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I had recently solved a problem.

A number 47_ _74 is multiple of at least two consecutive numbers. Find the numbers. The list of numbers may be of any length $\ge 2$.

I first saw that if they were multiples of 4 numbers then it must be divisible by 4 but it isn't so they are multiples of 2 or 3 numbers. Also all the two or three numbers must be 2-digit or 3-digit. I tried pairing consecutive numbers but no two consecutive numbers produced a result whose units digit was 4. So I tried for 3 number pairs. The only two pairs were $(*2, *3, *4)$ and $(*7, *8, *9)$. (Replace the stars with one-digit numbers). So now since $70\cdot 70\cdot 70=343000 \text{ and } 80\cdot 80\cdot 80 = 512000$. I tried $72\cdot 73\cdot 74$ and $77\cdot 78\cdot 79$ and $77\cdot 78\cdot 79$ produced a result of 474474 and fulfilled the result.

I want to know if my approach is practical. Is it correct? Can you suggest a better way of tackling this problem? I would love new answers. Can you suggest some 'elegant' proof?

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    $\begingroup$ I think that your approach is very good. Efficient and logical. $\endgroup$
    – Brian Cheung
    Mar 14, 2016 at 10:49
  • $\begingroup$ @user3313320 To me it doesn't seem 'elegant'. $\endgroup$
    – TheRandomGuy
    Mar 14, 2016 at 11:03
  • $\begingroup$ Maybe you can easy out the calculation by checking mod 10 on both sides. $\endgroup$
    – N.S.JOHN
    Mar 14, 2016 at 11:57
  • $\begingroup$ @N.S.JOHN I thought that's what I was doing. $\endgroup$
    – TheRandomGuy
    Mar 14, 2016 at 12:02
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    $\begingroup$ I think you solution is elegant. Since I'm lazy and I have a pocket calculator, after noticing, like you did, that it must be the product of $2$ or $3$ consecutive numbers, I would have computed the square and cube root of the upper bound $479974$, obtaining respectively $692$ and $78$ and I would have checked the neighborhood of these numbers. $\endgroup$
    – Giovanni Resta
    Mar 14, 2016 at 13:28

1 Answer 1

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I think there's not much of room for improvement in your approach. There's a little issue in you trying $72\cdot 73\cdot 74$ as it would be divisible by $4$.

By using modulo 5 arithmetics you can narrow down the possibilities quite narrowly. We know that $47..74\equiv4$. In modulo 5 arithmetics theres only three (or four) factorizations of $4\equiv 2\cdot 2\equiv 3\cdot 3\equiv 1\cdot 4\equiv 4\cdot 1$. Therefore it's rather obvious that two consecutive numbers can't equal $47..74$. For three consecutive to be equivalent to $4$ none of them can't be $0$ (because their product would be equivalent to $0$) - so that leaves only $1\cdot2\cdot3\equiv 1$ and $2\cdot3\cdot4\equiv 4$. Four consecutive are impossible as we already know (that we see by using modulo 4 arithmetics instead).

Next narrowing down is to use estimates to narrow down the possibilities, which also becomes quite narrow. By noting that a product $(n-1)n(n+1)$ has the property:

$$(n-1)^3 \le (n-1)n(n+1) \le (n+1)^3$$

and you have that the product should be in the range $[470074, 479974]$ you have that, $(n-1)^3 \le 479974$ which means $n-1 \le 78$ and $(n+1)^3\ge 470074$ which means $n+1\ge 78$, that is

$$77 \le n \le 79$$

But the restriction of none of $(n-1)$, $n$ and $(n+1)$ being divisible by four means that $n$ is even so that leaves only $n=78$ (or from the modulo 5 arithmetics we know that $n\equiv 3$, that is ends with $3$ or $8$), that is the possible candidate is $77\cdot78\cdot79 = 474474$

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