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?

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Answer 1 · 2015-05-16T15:23:31

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

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

Dividing all parts by 7 we get

$\frac{100}{7} < 1 + 9 k \leq \frac{999}{7}$ $100/7 < 1+9k <= 999/7$

Subtracting 1 from all parts we get

$\frac{93}{7} < 9 k \leq \frac{992}{7}$ $93/7 < 9k <= 992/7$

Dividing all parts by 9 we get

$\frac{93}{63} < k \leq \frac{992}{63}$ $93/63 < k <= 992/63$

So $2 \leq k \leq 15$ $2 <= k <= 15$

All of these values of $k$ $k$ give solutions.