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# Tricky algebra

The sum 7/12 + 9360/d is an integer. Compute the smallest positive integral value for d.

Note by Bohong Selalu
4 years, 6 months ago

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Write the second fraction as c.m/c.n where the greatest common divisor of m and n is 1. If 7/12 + m/n is to be some integer k, then 7n + 12m = 12nk ->12m = 12nk - 7n -> m = nk - 7 (n/12) . Therefore 12 divides n. We also have that 7n = 12nk - 12m -> 7 = 12k - 12m/n . But since m and n are relatively prime, we have that n divides 12. So n = 12. Therefore our goal is to have that when the fraction is reduced the denominator is 12. We need to find the smallest c so that the numerator 9360 has no factors of 12. Dividing 9360 by 12 two times shows that 9360 = 12^2 * 65. So c = 12^2 for the smallest value of d, and d = 12^2 12 = 1728. We must check that this is indeed an integer, and it is because 7/12 + 65/12 = 72/12 = 6. So d = 1728. CMIIW

- 4 years, 6 months ago

I think the answer is 1728 and the integer is 6

- 4 years, 6 months ago

12

- 4 years, 6 months ago

you should specify that d is positive

- 4 years, 6 months ago

it was specified so.

- 4 years, 6 months ago

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