Divisible problems 5

Number Theory Level 4

What is the remainder of (FACE) : (ACE) in base 16??

answer in BASE 10


The answer is 588.

Math Man by math man , 20, Indonesia

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2 solutions

Jared Low
Mar 22, 2015

FACE is 15 ∗ 1 6 3 + 10 ∗ 1 6 2 + 12 ∗ 16 + 14 = 64206 15*16^3+10*16^2+12*16+14=64206 and ACE is 10 ∗ 1 6 2 + 12 ∗ 16 + 14 = 2766 10*16^2+12*16+14=2766 in decimal, so the remainder from 64206 ÷ 2766 64206\div2766 is 588 \boxed{588}

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FACE/ACE=F000/ACE+1. F000=15 16^3. ACE=10 16^2+12 16+14=2766. F000/ACE=15 16^3/2766=22.21258134.
So remainder is 0.21258134*2766=588.

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