Sum of Digits

Find the sum of the digits in the expression 333333333\times 777777777.
Source: NCTM Mathematics Teacher, December 2005

SOLUTION
If we multiply the numbers by hand, we have the following addition
.\;\;\;\;\;\;\;\;\;2333333331
.\;\;\;\;\;\;\;\;2333333331
.\;\;\;\;\;\;\;2333333331
.\;\;\;\;\;\;2333333331
.\;\;\;\;\;2333333331
.\;\;\;\;2333333331
.\;\;\;2333333331
.\;2333333331
2333333331
———————————-
259259258740740741
The sum of the digits equals 81.

Answer: 81

Alternative solution
Let p=333333333 and q=777777777.
pq=(3/3)pq
=(3p)(q/3)
=(999999999)(259259259)
=(10^9-1)(259259259)
=259259259\cdot 10^9-259259259
We have the following subtraction
259259259000000000
-\;\;\;\;\;\;\;\;\;\;\;\;\;259259259
———————————
259259258740740741
If we add the digit 2 in the first column to the digit 7 in the tenth column and so on, we have
2+7\!=\!9,5+4\!=\!9,9+0\!=\!9,2+7\!=\!9,5+4\!=\!9,9+0\!=\!9,2+7\!=\!9,5+4\!=\!9,8+1\!=\!9
9(9)=81
This is true for any random 9-digit number, for example 187971192
187971192000000000
-\;\;\;\;\;\;\;\;\;\;\;\;\;187971192
———————————
187971191812028808
1+8\!=\!9,8+1\!=\!9,7+2\!=\!9,9+0\!=\!9,7+2\!=\!9,1+8\!=\!9,1+8\!=\!9,9+0\!=\!9,1+8\!=\!9
9(9)=81
In fact, the number 777777777 is not necessary in this problem; any 9-digit number that is a multiple of 3 will do.

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About mvtrinh

Retired high school math teacher.
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