Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Fill in the blanks of:

$$\square \;\square \times \square = \square \; \square \;\square =\square \times \square \;\square $$

With distinct numbers from the set $\{1,2,3,4,5,6,7,8,9\}$.

I was able to do it by trial-and-error, but I am looking for a more mathematical approach.

share|cite|improve this question
I don't see how this is linear algebra ... – user67258 Jul 18 '13 at 17:06
Trial-and-error can be a very powerful "mathematical approach"! – Shaun Ault Jul 18 '13 at 17:12
You have 9 numbers: call them $x_{1},x_{2},\dots,x_{9}$. The equations you can set up would be the following: $$(10x_{1}+x_{2})x_{3} =100x_{4}+10x_{5}+x_{6} =x_{7}(10x_{8}+x_{9})$$. The constraints are that $x_{j}\in \{1,2,\dots,9\},$ and $x_{i}\not=x_{j}$ for $i\not=j$. Not sure where to go from here. – Adrian Keister Jul 18 '13 at 17:13
Sometimes, trial-and-error is the only mathematical approach and sometimes the best! – Jeel Shah Jul 18 '13 at 17:13
The only thing slightly more mathematical than "trial and error" here is process of elimination. We know, for example, that $5$ cannot appear in any of the number's $1$'s digit. – Omnomnomnom Jul 18 '13 at 17:13
up vote 17 down vote accepted

All right: This is "mathematical". In other words, some reasoning, with a lot of case checking. So, we have that: $$ab\cdot c = de\cdot f = ghi$$

  • $b,c,f,e,i$ are not 5. Else requires a zero, or another 5.

  • If $ghi$ is odd, then we can see that $b,c,e,f$, and $i$ all have to be odd. This is impossible if none of them are a $5$. Therefore, $ghi$ is even.

  • $a$ and $d$ are not a $1$. Else, $g$ would also have be a $1$. We can also see that $c$ and $f$ are not $1$. $b$ and $e$ are not $1$ either. If $b$ was $1$, then $i = c$.

  • Thus, either $g$ or $h$ is a $1$.

Now, look at the number $5$. It is either $a,d$, $g$ or $h$.

  • First, assume that it is $g$. Then $ghi =51i$. And we can see that this is impossible.

  • Next, assume it is $h$. then $ghi = 15i$. $i$ is even. So where is the $9$? $9$ cannot be either $a$ or $d$. So assume it is $b$.

Then we have $a9\cdot c = de\cdot f = 15i$. Since $c$ is odd, $a$ cannot be $2$. If $a =3, c=4$. We have:

$$39\cdot 4 = 156 = 78\cdot 2$$

So if $5= h$, one answer.

Otherwise, $5$ is either $a$ or $d$. Without loss of generality, assume it is $a$, so that $5b\cdot c = de\cdot f = ghi$

Two cases, if $c$ is odd, $50\cdot c = 250$. We know that $g$ or $h$ is $1$. If $h$ is one, $b*c = 60$ something. That is impossible (only $7*9$ = 60 something, and that is odd).

So if $c$ is odd, $g = 1$. $c$ is then obviously $3$.

So we have $5b \cdot 3 = 1hi = de\cdot f$. b cannot be 2. Else, h is 5. If b is 4, we have 54*3 = 162 = ... impossible, as 7, 8, and 9 are left over.

As for $b =6$, we have $56\cdot 3 = 168$. Not possible. 2 instances of 6.

If b is 8. We have $58\cdot 3 = 174 = 29\cdot 6$

Final situation: if c is even.

$5b \cdot c = de \cdot f = ghi$

if $c$ is 2, then $g$ is 1, and $h$ is either 1 or 0. Impossible.

If $c$ is 4 or more, $g$ is not 1, so $h$ is 1.

We have

\begin{align} 5b \cdot c = g1i = de \cdot f \end{align}

If $c$ is 4, $b = 3$. Then $53 \cdot 4 = 212$. Impossible.

If $c$ is 6, $g$ is 3. So b is 2. $52 \cdot 6 = 312$. Impossible.

If $c$ is 8, $g$ is 4. b is 2. $52 \cdot 8=416$. No even numbers left. Therefore, impossible.

