Last digit in the expansion of 43^43=7
Last digit in the expansion of 33^33=3

Last digit of 43^43+33^33=0

Hence remainder=0

i'm curious if it can be solved like below.
The problem is to ask
43^43 + 33^33 mod 10

Modular Exponentiation
X^a (mod n).

43^43 mod 10
First,
43^1 mod 10 = 3
43^2 mod 10 = 3^2 mod 10 = 9
43^4 mod 10 = 9^2 mod 10 = 1
43^8 mod 10 = 1^2 mod 10 = 1
....
43^32 mod 10 = 1 mod 10 = 1

43 = 32 + 8 + 2 + 1
43^43 mod 10
= 43^32 mod 10 * 43^8 mod 10 * 43^2 mod 10 + 43^1 mod 10
= (1 * 1 * 9 * 3 ) mod 10
= 27 mod 10
= 7

Same procedure is done on 33
33^1 mod 10 = 3
33^2 mod 10 = 3^2 mod 10 = 9
33^4 mod 10 = 9^2 mod 10 = 1
.....
33^32 mod 10 = 1 mod 10 = 1

33 = 32 + 1
33^33 mod 10
=33^32 mod 10 * 33^1 mod 10
= (1 * 3 ) mod 10
= 3 mod 10
= 3

Therefore, their sum must be 3 + 7 which gives the unit digit 0.
The remainder is obviously 0.

chandra_adesh wrote:Last digit in the expansion of 43^43=7
Last digit in the expansion of 33^33=3

Last digit of 43^43+33^33=0

Hence remainder=0

how did you get the last digit? 8)

Thanks for this cool tip!

thats my suicide note none but this problem is responsible for it. :twisted:

haha i know such big numbers are justbubbles. i dealt with units 5 6 1 they are of course easy. but this with 3 as a unit must belong to ''tough'' category.

Stacey Koprince wrote:Whenever they ask you a "units digit" question with really high exponents, there will be a pattern and you will only have to follow the units digit through the problem (b/c it is all multiplication).

43^43:

3^1 = 3^2 = 3^3 = 23^4 = 83^5 = etc. so the units digit pattern is 3-9-7-1. The pattern repeats every 4th term. So 3^4, 3^8, 3^12, etc, will all have the units digit 1. 3^40 will be 1, 3^41 will be 3, 3^42 will be 9, 3^43 will be 7. Same pattern as above. 3^32 will be 1, 3^33 will be 3. units digit 7 + units digit 3 will equal units digit 0. Anything that ends in 0 will have a remainder of 0 when divided by 10.

WOW, beautiful trick. I totally missed that one 8) Thanks