Suppose that $a$ is a positive integer for which the least common multiple of $a+1$ and $a-5$ is $10508$. What is $a^2 - 4a + 1$

Question

4 years ago

10

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1 answer:

Nana76 [90] · Answer 1 · 2022-01-25T03:51:01+03:00

Answer:

21022.

Step-by-step explanation:

Find the prime factors of 10508:

2 ) 10508

2 ) 5254

37 ) 2627

71.

50208 = 2*2*37*71.

Now there is no integer value for a that would fit (a+ 1)(a - 5) = 10508 .

But we could try multiplying the LCM by 2:-

= 21016 = 2*2*2*37*71.

= 2*2*37 multiplied by 2 * 71

= 148 * 142.

That looks promising!!

a - 5 = 142 and

a + 1 = 148

This gives 2a - 4 = 290

2a = 294

a = 147.

So substituting a = 147 into a^2 - 4a + 1 we get:

= 21022.