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Number Theory Level 2

What is the remainder when : 1 × 1 0 1 + 2 × 1 0 2 + 3 × 1 0 3 + . . . . . . + 2018 × 1 0 2018 1×10^{1}+2×10^{2}+3×10^{3}+......+2018×10^{2018}

Is divided by 9 9 .


The answer is 3.

Akshat Sharda by Akshat Sharda , 21, India

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1 solution

Oscar Rojas
Jul 15, 2015

10 = 1 mod 9, after the secuence is 1+2+3+,,,,,,,,,,,,,,,,,2018, and we can see this 1+2+3+4+5+6+7+8+0+1+2............... but (1+8)*4= 36 and 36 = 0 mod 9 for this reason 2018 = 2 mod 9 then the remiander is 1+2= 3

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