Find the magical number!

Logic Level 3

$\large{\begin{array}{ccccccc} && & A & B& C & D&E\\ \times && & & & & &4\\ \hline & & & E & D& C & B&A\\ \hline \end{array}}$

Solve the cryptogram above, where $A,B,C,D$ and $E$ are not necessarily distinct digits. Submit your answer as the 5-digit integer $\overline{ABCDE}$ .

The answer is 21978.

2 solutions

Samara Simha Reddy
May 17, 2016

$ABCDE < \dfrac{1}{4} \times 100000$

$ABCDE < 25000$ So, A is 1 or 2.

Since, EDCBA is a multiple of 4, So it is even, $\color{#20A900}{\boxed{A = 2}}$

$4 \times ....E = ....2$ So either E is 3 or 8. But $4 \times 2....$ cannot be $3....$ So E is 8. $\color{#D61F06}{\boxed{E = 8}}.$

We have $2BCD8 \times 4 = 8DCB2$

and $4 \times BCD + 3 = DCB$

Lets look at it case by case, for the value of D.

$D = 1, B = 7\\ D = 2,B = 1\\ D = 3,B = 5\\ D = 4, B = 9\\ D = 5,B = 3\\ D = 6,B = 7\\ D = 7, B = 1\\ D = 8,B = 5\\ D = 9,B = 9\\ D = 0, B = 3$

But B cannot be greater than 2, or there would be a carry and B cannot be equal to 2, since A = 2. So $\color{#3D99F6}{\boxed{B = 1}}$ and D may be 2 or 4.

Again A is already 2, So $\color{#69047E}{\boxed{D = 7}}.$

So, we have $21C78 \times 87C12$

We now have $4 \times C + 3 = C \mod 10$

$3 \times C = 7 \mod 10$

$\therefore \color{#EC7300}{\boxed{C = 9}}.$

So the required number is $\large \boxed{\color{#20A900}{2} \color{#3D99F6}{1} \color{#EC7300}{9} \color{#69047E}{7} \color{#D61F06}{8}}$ .

Although your final answer is correct, your have one minor mistake of the part: A is already 2 so D = 7. The problem state that the digits are not necessary distinct.

The correct reasoning is that B*4 + something = D so D could not be 2, therefore D = 7

Tran Hieu - 5 years ago

Yeah sir! That's what I have stated right there!

Samara Simha Reddy - 5 years ago

@Tran Hieu is correct. The question stated that the digits are not necessarily distinct (meaning they can be the same). The explanation used will then be invalid. The correct way to eliminate this (besides changing the question) would be this:

If $B=2$ , then $4D+3 = 10n + 2$ for some integer $n$ . We see that this is impossible, since the LHS will have an odd integer, while the RHS will have an even integer. Therefore, the only option is $B=1$

Again, the explanation for the value of $D$ would be this:

If $D=2$ , to meet the thousands digit requirement, we will need $1 \times 4 + 8 = 12$ . This is impossible, since multiplication by $4$ only gives a maximum carry of $3$ , and we do not want a carry in the ten-thousands digit. Therefore, $D=7$

Surprisingly, even with the "not necessarily distinct digits" placed there, the solution is still unique.

Anyway, typos:

$D$ may be $2$ or $\color{#3D99F6}{7}$

And if I recall correctly, the proper notation for modular operations is $4C + 3 \equiv C \pmod{10}$

Hung Woei Neoh - 5 years ago

@Hung Woei Neoh – Wow! Great @Hung Woei Neoh

Samara Simha Reddy - 5 years ago

@Samara Simha Reddy – Ehhh......were you trying to tag me?

Hung Woei Neoh - 5 years ago

@Hung Woei Neoh – Yeah! Sorry Wrong tag. :D

Samara Simha Reddy - 5 years ago

Nice solution! (+1)

Swapnil Das - 5 years ago

Thank You!

Samara Simha Reddy - 5 years ago

Adarsh Mahor
Jul 29, 2016

A can be 1or 2 but 1 can't come in multiple of 4 at units place so A=2 4x2 =8 so E=8 B can always be 1 cause of E is 8 8+3=11 so at units place there is 1 so 4x7+1=31 D=7 So remaining is C which will always be more than 8 which can be 8 or 9 ..8 is already used so C=9 so ABCDE=21978

plz read carefully for getting all solution

Find the magical number!

The answer is 21978.

2 solutions

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