Sum of digits (Number theory)
Number Theory
Level
3
If the "final sum" of the digits of 147 is 1+4+7=12, than 1+2=3, so it is 3 (Sum until the result is a single digit number), what is the "final sum" of (52836289026)^58925?
9
6
1
8
3
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1 solution
The sum of the digits of the base is 6 (In counting, you can erase the 9's since adding a 9 does not change the sum of the digits of a number: ex. 3+9=12 -->1+2=3, the number we started with). 6^1=6, 6^2= 36, the "final sum" of 36 is 9, 6^3=216, the "final sum" of 216 is 9 and it will be 9 for every power of 6 bigger than 6^1, since we are just multiplying by 6 A number with 9 as its "final sum". So we get 9×6=54, 5+4=9, than multiply the number with 9 as its "final sum" by 6 and we still get a number with 9 as its "final sum" and so on. So the answer is 9 :)