If you were to produce a 3 digit multiple of 7 that has a final digital sum of 7 using the formula; 7(1+9k). How do you calculate what k is?

May 16, 2015

If I understand the question correctly, we get several constraints from the conditions:

$100 < 7 \left(1 + 9 k\right) \le 999$ to get a 3 digit number with final digit sum = 7.

Dividing all parts by 7 we get

$\frac{100}{7} < 1 + 9 k \le \frac{999}{7}$

Subtracting 1 from all parts we get

$\frac{93}{7} < 9 k \le \frac{992}{7}$

Dividing all parts by 9 we get

$\frac{93}{63} < k \le \frac{992}{63}$

So $2 \le k \le 15$

All of these values of $k$ give solutions.