Calculators are forbidden

Find the remainder when 1 1 + 2 2 + 3 3 + 4 4 + + 5 0 50 1^1 + 2^2 +3^3 +4^4 +\cdots +50^{50} is divided by 8.


The answer is 5.

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1 solution

Otto Bretscher
Dec 2, 2015

(All congruences will be mod 8.) We have 2 2 = 4 2^2=4 and k k 0 k^k\equiv 0 for even k 4 k\geq 4 . For odd k k we have k 2 1 k^2\equiv 1 so k k k k^k\equiv k . Thus our sum is 4 + 1 + 3 + 5 + . . . + 47 + 49 = 4 + 2 5 2 5 4+1+3+5+...+47+49=4+25^2\equiv \boxed {5}

Exactly the same method sir! For those who didn't understand the comment, k k k ( m o d 8 ) k^k\equiv k \pmod{8} for all odd k k ,consider this,let k = 4 n ± 1 k=4n\pm 1 ,then k k = k × ( 4 n ± 1 ) a n e v e n n u m b e r k^k=k\times (4n \pm 1)^{an\ even\ number} ,now consider this, ( 4 n ± 1 ) 2 = 16 n 2 + 1 ± 8 n 1 ( m o d 8 ) (4n\pm1)^2=16n^2+1\pm8n\equiv 1\pmod{8} .

Adarsh Kumar - 5 years, 6 months ago

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@Otto Bretscher

@Adarsh Kumar

Is there any alternate solution, using Euler's Totient Function?

Swapnil Das - 4 years, 11 months ago

Neat , short and elegant .Nice !

Raven Herd - 5 years, 6 months ago

Did same...

Dev Sharma - 5 years, 6 months ago

If I divide each odd number by 8, getting remainder 1, 3, 5 and 7. There are 6 times of 1,3,5 and 7 and only one 4 I got. In this I got 100. So i got 4 as answer. Please tell me where I went wrong

ramesh perumal - 4 years, 9 months ago

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