1) If n=b,b=0..9 is one-digit number. We need to check all of them.
n=0:n4−1=0−1=−1 is not divisible by 5.
n=1:n4−1=1−1=0 is divisible by 5.
n=2:n4−1=16−1=15 is divisible by 5.
n=3:n4−1=81−1=80 is divisible by 5.
n=4:n4−1=256−1=255 is divisible by 5.
n=5:n4−1=625−1=624 is not divisible by 5.
n=6:n4−1=1296−1=1295 is divisible by 5.
n=7:n4−1=2401−1=2400 is divisible by 5.
n=8:n4−1=4096−1=4095 is divisible by 5.
n=9:n4−1=6561−1=6560 is divisible by 5.
Among numbers from n=1 to n=9, any number n except 0 and 5 will satisfy the required property.
2) If n=10a+b, a=1..9;b=0..9, n is a two-digit number.
n4−1=(10a+b)4=(10a+b)2(10a+b)2−1=(100a2+20ab+b2)(100a2+20ab+b2)−1=10000a4+2000a3b+100a2b2+2000a3b+400a2b2+20ab3+100a2b2+20ab3+b4−1.
Each term in this sum is a multiple of 5 except b4−1 . We need to check all options for b . But this is the same as it was done in part 1).
Among numbers from b=1 to b=9, any number b except 0 and 5 will satisfy the required property. It means that the last digit of the number n can be any number except 0 and 5.
3) Similarly, one can show the same conclusion for a three-digit number and more.
Answer: in order for n4−1 to be divisible by 5, n must end in any of the digits 1, 2, 3, 4, 6, 7, 8, 9.
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