To determine the remainder of the division of 3302 + 7200 with 5.
Is it correct if I find the remainder of the division separately for 3302 and 7200 and then add the two of the remainders?
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1What do you mean by the 'rest' of the division? Are you referring to the remainder? – D.B. Jan 14 '19 at 00:22
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Yes the remainder, sorry. – Viktor Jan 14 '19 at 00:23
3 Answers
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The last digit of the powers of $3$ are $3,9,7,1,3,9,..$ they are cyclic and repeating every four times, so the last digit of $3^{302}$ is the same as the last digit of $3^{302 (mod 4)}=3^2=9$.
Same argument for the last digit of the powers of $7$: $7,9,3,1,7,9...$. Therefore the last digit of $7^{200}$ is the same as the last digit of $7^0=1$.
Thus the last digit of $3^{302}+7^{200}$ is $9+1=10=0 \ mod \ 5$
user289143
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Oh sorry. Anyway $302 mod 4 = 202 mod 4 = 2 mod 4$ so the result is the same – user289143 Jan 14 '19 at 00:39
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The remainder of 3^302 mod 5 is 4 and the remainder of 7^200 mod 5 is 1. So the result is 4 + 1 = 5. But 5 mod 5 is zero so the remainder is 0? Can I go this way? – Viktor Jan 14 '19 at 00:46
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It's an easier way to calculate for me. I only needed to be sure if I could find the remainders separately. Thank you. – Viktor Jan 14 '19 at 00:55
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$\begin{align}\bmod 5\!:\ \ \ \ &\ \ \ 3^{302}+7^{200}\\ \equiv &\ \ \ 9^{151}+49^{100}\\ \equiv &\ (-1)^{151}\!+(-1)^{100}\\ \equiv &\ \ \ {-1}\ \ +\ \ 1 \end{align}$
using standard congruence rules, including the Sum Rule (which answers your query affirmatively).
Bill Dubuque
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In general, you would add the remainders of $3^{302}/5$ and $7^{200}/5$ modulo $5$.
D.B.
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The remainder of 3^302 mod 5 is 4 and the remainder of 7^200 mod 5 is 1.But 5 mod 5 is zero so the remainder is 0? – Viktor Jan 14 '19 at 00:47
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