Brilli picnic

Since $15 \times 6 = 90$ , there are more than 15 ants. Since $17 \times 6 = 102$ , there are less than 17 ants.

If there are 16 ants, then there are $16 \times 6=96$ ant legs, which fulfills the requirements. Hence, there are 16 ants.

Solution 1 (modulus)

Suppose there are $n$ total ant legs. We are given $91 \le n \le 99$ and $n \equiv 0 \pmod{6}$ .

Let $a$ be the ones digit of $n$ . Then, $n = 90 + a \equiv 0 \pmod{6}$ . Because $6 | 90$ , then $a \equiv 0 \pmod{6}$ .

$a$ is a single digit greater than 1, so $a = 6$ . Hence, $n = 96$ , and there are $96 / 6 = 16$ ants.

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Solution 2 (divisibility properties)

Suppose there are $n$ total ant legs. Each ant has 6 legs, so $n$ must be divisible by 6. A number is divisible by 6 only if it is even and the sum of its digits is a multiple of 3.

We are given that $91 \le n \le 99$ . So, the first digit of $n$ is $9$ . Because $9$ is a multiple of 3, the second digit also must be a multiple of 3.

So, the second digit must be positive, even, and a multiple of 3. Of the positive even digits 2, 4, 6, and 8, only $6$ is a multiple of 3. Hence, $n = 96$ , and there are $96 / 6 = 16$ ants.