Final

[cinelli]

If you evaluate 19^92 and write it in the usual decimal form, what are the final six digits?

Cinelli

[Gengulphus]

If you evaluate 19^92 and write it in the usual decimal form, what are the final six digits?

Spoiler...

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The final 6 digits of a number x are just x MOD 1000000 (i.e. the remainder when it's divided by 1000000), with leading zeros added if necessary, and the MOD function has the property that (xy) MOD N = ((x MOD N) * (y MOD N)) MOD N.

That gives us in succession, starting with 19 MOD 1000000 = 19 and using a calculator:

19^2 MOD 1000000 = 19*19 MOD 1000000 = 361
19^3 MOD 1000000 = 19*361 MOD 1000000 = 6859
19^5 MOD 1000000 = 361*6859 MOD 1000000 = 2476099 MOD 1000000 = 476099
19^10 MOD 1000000 = 476099*476099 MOD 1000000 = 226670257801 MOD 1000000 = 257801
19^20 MOD 1000000 = 257801*257801 MOD 1000000 = 66461355601 MOD 10000 = 355601
19^23 MOD 1000000 = 355601*6859 MOD 1000000 = 2439067259 MOD 1000000 = 67259
19^46 MOD 1000000 = 67259*67259 MOD 1000000 = 4523773081 MOD 1000000 = 773081
19^92 MOD 1000000 = 773081*773081 MOD 1000000 = 597654232561 MOD 10000 = 232561

So the answer is that the last six digits are 232561.

Gengulphus

[cinelli]

Well done to Gengulphus for the correct answer. He did well to do those multi-digit multiplications using a calculator – mine can handle only ten digits. I was aware of working in modulo million but I have another method. If you have two different methods which produce the same answer, you can be pretty sure it is correct. Or in this case with a probability of 999999/1000000. So some spacing in case anyone else would like come up with an alternative method.

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Write 19^92 as (20-1)^92 and expand by the binomial theorem:

19^92 = (20-1)^92 = 20^92 – 92*20^91 + ... + 92Cr*20^(92-r)*(-1)^r + ... +1

How does this help, you might ask, introducing a sum with 93 terms? It is because most of those 20s raised to some power end in several zeros and contribute nothing to the final six digits of the sum, the part we are interested in. In fact only the final six terms make a contribution. We can write

(20-1)^92 = m(1000000) – 92C5*20^5 + 92C4*20^4 – 92C3*20^3 + 92C2*20^2 - 92C1*20 + 1

where m(1000000) denotes an integer multiple of 1000000. This expression is

m(1000000) – 49177128*3200000 + 2794155*160000 – 125580*8000 + 4186*400 – 92*20 + 1
= m(10^6) - 157366809600000 + 447064800000 – 1004640000 + 1674400 – 1840 + 1
= m(10^6) -600000 + 800000 – 640000 + 674400 – 1840 + 1
= m(10^6) + 232561

leaving 232561 as the final six digits, as before.

Cinelli