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i think it is logical or tricky question. But i didnot get into to solve.
the question is
what is the sum of the sum of the sum of the digits(last sum) fo the no.
4444^4444
plz help me to solve that
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The interesting thing about finding the sum of digits is that what you're essentially doing is taking away lots of multiples of 9.
Take 4444. The sum of its digits is 4+4+4+4 = 16.
However, a longer way of writing out 4444 would be 4*1000 + 4*100 + 4*10 + 4.
Take away 4*999 + 4*99 + 4*9 (a multiple of 9), and you're left with 4+4+4+4.
You can do a similar thing to find the sum of 16's digits. 16-9 = 1+6 = 7.
Now, by the rules of modulo arithmetic, a^n (mod 9) = (a+9k)^n (mod 9).
That means that the answer you're looking for is the sum of the sum of the sum of... the digits of 7^4444.
Starting off at 7^1, 7^2, 7^3, ... produces the pattern 7, 4, 1, 7, 4, 1...
Realising that 4444 is one more than a multiple of 3, that means that the last sum of the digits of 4444^4444 is 7.
Why did the vector cross the road?
It wanted to be normal.
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thanks mathsyperson
u r great
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