A special property for a number

$k = 1024$ .

For some general $q$ -sum $x$ , write $x$ as $x = (z+1)+(z+2)+(z+3)+...+(z+q) = qz + \frac{q(q+1)}{2}$ . We consider the residue of $x$ mod $q$ :

$x\equiv qz + \frac{q(q+1)}{2} \equiv \frac{q(q+1)}{2} \mod q$

• Case 1: $q\equiv 1 \mod 2$ :

Write $q$ as $q = 2p+1$ for some positive integer $p$ . Then:

$x\equiv \frac{q(q+1)}{2} \equiv \frac{(2p+1)(2p+2)}{2} \equiv (p+1)(2p+1) \equiv 0 \mod 2p+1$

• Case 2: $q\equiv 0 \mod 2$ :

Write $q$ as $q = 2p$ for some positive integer $p$ . Then:

$x\equiv \frac{q(q+1)}{2} \equiv \frac{(2p)(2p+1)}{2} \equiv p(2p+1) \equiv 2p(p) + p \equiv p \mod 2p$

We have shown that if the residue of $x$ mod $q$ is not consistent with the cases above, then $x$ is not a $q$ -sum.

Now, since $x$ is a $2017$ -sum, it must take the form $x = \frac{2017(2018)}{2} + 2017d = 2017(1009+d)$ for some positive integer $d$ . We let $k = 1009 + d$ .

Consider the following restrictions on $x = 2017k$ , as given by the residue cases above:

$x \not\equiv 1 \mod 2$

$x \not\equiv 0 \mod 3$

$x \not\equiv 2 \mod 4$

$x \not\equiv 0 \mod 5$

$x \not\equiv 3 \mod 6$

...

$x \not\equiv 0 \mod 2015$

$x \not\equiv 1008 \mod 2016$

Note that because 2017 is prime, it has an inverse modulo $2,3,4,...,2016$ , meaning that the same restrictions apply to $x$ as to $k$ :

$k \not\equiv 1 \mod 2$

$k \not\equiv 0 \mod 3$

$k \not\equiv 2 \mod 4$

$k \not\equiv 0 \mod 5$

$k \not\equiv 3 \mod 6$

...

$k \not\equiv 0 \mod 2015$

$k \not\equiv 1008 \mod 2016$

Now, we wish to find a minimal such $k$ . Note that if $t$ is odd, $k \not\equiv 0 \mod t$ . This means that, since all primes except for 2 are odd, $k$ is not a multiple of any primes except for 2. In other words, the prime factorization of $k$ contains only factors of 2, or $k$ is a power of 2. Since $k$ is minimal, a power of 2, and greater than or equal to $1009$ , we have $k = 1024$ .

Note: This problem was inspired by a problem on the 2017 William Lowell Putnam Exam.