# Intermediate Maths Olympiad( Year 10) question on numeric manipulation

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

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

• For three consecutive numbers often better to write $(n-1)\cdot n\cdot (n+1)=n^3-n$ Mar 11, 2016 at 21:50
• Agree! I will edit my answer. Thanks| Mar 11, 2016 at 21:51
• THank you for your answer, but im a bit confused by your working here; Mar 13, 2016 at 17:46
• how did you go from 470074<a^3 -a) 479974 to 77<a<78? Mar 13, 2016 at 17:47
• $470074 \leq (a-1)a(a+1) \leq (a+1)^3$. Thus, $470074^{1/3} \leq a+1$, or $78 \leq a+1$, or $77 \leq a$. Similarly, $(a-1)^3 \leq (a-1)a(a+1) \leq 479974$, thus $a \leq 78$. Then $a$ must be either $77$ or $78$ Mar 13, 2016 at 17:59