# Thread: modular arithmetic

1. ## modular arithmetic

Determine the last digit (the ones digit) in the base 8 expansion of 9^1000, 10^1000, and 11^1000.

I'm so confused is there a trick to doing this problem?

2. Originally Posted by demon1
Determine the last digit (the ones digit) in the base 8 expansion of 9^1000, 10^1000, and 11^1000.

I'm so confused is there a trick to doing this problem?
9 = 8+1

so look at (8+1)^1000, all the terms in the binomial expansion of this are
multiples of powers of 8 except for the last one, and so do not contribute to
the least significant digit base 8 of the result. So only 1^1000 contributes to
the ;east significant digit, but this is 1, so the least significant digit is 1.

11= 8+3

For the same reasons as above the least significant digit of 11^1000 is 3^1000
in base 8. But 3^1000 = 9^500, so in base 8 3^1000 = 9^500 which has least
significant digit 1 for the same reason as in the first part.

10=8+2

So using the same sort of argument as in the first part we know that the least
lsignificant digit base 8 is the least significant digit base 8 of 2^1000 base 8.

2^1000 = 2.2^999 = 2.8^333

so the least significant digit of 2^1000 in base 8 is 0.

RonL

RonL

3. Hello, demon1!

Determine the last digit (the ones digit) in the base 8 expansion of:
. . (a) 9^1000 . . (b) 10^1000 . . (c) 11^1000

(a) Write nine in base-eight: .9 = 11
8

If a number ends in 1, powers of the number will also end in 1.
. . 11² = 121
8, .11³ = 13318, .11^4 = 146418, . . .

Therefore: 9^1000 =(11
8)^1000 ends in 1.

(b) Write ten in base-eight: .10 = 12
8

Then: .(12
8)² = 1448, .(128)³ = 17508

Thereafter, all powers of 12 will end in 0.

Therefore: 10^1000 = (12
8)^1000 ends in 0.

(c) Write eleven in base-eight: .11 = 13
8

Then: .13 = 13, .13² = 171
8, .13³ = 24638, .13^4 = 344618, . . .

The ones-digits alternate: 3, 1, 3, 1, . . .

Therefore: 11^1000 = (13
8)^1000 ends in 1.