Number Theory

There is no need to calculate till the 24th power. Actually there is a pattern to solve this question.

If you find the solution till the 8th power you will know what I want to say.

$1^{1} +$ $2^{1} +$ $3^{1} +$ $4^{1} = 10$ (divisible by 5, Yes)

$1^{2} +$ $2^{2} +$ $3^{2} +$ $4^{2} = 30$ (Yes)

$1^{3} +$ $2^{3} +$ $3^{3} +$ $4^{3} = 100$ (Yes)

$1^{4} +$ $2^{4} +$ $3^{4} +$ $4^{4} = 354$ (No)

$1^{5} +$ $2^{5} +$ $3^{5} +$ $4^{5} = 1300$ (Yes)

$1^{6} +$ $2^{6} +$ $3^{6} +$ $4^{6} = 4890$ (Yes)

$1^{7} +$ $2^{7} +$ $3^{7} +$ $4^{7} = 18700$ (Yes)

$1^{8} +$ $2^{8} +$ $3^{8} +$ $4^{8} = 72354$ (No)

You can see that every 4th term power (multiple of 4) is not divisible by 5.

That means the values of n are 4, 8, 12, 16, 20 and 24 (less than 25).

Sum of integers $= 4 + 8 + 12 + 16 + 20 + 24 = \boxed{84}$

This is not a solution.

As a clarification, this question was initially set without the word "not". As such, the answer should have been 216, and I updated it accordingly. Those who entered in 84 were thus marked wrong. Later on, Satvik added in the word "not", which made the answer become 84 again. I updated the answer again, and those who entered 84 were thus marked correct.

This would explain why those who entered 84 got confusing emails of "Your answer is wrong" and then "Your answer is correct". Those who entered 216 before i posted this were not marked wrong (However, note that as the answers were changed multiple times, it is possible that some of you do not see an accurate display.)

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>>> li=[]
>>> for n in range(25):
    if ((1**n+2**n+3**n+4**n)%5 != 0):
        li.append(n)


>>> sum(li)

It prints 84

Just test the last digit of the given set for n=1to 4.... as this sequence repeats after 4 terms...... now it is found that for n=4 the given set is not divisible by 5 so we have the values of n as 4,8, 12, 16, 20, 24 so there sum is 84