Question

What is the smallest integer k>2000 such that both 17k/66 and 13k/105} are terminating decimals?

Answer 1

17k/66 and 13k/105 must reduce to fractions with a denominator that only consists of powers of 2 or 5.

For example, some fractions with terminating decimals are

1/2 = 0.5

1/4 = 1/2² = 0.25

1/5 = 0.2

1/8 = 1/2³ = 0.125

1/10 = 1/(2•5) = 0.1

1/16 = 1/2⁴ = 0.0625

and so on, while some fractions with non-terminating decimals have denominators that include factors other than 2 or 5, like

1/3 = 0.333…

1/6 = 1/(2•3) = 0.1666…

1/7 = 0.142857…

1/9 = 1/3² = 0.111…

1/11 = 0.09…

1/12 = 1/(2²•3) = 0.8333…

etc.

Since 66 = 2•3•11, we need 17k to have a factorization that eliminates both 3 and 11.

Similarly, since 105 = 3•5•7, we need 13k to eliminate the factors of 3 and 7.

In other words, 17k must be divisible by both 3 and 11, and 13k must be divisible by both 3 and 7. But 13 and 17 are both prime, so it's just k that must be divisible by 3, 7, and 11. These three numbers are relatively prime, so the least positive k that meets the conditions is LCM(2, 7, 11) = 231, and thus k can be any multiple of 231.

If you're familiar with modular arithmetic, this is the same as solving for k such that

13k ≡ 0 (mod 3)

17k ≡ 0 (mod 3)

17k ≡ 0 (mod 7)

13k ≡ 0 (mod 11)

and the Chinese remainder theorem says that k = 231n solves the system of congruences, where n is any integer.

Now it's just a matter of finding the smallest multiple of 231 that's larger than 2000, which easily done by observing

2000 = 8•231 + 152

and so k = 9•231 = 2079.