Question 1208429: Find the last two digits of the number 3^123 + 7^123 + 9^123.
Found 3 solutions by greenestamps, math_tutor2020, ikleyn:
Answer by greenestamps(13203)   (Show Source):
You can put this solution on YOUR website!
Look for the repeating pattern of the last two digits of 3, 7, and 9 to increasing powers.
3: 03, 09, 27, 81, 43, 29, 87, 61, 83, 49, 47, 41, 23, 69, 07, 21, 63, 89, 67, 01, 03...
That pattern repeats with a cycle length of 20. 123 mod 20 = 3, so the last two digits of 3^123 is the 3rd number in the pattern: 27
7: 07, 49, 43, 01, 07...
That pattern repeats with a cycle length of 4. 123 mod 4 = 3, so the last two digits of 7^123 is the 3rd number in the pattern: 43
9: 09, 81, 29, 61, 49, 41, 69, 21, 89, 01, 09 ...
That pattern repeats with a cycle length of 10. 123 mod 10 = 3, so the last two digits of 9^123 is the 3rd number in the pattern: 29
27+43+29 = 99
ANSWER: 99
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
Greenestamps has a great solution. I'll provide another route.
Use the process described on this page to evaluate the following
3^123 = 27 (mod 100)
7^123 = 43 (mod 100)
9^123 = 29 (mod 100)
Each item can be verified with technology. Some examples are WolframAlpha, GeoGebra, spreadsheet, etc.
The results we got were 27, 43, and 29.
They are the last two digits of 3^123, 7^123, and 9^123 in that exact order.
Therefore, 27+43+29 = 99 are the last two digits of 3^123 + 7^123 + 9^123.
Verification with WolframAlpha
https://www.wolframalpha.com/input?i=%283%5E123+%2B+7%5E123+%2B+9%5E123%29+mod+100
Answer by ikleyn(52835)  (Show Source):
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