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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
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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
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OMG thank you so much. I understand how you did that now.

hhhhh  May 3, 2022
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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

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