What is the remainder when (22^33 + 10^35) is divided by 45?

1) 8
2) 11
3) 2
4) None of these

• Powers of 10 when divided by 45 will always leave a remainder of 10. When 22^1, 22^2, 22^3, 22^4 and 22^5 are divided by 45, the remainders are 22, 34, 28, 31 and 7 respectively. In this manner, 22^10 will leave remainder 49, which will reduce to 4. So, 22^33 = 22^10 × 22^10 × 22^10 × 22^3 when divided by 45 will leave a remainder of 4 × 4 × 4 × 28 = 1792. This can be further reduced to 37. So the effective remainder will be 37 + 10 = 47 which can be further reduced to 2.
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