(a) Observe that 7524 = 71000 + 5100 + 210 + 4 = 7(999 + 1) + 5(99 + 1) + 2(9 + 1) + 4 = (7999 + 7) + (599 + 5) + (29 + 2) + 4 = (7999 + 599 + 29) + (7 + 5 + 2 + 4) = (71119 + 5119 + 29) + (7 + 5 + 2 + 4) = (7111 + 511 + 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 divis- ible 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 divis- ible by 9, then n is divisible by 9.

(b)Given a positive integer n written in decimal form, the alter- nating sum of the digits of n is obtained by starting with the right-mostdigit,subtractingthedigitimmediatelytoitsleft, adding the next digit to the left, subtracting the next digit, and so forth. For example, the alternating sum of the digits of 180,928 is 8 2 + 9 0 + 8 1 = 22. Justify the fact that for any nonnegative integer n, if the alternating sum of the digits of n is divisible by 11, then n is divisible by 11.