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?

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

Step-by-step 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

the answer is 1/4

step-by-step explanation: