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Let \(a\equiv (3^{-1}+5^{-1}+7^{-1})^{-1}\pmod{11}\). What is the remainder when a is divided by 11?

 May 3, 2022
edited by hhhhh  May 3, 2022
 #1
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You can compute that a = 27, so the remainder is 5.

 May 3, 2022
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Thank you for attempting to help, but the answer is actually 10, as i just found.

hhhhh  May 3, 2022
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Note that \(11 \times 1 + 1= 12 = 3 \times 4\), so 3 and 4 are multiplicative inverses of each other mod 11.

Similarly, we have \(11 \times 4 + 1 = 45 = 5 \times 9\) so 5 and 9 are multiplicative inverses of each other mod 11.

Also, \(11 \times 5 + 1 = 56 = 7 \times 8\) so 7 and 8 are multiplicative inverses of each other mod 11.

 

Therefore, \(a \equiv (3^{-1} + 5^{-1} + 7^{-1})^{-1} \equiv (4 + 9 + 8)^{-1} \equiv 21^{-1} \pmod{11} \equiv -1^{-1} \equiv 10 \pmod{11}\).

 May 3, 2022

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