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| Math Central | Quandaries & Queries |
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Question from Randy, a student: A positive integer n is chosen. Then the product n ( n + 1 ) is computed and two digits are appended to the end of the product. The resulting number is the square of an integer. Show that it is always possible to complete the process above or give a positive integer n for which the process cannot be completed. |
We have two responses for you
Randy,
Try this experiment: Pick an integer n, for example say n = 13, and extend n by an extra digit by putting a 5 after it (to get 135 in my example). Now compare the square of that number with n(n+1). (In my example you would compare 13*14 with 1352.)
Chris
Randy,
So you want n(n+1)x100 + T to be a square, where T is a two digit number. Expanding, you get 100n2 + 100n + T, that is (10n)2 + 2 × (10n) × 5 + T, which you want to be a square (A+B)2 = A2 + 2 × A × B + B2.
Claude
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