Observe that Since the sum of the digits of 7524 is

QUESTION:

Observe that

\(\begin{aligned}

7524 &=7 \cdot 1000+5 \cdot 100+2 \cdot 10+4 \\

&=7(999+1)+5(99+1)+2(9+1)+4 \\

&=(7 \cdot 999+7)+(5 \cdot 99+5)+(2 \cdot 9+2)+4 \\

&=(7 \cdot 999+5 \cdot 99+2 \cdot 9)+(7+5+2+4) \\

&=(7 \cdot 111 \cdot 9+5 \cdot 11 \cdot 9+2 \cdot 9)+(7+5+2+4) \\

&=(7 \cdot 111+5 \cdot 11+2) \cdot 9+(7+5+2+4) \\

&=(\text { an integer divisible by } 9) \\

& \quad+(\text { the sum of the digits of } 7524)

\end{aligned}\)

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.

Text Transcription:

7524 =7 cdot 1000 + 5 cdot 100 + 2 cdot 10 + 4

         =7(999+1) + 5(99 + 1) + 2(9 + 1) + 4

         =(7 cdot 999 + 7) + (5 cdot 99 + 5) + (2 cdot 9 + 2) + 4

         =(7 cdot 999 + 5 cdot 99 + 2 cdot 9) + (7 + 5 + 2 + 4)

         =(7 cdot 111 cdot 9 + 5 cdot 11 cdot 9 + 2 cdot 9) + (7 + 5 + 2 + 4)

         =(7 cdot 111 + 5 cdot 11 + 2) cdot 9 + (7 + 5 + 2 + 4)

         =( an integer divisible by 9)

                                +( the sum of the digits of 7524)