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Question 1 of the first ever International Maths Olympiad

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Edited by Richard Walker, Friday 24 July 2026 at 00:21

From the first international Maths Olympiad 1959

Question 1

Prove that for every natural number n the fraction

21 times n plus four divided by 14 times n plus three

is already in lower terms and cannot be canceled further.

Solution

If we have a pair of natural numbers open a comma b close with b greater than a then any common factor of a and b must also be a common factor of left parenthesis b minus a comma b right parenthesis .

Any common factor of left parenthesis 21 times n plus four comma 14 times n plus three right parenthesis must be a common factor of left parenthesis two times left parenthesis 21 times n plus four right parenthesis comma three times left parenthesis 14 times n plus three right parenthesis right parenthesis equals left parenthesis 42 times n plus eight comma 42 times n plus nine right parenthesis and so of left parenthesis left parenthesis 42 times n plus nine right parenthesis minus left parenthesis 42 times n plus eight right parenthesis comma 42 times n plus nine right parenthesis equals left parenthesis one comma 42 times n plus nine right parenthesis .

But the highest common factor of left parenthesis one comma 42 times n plus nine right parenthesis is one and one is therefore the highest common factor of the numerator and denominator of the given fraction, which is therefore already in lowest terms.

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