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This question was on the Maths olympiad a few days ago, and I have been stumped by it for a few days now. The question goes as follows:

I multiply two or more consecutive integers to obtain a six digit number, n. The first two digits of n are 4 and 7, and the last two are 7 and 4. (4 7 _ _ 7 4).
Work out n.

What I found especially hard was characterising the integers that multiply to form n. there are two or more integers, so how would you characterise this in a GCSE format, considering this is a year 10 paper.
Any help would be greatly appreciated

Mathmo123
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Couldnotthinkofadisplayname
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1 Answers1

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If there were $4$ consecutive numbers, then there must be $2$ even numbers, which make the product divisible by $4$, which is not the case (as the last $2$ numbers of $n$ are $74$). So you must have at most $3$ numbers.

Consider the case where there are $2$ consecutive numbers. By trial and error, there doesn't exist any such pair such that the last digit of their product is $4$.

Consider the case where there are $3$ consecutive numbers.

$470074 \leq (a-1)a(a+1) = a^3-a \leq 479974 \Rightarrow 77\leq a \leq 78 \Rightarrow a = 77$.

Then $77 \times 78 \times 79 = 474474$.

SiXUlm
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