JEE Novice - (9)

We define $f_{n} = \sum_{i = 1}^4 i^{n}$ , first of all check for n = 1,2,3,4. In n = 4 the unit digit is not 0. n =4k +1, 4k+2 or 4k+3 now for odd cases of n $f_{n} = 1^n + 3^{n} + 2^{n} [2^{n} + 1]$ , $1^{n} + 3 ^ {n} = 4 ×(3^{n-1}+ ...... + 3 ^{0})$ now note that there are n terms in the just above stated sum All of them are odd as $3^{j}$ is always odd , one pair of the odd terms when added becomes even but as there are n terms and n is odd one odd term will be lesft at last and odd + even is odd. So here maximum power of 2 dividing $f_{n}$ is 2 as only 4 divides $1^{n} + 3^{n}$ and so maximum zeroes at the end is 2. If we proceed similarly for the only even case n = 4k + 2 . We get maximum power of 2 dividing $f_{n}$ is 1. Since we had to obtain maximum no of zeroes at the end of $f_{n}$ our answer is of the odd case i.e. $\boxed{2}$

let's define $a _n = 1 ^ n + 2^ n + 3^n + 4^n$

we can see that for $n> 2, a_n = 2$ $(\mod 8)$ or $a_n = 4 (\mod 8 )$

So $a_n$ has a maximum of two trailing zeros for n> 2.

and since $a_0 = 1, a_1 = 10, a_2 = 30, a_3 = 100.$

The result is 2

Python 2.7:

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from time import time
def f(n):
    return 1**n + 2**n + 3**n + 4**n
def trailing_zeroes(n):
    count = 0
    for digit in str(n)[::-1]:
        if digit == '0':
            count += 1
        else:
            break
    return count
largest = 0
n = 0
seconds = 30
end = time() + seconds
while time() < end:
    count = trailing_zeroes(f(n))
    if count > largest:
        largest = count
    n += 1
print "Answer:", largest
print "Ran for", seconds, "seconds"
print "Tested up to n =", n-1

First note that n can't be divisible by 4 as in that case the expression is congruent to 4 modulo 10 and so it won't end in 0. So, if n is even, the highest power of 2 which will divide the expression is 1. So, we need to consider the case only for odd n whether it can end in more than 1 zeroes. Now, as n is odd, we can easily use LTE (Lifting The Exponent Lemma) on $3^n + 1^n$ and $4^n + 2^n$ separately and we get the answer as 2.