A problem by Scrub Lord
Level
pending
What is the second smallest integer n, such that the last (rightmost) three digits of n^2 are 444.
Hint : the smallest such number is 38.
The answer is 462.
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1 solution
Let the desired number be x. Then
x = 38 + n
x^2 = 38^2 + 76n + n^2
Now, let n(76 + n) be the smallest possible multiple of 1000. That means that n(76 + n) is divisible by both 5^3 and 2^3. First we observe that either n or 76 + n is going to be divisible by 5. Then, both n and n + 76 leave the same remainder when divided by 4, which means that if n is divisible by 4, then so is n + 76. One of the factors has to be divisble by 125, and both have to be divisible by 4 to meet the above criteria. To minimize,
n + 76 = 125 * 4 = 500
n + 38 = 462 = x