Question

The last three digits of the number $(81{)}^{25}$ are x, y and z. Find x+y+z

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Answer 1

When we get a number like 81, if we write it as 80+1 and expand $(81{)}^{25}$, we can find the last three digits easily. This is because, 80 is multiple of 10 and in the expansion, there will be a lot of terms which are multiples of 10. It will be easy to decide the last digits.

Now consider $(81{)}^{25}$ = $(80+1{)}^{25}$

$(80+1{)}^{25}$ = ${}^{25}{C}_0(80{)}^{25}+{}^{25}{C}_1(80{)}^{24}+{}^{25}{C}_2(80{)}^{23}............{}^{25}{C}_{24}(80)+{}^{25}{C}_{25}$

= ${}^{25}{C}_0(80{)}^{25}+{}^{25}{C}_1(80{)}^{24}+{}^{25}{C}_2(80{)}^{23}............25\times (80)+1$

= ${}^{25}{C}_0(80{)}^{25}+{}^{25}{C}_1(80{)}^{24}+{}^{25}{C}_2(80{)}^{23}............2000+1$

= ${}^{25}{C}_0(80{)}^{25}+{}^{25}{C}_1(80{)}^{24}+{}^{25}{C}_2(80{)}^{23}............2001$

Last two terms gives 2001 and the remaining terms will be a multiple of 1000 (Because there will be one 25 and at least two 80s in it)

So the last three digits will be 001

Sum of last three digit = 0 + 0 + 1 = 1