Bonus) Let n be an integer whose sum of digits equals 113. Can n be the square of an integer? If yes, give an example. If no, explain. (Hint: Use arithmetic mod 9?)
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Observe that 7524 = 7·1000 + 5·100 + 2·10 + 4 = 7(999 + 1) + 5(99 + 1) + 2(9 + 1) + 4 = (7·999 + 7) + (5·99 + 5) + (2·9 + 2) + 4 = (7·999 + 5·99 + 2·9) + (7 + 5 + 2 + 4) = (7·111·9 + 5·11·9 + 2·9) + (7 + 5 + 2 + 4) = (7·111 + 5·11 + 2)·9 + (7 + 5 + 2 + 4) = (an integer divisible by 9) + (the sum of the digits of 7524). Since the sum of the digits of 7524 is divisible by 9, 7524 can be written as a sum of two integers each of which is divisible by 9. It follows from exercise 15 that 7524 is divisible by 9. Generalize the argument given in this example to any nonnegative integer n. In other words, prove that for any nonnegative integer n, if the sum of the digits of n is divisible by 9, then n is divisible by 9.
Madhur L.
(b) An integer is divisible by 3 if and only if the sum of its digits (in decimal) is divisible by 3 (for example, 68967 is divisible by 3 if and only if 6+8+9+6+7=36 is divisible by 3, if and only if 3+6=9 is divisible by 3). Prove rigorously that this rule holds for all integers. (c) State the analogous divisibility rule for 9 and explain (informally, referring to your proof in the previous part) why this rule always holds.
Adi S.
a. Substitute each of the integers from 1 to 9 for $n$ in the expression $n^{2}+n+11 .$ b. Using inductive reasoning. guess what kind of number you will get when you substitute any positive integer for' $n$ in the expression $n^{2}+n+11$ . c. Test your guess by substituting 10 and 11 for $n .$
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