Problem Solving

What is the remainder when 43^43+ 33^33 is divided by 10?

1
3
9
7
0

OA: E

GmatKiss wrote:What is the remainder when 43^43+ 33^33 is divided by 10?

1
3
9
7
0

OA: E

we know 43/10 = R(3), similarly 43^4 /10 =R(1) hence 43^40/10 = R (1) therefore 43^43/10 = R(7)
similarly 33/10 = R(3), similarly 33^4/10 = R(1) hence 33^32/10 = R(1) therefore 33^33/10 = R(3)
so, 43^43 + 33^33 / 10 => R(7) + R(3) / 10 => R (0)

Hence E!

n@resh, interesting way of approaching this q., but clearly not systematic

my take is similar in that it utilizes the last digit (unit's digit) method. I separate the unit's digit of each number from the very beginning. Any number is divisible with no remainder or remainder=0 by 10, if the last digit (unit's digit) of the number is 0. In all other cases the last digit (unit's digit) distinct from 0 is the remainder ...

43^43 will give us the last digit of 3^43 with the cycle applications of power for 3
below i illustrate the comprehensive cycle data
3^1=3
3^2=9
3^3=27
3^4=1, every 4 units in power the cycle repeats for 3. Hence 43/4 will give us 10*4 and 3 additional units of power for 3, i.e. 3^43 has the last digit 7

the same should be applied to 33^33 which has the last digit (unit's digit) of 3 (33/4=8*3+1)

two last digits 7+3 will result in unit's digit 0

answer must be e

cycles 1 and 3 resulting in the unit's digit 0

n@resh wrote:

GmatKiss wrote:What is the remainder when 43^43+ 33^33 is divided by 10?

1
3
9
7
0

OA: E

we know 43/10 = R(3), similarly 43^4 /10 =R(1) hence 43^40/10 = R (1) therefore 43^43/10 = R(7)
similarly 33/10 = R(3), similarly 33^4/10 = R(1) hence 33^32/10 = R(1) therefore 33^33/10 = R(3)
so, 43^43 + 33^33 / 10 => R(7) + R(3) / 10 => R (0)

Hence E!

(3+40)^43 + (3+30)^33

only 3^43 and 3^33 will have remainder if divided by 10

for both the remainder pattern is [3, 9, 7, 1]
43/4: remainder is 3 refers 3rd position=7
33/4: remainder is 1 refers 1st position=3
7+3=10
Therefore, remainder is 0.