Basically, that leaves the two answers.

share|cite|improve this answer
Thanks for editing the answer. – user86828 Jul 18 '13 at 19:16
it would be great if you could finish the edit :) – nbubis Jul 18 '13 at 19:18
Merr, all right. – user86828 Jul 18 '13 at 19:28
@user86828 Great! Thanks! – Mahdi Khosravi Jul 18 '13 at 21:12

A number of observations can be made that narrow down the number of "guesses" that need to be made:

  • Neither $x_2$, $x_3$, $x_7$ nor $x_8$ can be a $5$, because this would result in either a $0$ or another $5$ in the evaluation of the product. For the same reason, $x_6$ (the final digit of the product) cannot be $5$.
  • Neither $(x_2,x_3)$ nor $(x_7,x_9)$ can be a pair of numbers such that the ones place of their product is one of the multipiers. That is, given a pair of numbers $m$ and $n$, $m\not\equiv mn \pmod{10}$ and $n\not\equiv\pmod{10}$.
  • Neither $x_3$ nor $x_7$ can be a $1$, which would result in a $2$ digit product (and a repeat of each digit in the corresponding multiplier).
  • You probably want the product to be a multiple of $6.$ It is not guaranteed, but gives more flexibility.
  • The only digit pair that multiplies to $1$ is $3\times 7$, so the three-digit number cannot end in $1$. Similarly $3\equiv 7\times 9$ only, $7\equiv 3\times 9$ only, and $9$ is not the product of distinct digits. So the result is even.
share|cite|improve this answer
More than just $x_3$ and $x_7$; a $5$, if it occurs, can only occur in $x_1$, $x_4$, $x_5$, or $x_8$; $5$ in any last digit is forbidden. Similarly, $x_2$ and $x_9$ can't be $1$s, either. Further, we know that $x_6$ must be even - if it were odd then all five of the last digits would be odd, and since there are only four odd numbers to go around (remember, we're not allowed to use $5$ there) then the product must be even. – Steven Stadnicki Jul 18 '13 at 18:30

For general reference, a brute force approach leads to there only being two distinct solutions: $$29 \cdot 6 = 174 = 58 \cdot 3$$ $$39 \cdot 4 = 156 = 78 \cdot 2$$

Edit: If you want to see this for yourself:

import itertools
l = itertools.permutations(range(1,10))
for x in l:
    a, b, c = (10*x[0] + x[1]) * x[2], 100*x[3] + 10*x[4]+ x[5], (10*x[6] + x[7])*x[8]
    if (a == b and b == c): 
        print 10*x[0] + x[1], "*", x[2], "=", b, "=", 10*x[6] + x[7], "*", x[8]
share|cite|improve this answer
How do we know that these are the only two? – Ataraxia Jul 18 '13 at 18:32
@ZettaSuro - brute force. – nbubis Jul 18 '13 at 18:33
Of course it can be done by programming , But the question is mathematical approach – Harish Kayarohanam Jul 18 '13 at 18:37
@HarishKayarohanam - I know, but I thought the actual solutions may be useful, both for the OP and also for inspiration as to a "mathematical solution", assuming such a one exists. – nbubis Jul 18 '13 at 18:39
Useful to check. – user86828 Jul 18 '13 at 19:49

Elimination, elimination. $$\overline{x_1x_2} \cdot x_3 = \overline{x_4x_5x_6}= x_7 \cdot \overline{x_8x_9}$$
Let's call : $$\overline{x_1x_2} = a$$ $$\overline{x_4x_5x_6} = b$$ $$\overline{x_8x_9} = c$$ So $$a \cdot x_3 = b = x_7 \cdot c $$

  • What we know:
    1. $x_3$ or $x_7$ cant't be $1$
    2. $a$ or $x_3$ can't be both primes. Also $c$ and $x_7$.
    3. $b$ cant't be prime
    4. $ 123\le c\le 776 = 98 *7 $
    5. Last digit of $a$ or $c$ can't be $1$ because $\color{red}{x_3} \cdot \overline{x_1 1} = \overline{x_4x_5 \color{red}{x_3}}$
    6. The last digit of a,b or c can't be $5$

      This narows c to 500 numbers. If i have more ideas i will post here.
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.