n = \boxed{11}$ . In any sequence of 10 or more successive numbers, there are at least two multiples of 5, so that condition is fulfilled.

@Brian Charlesworth Nice one this problem... seems easy but is a great application of the greedy algorithm. We just need to check the shortest distance between multiples of 31^2 and 13^2 and I believe the answer would be less than 11 if there was no condition on the number of digits.