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Project Euler 33

Project Euler

Problem33を解き直しました.

  • Digit cancelling fractions

The fraction 49/98 is a curious fraction, as an inexperienced mathematician in attempting to simplify it may incorrectly believe that 49/98 = 4/8, which is correct, is obtained by cancelling the 9s.

We shall consider fractions like, 30/50 = 3/5, to be trivial examples.

There are exactly four non-trivial examples of this type of fraction, less than one in value, and containing two digits in the numerator and denominator.

If the product of these four fractions is given in its lowest common terms, find the value of the denominator.

https://projecteuler.net/problem=33

問題は「49/98 => 4/8のように,分子・分母の共通数字を消すという誤った約分をしても問題がない「1より小さく分子・分母がともに2桁の数になる分数」は4個ある(ただし30/50 => 3/5というようなものは除く).これらの4個の分数の積が約分されたときの分母の値を求めよ」.

# (10 * a + b) / (10 * b + c) = a / c だけ考えれば良い
ansNume = 1
ansDeno = 1
for a in 1..9 do
	for b in 1..9 do
		for c in 1..9 do
			nume = 10 * a + b
			deno = 10 * b + c
			g1 = nume.gcd(deno)
			g2 = a.gcd(c)
			if nume / g1 == a / g2 && deno / g1 == c / g2 && nume < deno then
				ansNume *= (nume / g1)
				ansDeno *= (deno / g1)
			end
		end
	end
end
p ansDeno / ansNume.gcd(ansDeno)

プログラムのコメントにもありますが,
数学的な解析から,
(10 * a + b) / (10 * b + c) = a / c だけ考えれば良いです.