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# NT problem

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The number $$2^{1993}+3^{1993}$$ is a multiple of What is the units digit of the quotient $$\frac{2^{1993}+3^{1993}}{5}?$$

May 3, 2022
edited by hhhhh  May 3, 2022

#1
-3

The units digit of the quotient is 1.

May 3, 2022
#2
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I'm assuming that the expession ((2^1993 + 3^1993) is a multiple of 5.

(2^1993 + 3^1993)/5  mod 10^10==1964726983 - these are the last 10 digits of the quotient.

May 3, 2022
#3
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OMG thank you so much. I understand how you did that now.

hhhhh  May 3, 2022
#4
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If you can determine the last two digits of $$2^{1993} + 3^{1993}$$, it suffices to determine the last digit of $$\dfrac{2^{1993} + 3^{1993}}5$$.

$$2^{1993} + 3^{1993}$$ is divisible by 5 because $$2^{1993} + 3^{1993} \equiv 2^{1993} + (-2)^{1993} \equiv 2^{1993} - 2^{1993} \equiv 0 \pmod 5$$.

Note that $$2^{1993} \equiv 2^{1993 \operatorname{mod}40}\equiv 2^{33} \pmod{100}$$  by Euler's theorem.

Also $$2^{33} \equiv 0 \pmod 4$$ and $$2^{33} \equiv 2^{33\operatorname{mod}20} \equiv 2^{13} \equiv 8192 \equiv 17\pmod {25}$$ again by Euler's theorem.

By Chinese remainder theorem, $$2^{1993} \equiv 2^{33} \equiv 92 \pmod {100}$$.

Similarly with $$3^{1993}$$, we have $$3^{1993} \equiv 23\pmod{100}$$

Therefore, $$2^{1993} + 3^{1993} \equiv 92 + 23 \equiv 115 \equiv 15 \pmod{100}$$.

Then $$2^{1993} + 3^{1993} = 100k + 15$$ for some integer k.

Then $$\dfrac{2^{1993} + 3^{1993}}5 = 20k + 3$$ for some integer k. Therefore, the last digit is 3.

May 3, 2022