1962 AHSME Problems/Problem 38
The population of Nosuch Junction at one time was a perfect square. Later, with an increase of , the population was one more than a perfect square. Now, with an additional increase of , the population is again a perfect square.
The original population is a multiple of:
Let original population count, the second population count, and the third population count We first see that or . We then factor the right side getting . Since we can only have an nonnegative integral population, clearly and both factor . We factor into There are a few cases to look at: and . Adding the two equations we get or , which means . But looking at the restriction that the second population + third population... a perfect square.
and . Adding the two equations we get or , which means . Looking at the same restriction, we get + + + , which is NOT a perfect square.
Finally, and . or , which means . Looking at the same restriction, we get + + + . Thus we find that the original population is . Or is a multiple of
Let original population. Translating the word problem into a system of equations, we got: for some positive integers , and . Now, by subtracting from (i.e. ), we got: Since y and z are both positive integers and 101 is a prime, by factoring, the only working solution for us is and . Plugging that back to and simplify, we got , a multiple of . Therefore, the answer is . -nullptr07