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Little Johnny Red forgot to write the multiplication sign between two 3-digit numbers $x$ and $y$ and simply wrote them as one number. When Johnny's teacher graded the homework, it turned out that the 6-digit number written by Johnny was three times the correct product $xy$.

Find the positive integers $x$ and $y$!

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  • 3
    $\begingroup$ The phrase "three times greater than" is problematic. I generally take its intent to be "three times as much as", but the literal meaning should be "four times as much as", because 3x + x =4x. Please clarify the meaning. $\endgroup$ – Monty Harder Oct 5 '15 at 18:14
  • $\begingroup$ @Gamow x is three times x = 3y greater than y y + x is three times greater than y x = y + 3x if you can read that. $\endgroup$ – warspyking Oct 5 '15 at 19:28
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The only answer is

$167$ and $334$.

I formulated the question as follows:

If the numbers are $x$ and $y$, then $1000x + y = 3xy$.

This can be rewritten as

$y=\frac{1000x}{3x-1}$

If

If $3x-1$ is to completely divide $1000$, and $x$ is a three-digit number, then $3x-1$ can be only equal to $500$. Then $x$ solves to $167$ and $y$ is $334$.

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    $\begingroup$ Note that $(3x-1)$ and $x$ won't share any prime factors in common. Therefore if $(3x-1)$ divides evenly into $1000x$, then $(3x-1)$ must divide $1000$. $\endgroup$ – mathmandan Oct 5 '15 at 15:26
  • $\begingroup$ Once you get here $y=\frac{1000x}{3x-1}$, given that $100 \le x \le 999$, $y = 334$. $\endgroup$ – ricksmt Oct 5 '15 at 18:44
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Because who doesn't like brute force? (And because the answer is already here)

python2:

>>> for i in range(100,1000):
...     for j in range(100, 1000):
...             if 1000*i + j == 3*i*j:
...                     print i, j
... 

167 334

For numbers this size it's quicker than working it out 'properly'

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  • $\begingroup$ I think the intent was to do it without a program, given the "no-computers" tag. $\endgroup$ – dpwilson Oct 6 '15 at 14:18
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    $\begingroup$ @dpwilson, that'll teach me to not read tags $\endgroup$ – Holloway Oct 6 '15 at 14:19
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One answer would be

$167\times334\ (\times3 = 167334)$

How I found it:

The easiest way to have at least one of the numbers $x$ in the product with $3$ is if the other number $y$ multiplied with $3$ is a bit over $1000$. The smallest such $y$ is $334$ which gives, multiplied with $3$, the product $1002$. Afterwards we only need an $x$ that multiplied with $2$ gives us $334$; therefore $x = 167$.

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Let's say that $x$ and $y$ are the 3 digit numbers, then

$3*x*y = 1000x + y$
$3y = 1000 + y/x$

That number $y/x$ is positive and less than 10, because $y<=999$ and $x>=100$, so we can rewrite this equation as:

$1000 < 3y < 1010$
$333.3 < y < 336.7$

So $y$ can be 334, 335 or 336

Substitute these values to our equation and we get the following:

$3x*334 = 1000x + 334$
$1002x-1000x = 334$
$x = 167$

So 334 and 167 is a match

$3x*335 = 1000x + 335$
$5x = 335$
$x = 67$

67 is a 2 digit number, so y can't be 335

$3x*336 = 1000x + 336$
$8x = 336$
$x = 42$

42 is also not a 3 digit number, so y can't be 336

Therefore the only answer is:

334 and 167

Check that this answer is valid:

$334*167 = 55778$
$55778 * 3 = 167334$

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