rom help

1    5,   

2    25,

3    125,           odd

4    625,             even

5    3125,           odd

6    15625           even

7     78125             odd

8    390625           even

ok I have a pattern

5^137 = 1000n+ 125 = 125*8*n  +   17*8     +  1  =  8(1000n+17) +1       n has a specific integer value

Yes I made a small stupid numeric error that i am quite sure hatchet288 was smart enough to correct.

This last line should have been

5^137 = 1000n+ 125 = 125*8*n  +   15*8     + 5  =  8(1000n+15) +5       n has a specific integer value

so

\(5^{137}\div 8 \)        has a remainder of 5

It was no big deal, any question asker should have been able to pick up that little error for themsleves.

BUT Good effort Tertre, it was a good catch.

Melody, you're on the right track, but here is my attempt:

You're right about the odd and even switching,

5^3=125

5^4=625

5^5=last three digits of 125

5^6=last three digits are 625

So, for odd powers of five, the last three digits are 125 and for even powers five, the last three digits are 625. By the divisibility rule for 8, and we have \(\frac{125}{8}\) since 137 is odd. Thus, the remainder is \(\boxed{5}.\)

Here's how I view this....

5^1 / 8 = (8 - 3)^1 / 8   

Every term in the expansion is a multiple of 8 except the last  which is

(-3)^1 / 8  = (-3)^1  mod 8  = 5

Similarly

5^2 /8  =  (8 - 3)^2 / 8

Again we only need consider the last term which is

(3)^2/8  =  (3)^2 mod 8  =   1

etc....

So...we only need to consider  the last term in each subsequent power expansion and note that

(3)^n mod 8 =  1      and

(-3)^n mod 8 =  5

So

5^137 / 8    =  (8 - 3)^137 / 8      will  have (-3)^137 / 8  as a last term

And the remainder is  (-3)^137 mod 8  =   5

Working to mod 8 ,

\(5^{137}\equiv5.5^{136}\equiv5.25^{68}\equiv5.1^{68}\equiv 5.\)

So the remainder will be 5.