The product is 1. Not sure what the sum is, though!

Number Theory Level 3

I have two positive integers a a and b b , both less than 27 27 . Their product leaves a remainder of 1 1 when divided by 27 27 . Find the sum of all possible values of [ ( a + b ) m o d 27 ] \left[\left(a+b\right)\,\mathrm{mod}\,27\right] .


The answer is 54.

Francis Gerard Magtibay by Francis Gerard Magtibay , 28, Philippines

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1 solution

I only have a brute-force solution at the moment, as I have committed all possible pairs ( a , b ) \left(a,b\right) to memory. Here are the integer pairs that satisfy the conditions of the problem, along with the corresponding values for a b ab , ( a b ) m o d 27 \left(ab\right)\,\mathrm{mod}\,27 , a + b a+b , and ( a + b ) m o d 27 \left(a+b\right)\,\mathrm{mod}\,27 .

a a b b a b ab ( a b ) m o d 27 \left(ab\right)\,\mathrm{mod}\,27 a + b a+b ( a + b ) m o d 27 \left(a+b\right)\,\mathrm{mod}\,27
1 1 1 1 2 2
2 14 28 1 16 16
4 7 28 1 11 11
5 11 55 1 16 16
8 17 136 1 25 25
10 19 190 1 29 2
13 25 325 1 38 11
16 22 352 1 38 11
20 23 460 1 43 16
26 26 676 1 52 25

The sum of the distinct entries in the ( a + b ) m o d 27 \left(a+b\right)\,\mathrm{mod}\,27 column is 2 + 11 + 16 + 25 = 54 2+11+16+25=54 .

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but is there any simpler way of doing this

durgaprasad sahu - 3 years, 3 months ago

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I'm not sure :(

Francis Gerard Magtibay - 3 years, 3 months ago

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