Question

Mr. Scott likes to pose interesting problems to his Mathematics classes. Today,

he started with the expression 6^2020 + 7^2020. He stated that the expression was
not equivalent to 13^2020 and that he was not interested in the actual sum. His
question to his class and to you is, “What are the final two digits of the sum?

Answer 1

The last two digits of 6²⁰²⁰ + 7²⁰²⁰ is 7, 7

Explanation:

The given question is to find the value of the sum of two numbers 6²⁰²⁰ + 7²⁰²⁰, where, 6²⁰²⁰ + 7²⁰²⁰ ≠ 13²⁰²⁰

We have;

The last two digits of the powers of 6 are;

6¹ = 6, last two digits = 06

6² = 36, last two digits = 36

6³ = 216, last two digits = 16

6⁴ = 1,296, last two digits = 96

6⁵ = 7,776, last two digits = 76

6⁶ = 46,456, last two digits = 56

6⁷ = 279,936, last two digits = 36

6⁸ = 1,679,616, last two digits = 16

6⁹ = 10,077,696, last two digits = 96

6¹⁰ = 60,466,176, last two digits = 76

6¹¹ = 362,767,056, last two digits = 56

Therefore, the last two digits repeat itself every 5 times

For 6²⁰²⁰, the last two digits will be for the 2020/5 = 404 complete times, which will be 76

The last two digits of the powers of 7 are;

7¹ = 7, last two digits = 07

7² = 49, last two digits = 49

7³ = 343, last two digits = 43

7⁴ = 2,401, last two digits = 01

7⁵ = 16,807, last two digits = 07

7⁶ = 117,649, last two digits = 49

7⁷ = 823,543, last two digits = 43

7⁸ = 5,764,801, last two digits = 01

Therefore, the last two digits repeat itself every 4 times

For 7²⁰²⁰, the last two digits will be for the 2020/4 = 505 complete times, which will be 01

Therefore, the last two digits of 6²⁰²⁰ + 7²⁰²⁰ is 76 + 01 = 77, which is 7, 7