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Don't lie. It's less concise than this.

Think what you mean by

We can always find a and b to satisfy ax + bx = d where d = gcd(x, y)

is

We can always find a and b to satisfy ax + by = d where d = gcd(x, y)

B

You are right!

Q

thanks so much.I'll try your idea

Same as mine but I condensed it down to one line: return (Math.log(num) / Math.log(4)) % 1 == 0;

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