$\square\square\times\square =\square\square\square =\square\times\square\square\,\,\,$ fill blanks with distinct numbers from$\{1,2,3,4,5,6,7,8,9\}$

Question

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.

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. — Ben Grossmann, Jul 18 '13 at 17:13
Well you can at least say that $x_1x_3 \ge 10$ and $x_7x_8\ge 10$ — xavierm02, Jul 18 '13 at 17:14
Note that neither the third nor the seventh digit can be $1$, since that could not produce a three-digit product. — dfeuer, Jul 18 '13 at 17:15
Observe there is more than one way of doing it, consider swapping the first blank with the second to last, the second blank with the last blank, and the third blank with the third to last blank. — George V. Williams, Jul 18 '13 at 17:15
The 1 has to appear in the three-digit number, since as a one-digit it could not multiply to a three-digit, and as the second digit then $n\times(10+m)$ would be less than 200. — Stefan Hamcke, Jul 18 '13 at 17:16
Another thing is that the x1, x2x3 , x7 , x8x9 are factors of x4x5x6 — Harish Kayarohanam, Jul 18 '13 at 17:27
@StefanH.: Why couldn't $1$ appear in the units place of a two-digit number? — dfeuer, Jul 18 '13 at 17:29
@dfeuer: If 1 were a first digit in a two-digit, then $n\times(10m+1)$ would give $n$ as the first digit, so $n$ would appear twice. If 1 were one of the one-digit numbers, then $1\times(10n+m)$ would be less than 100. If 1 were a second digit, then $n\times(10+m)$ would be less than 163. If 1 — Stefan Hamcke, Jul 18 '13 at 17:30
vertical space I can't edit my comment since the edit link is covered by one of the Related questions. lol — Stefan Hamcke, Jul 18 '13 at 17:32
Note that $n\times(10+m)$ is at most 162. $n\times(10m+1)$ would lead to $n$ appearing twice. And $1\times(10n+m) would be less than 100. — Stefan Hamcke, Jul 18 '13 at 17:37
@xavierm02 What if there's a carry from the first multiplication that brings the middle $\ge100$? E.g. $14\times9$. — rurouniwallace, Jul 18 '13 at 18:01
There are four solutions. I doubt that the search tree can be placed on one math.stackexchange page. — Christian Blatter, Jul 18 '13 at 18:13
@ZettaSure : Yeah right. For some reason I thought that because they were digits, that wouldn't happen. — xavierm02, Jul 18 '13 at 18:33

user86828 · Accepted Answer · 2013-07-18T19:28:15.620

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.

Thanks for editing the answer. – user86828 Jul 18 '13 at 19:16 — user86828, Jul 18 '13 at 19:16
it would be great if you could finish the edit :) – Nathaniel Bubis Jul 18 '13 at 19:18 — Nathaniel Bubis, Jul 18 '13 at 19:18
@user86828 Great! Thanks! – Mahdi Khosravi Jul 18 '13 at 21:12 — Mahdi Khosravi, Jul 18 '13 at 21:12

rurouniwallace · Answer 2 · 2013-07-18T19:09:55.047

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.

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

Nathaniel Bubis · Answer 3 · 2013-07-18T18:35:14.840

2

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]

edited Jul 18 '13 at 18:35

answered Jul 18 '13 at 18:30

Nathaniel Bubis

33,018

1

How do we know that these are the only two? – rurouniwallace Jul 18 '13 at 18:32
@ZettaSuro - brute force. – Nathaniel Bubis Jul 18 '13 at 18:33
4

Of course it can be done by programming , But the question is mathematical approach – Harish Kayarohanam Jul 18 '13 at 18:37
3

@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. – Nathaniel Bubis Jul 18 '13 at 18:39
I really doubt what inspiration it gains . The person @Maths lover has told that he got the solution by trial and error. So solution is not the problem but how we can get it is the problem – Harish Kayarohanam Jul 18 '13 at 18:42
I hope this answer can be a comment . The owner @Maths lover should remove this from it being an accepted answer . – Harish Kayarohanam Jul 18 '13 at 18:44
Well, at the very least it corroborates Ross Milikan's observation that it would be a good idea to have the product divisible by $6$. – rurouniwallace Jul 18 '13 at 18:46
2

Useful to check. – user86828 Jul 18 '13 at 19:49

score 1 · Answer 4 · answered Jul 18 '13 at 19:18

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.

$\square\square\times\square =\square\square\square =\square\times\square\square\,\,\,$ fill blanks with distinct numbers from$\{1,2,3,4,5,6,7,8,9\}$

4 Answers4