What is the largest number of 0's that can occur at the end of 1^n+2^n +3^+4^n for every positive integer n

Answer 1

For n = 1, you have 10 (1 zero)
For n = 2, you have 30 (1 zero)
For n = 3, you have 100 (2 zeroes)
For n = 4, you have 354 (no zeroes)
For n = 5, you have 1300 (2 zeroes)
For n = 6, you have 4,890 (1 zero)
For n = 7, you have 18,700 (2 zeroes)
For n = 8, you have 72,354 (no zeroes)
For n = 9, you have 282,340 (1 zero)
For n = 10, you have 1,108,650 (1 zero)
For n = 11, you have 4,373,500 (2 zeroes)
For n = 12, you have 17,312,754 (no zeroes)
For n = 13, you have 68,711,380 (1 zero)
etc.

Answer 2

The max number of zeros is two,since if it were three, then
8 would divide your number, which it never does. To check this note that it fails for n=1,2. so if 8 divides then it has to be for n>2. At that point 8 divides 2^n + 4^n so it would have to
divide 1^n + 3^n also. But 3^n leaves remainders 3,1 mod8
and 1^n leaves a remainder of 1 mod 8. When you add these they sum to a remainder of 2 or 4 mod8. Therefore 8 does
not divide 1^n+3^n and also does not divide
1^n+2^n+3^n+4^n and therefore since 8 divides 1000,
it's not possible for 1000 to divide your number and therefore
it can't end in three zeros. As Puzzling showed, it can end in
2 zeros so that 2 zeros is the max.

Answer 3

THE END OF
1^n+2^n +3^n + 4^n = 0 for any value of "n" .
The largest number of zero's that can occur at the end that is unit's place is "one' {1} only.

Answer 4

1^n+2^n+3^n+4^n=10*n

Find all points where they intersect