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Question 1206604: Find the units digit for the sum 13^35 + 57^30 + 34^33
Found 2 solutions by greenestamps, math_tutor2020:
Answer by greenestamps(13215)   (Show Source):
You can put this solution on YOUR website!
Since we are only looking for the units digit of the expression, we are only interested in the units digits of any calculations. In other words, we are doing our calculations mod 10.
13^35: The repeated pattern for successive powers of 3 is (3, 9, 7, 1). 35 divided by 4 leaves remainder 3, so the units digit of 13^35 is the 3rd number in the pattern, which is 7.
57^30: The repeated pattern for successive powers of 7 is (7, 9, 3, 1). 30 divided by 4 leaves remainder 2, so the units digit of 57^30 is the 2nd number in the pattern, which is 9.
34^33: The repeated pattern for successive powers of 4 is (4, 6). 33 divided by 2 leaves remainder 1, so the units digit of 34^33 is the 1st number in the pattern, which is 4.
Then the units digit of the expression is the units digit of 7+9+4=20, which is 0.
ANSWER: 0
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For other tutors who might be looking at this problem....
When I first tried to post a response to this problem, some weird stuff happened in the middle of writing my response, and I got kicked out of the site. Possibly whatever happened to cause that caused the statement of the problem to be corrupted.
As I hope you can see now, the expression for which we were to find the units digit is
13^35 + 57^30 + 34^33
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
Edit: It appears that another tutor has changed the question, which seems strange. For me the initial question simply had 13^35 and nothing else. I'll keep my solution for anyone who may need it.
I'm not sure what you mean by "sum" when there's only one item mentioned here. It appears you want to find the units digit for 13^35
When we divide by 10 and look at the remainder, that's the same as focusing only on the units digit.
Example: 108/10 = 10 remainder 8
This is where the modulo operator comes in handy.
Let's look at powers of 13 mod 10.
13^1 = 13 = 3 (mod 10)
13^2 = 169 = 9 (mod 10)
13^3 = 13*13^2 = 3*9 = 27 = 7 (mod 10)
13^4 = 13*13^3 = 3*7 = 21 = 1 (mod 10)
13^5 = 13*13^4 = 3*1 = 3 (mod 10)
In short,
13^1 = 3 (mod 10)
13^2 = 9 (mod 10)
13^3 = 7 (mod 10)
13^4 = 1 (mod 10)
13^5 = 3 (mod 10)
The pattern is 3,9,7,1 which repeats forever.
This repeating block consists of 4 items. It will mean we divide 35 by 4 to look at the remainder.
35/4 = 8 remainder 3
The "remainder 3" leads to exponent 3, so 13^35 and 13^3 have the same remainder when dividing by 10.
13^35 = 13^3 = 7 (mod 10)
Therefore, 13^35 is some very large number that ends with a 7.
Verification using WolframAlpha
https://www.wolframalpha.com/input?i=13%5E35+mod+10
and
https://www.wolframalpha.com/input/?i=13%5E35
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