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What is the remainder when 23

^{11}+19^{13}+17^{22}is divided by 5?Correct Answer:

**A**The exponents of each number form a pattern.

- The key to this problem is finding the patterns of final digits for powers of 23, 19, and 17.
- The remainder when a number is divided by 5 is determined entirely by the number's final digit. The final digit of an integer power is not effected by any digits of the base number except for the last digit. These facts vastly simplify the problem.
- Because the last digit is the only one that is relevant in this problem we can concern ourselves with the powers of 3, 9, and 7 in place of 23, 19, and 17.
- Next we look at the sequences of final digits for powers of these numbers.

3^{1}ends in 3, 3^{2}ends in 9, 3^{3}ends in 7, 3^{4}ends in 1, 3^{5}ends in 3, 3^{6}ends in 9

Since the pattern repeats after 4 terms, it appears that whenever the exponent is a multiple of 4, the power of 3 ends in 1.

9^{1}ends in 9, 9^{2}ends in 1, 9^{3}ends in 9

Since the pattern repeats after 2 terms, it appears that whenever the exponent is a multiple of 2, the power of 9 ends in 1.

7^{1}ends in 7, 7^{2}ends in 9, 7^{3}ends in 3, 7^{4}ends in 1, 7^{5}ends in 7

Since the pattern repeats after 4 terms, it appears that whenever the exponent is a multiple of 4, the power of 7 ends in 1. - Using these patterns, we can find the final digits of the relevant powers. Because 11 is 3 more than a multiple of 4, 23
^{11}should end in the digit that comes 3 terms after 1 in the 3 sequence, which is 7. Because 13 is 1 more than a multiple of 2, 19^{13}should end in the digit that comes 1 terms after 1, in the 9 sequence, which is 9. - Because 22 is 2 more than a multiple of 4, 17
^{22}should end in the digit that comes 2 terms after 1 in the 7 sequence, which is 9. - Because we are adding numbers ending in 7, 9, and 9, we will get a sum that ends in the same digit as 7+9+9=25. The sum must thus end in 5 and be divisible by 5.
- The remainder when dividing by 5 will this be 0.
- Answer A is correct.