NT

Number Theory Level pending

x x and y y are positive integers that are non-multiples of 3 greater than 0. Find the sum of the numbers that can be the remainders when x 3 + y 3 x^{3}+y^{3} is divided by 9.


The answer is 9.

Vilakshan Gupta by Vilakshan Gupta , 19, India

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1 solution

Marta Reece
May 9, 2017

The numbers are not divisible by 3, so they can be in one of two formats: 3 n + 1 3n+1 and 3 n + 2 3n+2

The third powers of these numbers are then:

( 3 n ) 3 + 3 × ( 3 n ) 2 + 3 × ( 3 n ) + 1 (3n)^3+3\times(3n)^2+3\times(3n)+1

( 3 n ) 3 + 3 × ( 3 n ) 2 × 2 + 3 × ( 3 n ) × 4 + 8 (3n)^3+3\times(3n)^2\times2+3\times(3n)\times4+8

The first three terms of both of these expressions are divisible by 9, so the remainders are 1 and 8.

Any combinations of these remainders can appear, giving us the possibilities of 1 + 1 = 2 , ( 1 + 8 ) m o d 9 = 0 , ( 8 + 8 ) m o d 9 = 7 1+1=2, (1+8)\mod 9 =0, (8+8)\mod 9 =7

Finally 2 + 0 + 7 = 9 2+0+7=\boxed{9}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Marta, I think it would be slightly clearer if instead of presenting the generic forms 3n +1, 3n +2 which gives the appearance that the numbers are consecutive, that we consider x = 3n +1, y = 3m + 1 and x = 3n + 1, y = 3m + 2, x = 3n + 2, y = 3m + 2; just a thought, Ed Gray

Edwin Gray - 2 years, 9 months ago

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