Slick Cryptarithm

We have $12\overline{ABCDE}=\overline{CDE0AB}$ so $12(1000 \overline{AB} +\overline{CDE})=1000 \overline{CDE}+\overline{AB} \\ 12000 \overline{AB} +12 \overline{CDE}=1000 \overline{CDE}+\overline{AB} \\ 11999 \overline{AB}=988 \overline{CDE} \\ \frac{\overline{AB}}{ \overline{CDE}}=\frac{76}{923}$ We are done!

Let $x=\overline { ABCDE }$

We have $x={ 10 }^{ 4 }A+{ 10 }^{ 3 }B+{ 10 }^{ 2 }C+10D+E$

Multiplying 1000 to both sides, $1000x={ 10 }^{ 7 }A+{ 10 }^{ 6 }B+{ 10 }^{ 5 }C+{ 10 }^{ 4 }D+{ 10 }^{ 3 }E\qquad \qquad \left( 1 \right)$

By the question we have, $12x=\overline { CDE0AB }$

Which gives us $12x={ 10 }^{ 5 }C+{ 10 }^{ 4 }D+{ 10 }^{ 3 }E+10A+B\qquad \qquad \qquad \left( 2 \right)$

Subtracting $\left( 2 \right)$ from $\left( 1 \right)$ we get $988x={ 10 }^{ 7 }A+{ 10 }^{ 6 }B-10A-B\qquad$

Factorising the $R.H.S$ , $988x=\left( { 10 }^{ 6 }-1 \right) \left( 10A+B \right)$ $\Rightarrow x=\frac { \left( { 10 }^{ 6 }-1 \right) \left( 10A+B \right) }{ 988 } \\ \Rightarrow x=\frac { 76923\left( 10A+B \right) }{ 76 }$ (Note that the above fraction is in its simplest form as 76923 and 76 don't share common factors.)

Since $x$ is an integer, $10A+B$ should be a multiple of 76, and since $10A+B$ is a 2 digit number, it has to be 76 only. (For $A=7$ , $B=6$ )

So we get $x=\frac { 76923\left( 76 \right) }{ 76 }$

Which gives us $x=\boxed { 76923 }$

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Verification

For $x=76923$ , we have $A=7$ , $B=6$ , $C=9$ , $D=2$ , $E=3$ .

By plugging in values we have $\frac { \begin{matrix} & 7 & 6 & 9 & 2 & 3 \\ \times & & & & 1 & 2 \end{matrix} }{ 9\quad 2\quad 3\quad 0\quad 7\quad 6 }$

And $76923 \times 12$ is indeed $923076$

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Please upvote if you like the solution.

"New Math solution"

120000A+12000B+1200C+120D+12E= 100000C+10000D+1000E+10A+B

Simplify to the positive side

119990A+11999B=98800C+9880D+988E

Factor:

11999(10A+B)=988(100C+10D+E)

Common factor 13:

(923×13)(10A+B)=(76×13)(100C+10D+E) factor out the 13 on each side and you are left with... 923(10A+B)=76(100C+10D+E)

Using the proof YX=XY you get

10A+B=76 and 923=100C+10D+E

A becomes the 10s digit in the left equation with B in the 1s.

C becomes the 100s digit, D becomes the 10s digit and E in the 1s for the right equation.

AB=76 and 923=CDE

So: ABCDE=76923

Verify: 76923×12=923076=CDE0AB

12 ABCDE = 10ABCDE + 2ABCDE = CDE0AB

           2A  2B   2C  2D  2E

 + A     B   C    D      E     0
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This method is the best

let x=AB, y=CDE, 0<x<100 we have : 12 (1000x+y)=1000y+x <=> 923x=76y, both x and y are integers, so x must be divisible by 76, but x<100 so x=76, and y=923 we have ABCDE= 76923, lets try again, ABCDE x12= CDE0AB and it's right, so the answer is 76923

I used code to solve this (like usual):

Let $x=\overline{AB}$ and $y=\overline{CDE}$ . Then the problem is restated as $12(1000x+y)=1000y+x$ Regroup x and y $11999x=988y$ Divide out $\gcd(11999, 988)=13$ $923x=76y$ The numerical values being coprime, we must have $x=76n \text{ and }y=923n$ for some integer value n. Because $10\leq x \leq 99$ , we infer that $n=1$ so that $x=\overline{AB}=76, y=\overline{CDE}=923$ . Check: $76923×12=923076$

$12(10000A+1000B+100C+10D+E)=100000C+10000D+1000E+10A+B$

$11ABCDE=99900C+9990D+999E-9990A-999B$

$11ABCDE=999(100C+10D+E-10A-B)$

Therefore, $ABCDE$ is a multiple of 999,since 999 is not divisible by 11.Then,

$AB+CDE=999$ and $CDE-AB=11(\frac{ABCDE}{999}$ )

$2AB/11$ has a remainder of 9 ,so $AB/11$ has a remainder of 10.Therefore A-B=1.Let's test them out one by one.

10989x12 =131868,21978x12=263736,32967x12=395604,43956x12=527472,54945x12=659340,65934x12=791208, 76923x12 =923076 ,87912x12=1054944,98901x12=1186812.The answer is 76923.