Question

suppose that $a$ is a positive integer for which the least common multiple of $a 1$ and $a-5$ is $10508$. what is $a^2 - 4a 1$?

Answer 1

21022

Explanation:

Since 10508 is the LCM of a + 1 and a - 5, then the product of a + 1 and a - 5 must be a multiple of 10508, or 10508k, for any integer k.

(a + 1)(a - 5) = 10508k

Let k = 1

(a + 1)(a - 5) = 10508

Let a = 104, then (a + 1)(a - 5) = 10395, too low

Let a = 105, then (a + 1)9a - 5) = 10706, too high

Try k = 2

(a + 1)(a - 5) = 21016

Let a = 146, then (a + 1)(a - 5) = 20727, too low

Let a = 147, then (a + 1)(a - 5) = 21016, exactly correct.

We need k = 2, and a = 147.

Then for a = 147,

a² - 4a + 1 = 147² - 4(147) + 1 = 21022

Answer: 21022