Thanks to Tyler and Amit for suggestion! Thanks Rolan for alerting me of a typo in an earlier version!

Here’s a nice logical puzzle. If ABCDE times 4 equals EDCBA, and each letter is a different digit from 0 to 9. What is the value of each letter?

As usual, watch the video for a solution.

ABCDE Times 4 Equals EDCBA

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“All will be well if you use your mind for your decisions, and mind only your decisions.” Since 2007, I have devoted my life to sharing the joy of game theory and mathematics. MindYourDecisions now has over 1,000 free articles with no ads thanks to community support! Help out and get early access to posts with a pledge on Patreon.

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Answer To ABCDE Times 4 Equals EDCBA

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

Let’s start with A. Since 30000×4 = 120000 we cannot have A being a value 3 or larger since the resulting number is too large: we need the result to be a 5 digit number. So A = 0, 1, 2.

Since D times 4 results in a units digit A, we know A is even. So A = 0 or 2. If A = 0, then let’s consider the times table for 4.

4×0 = 0
4×1 = 4
4×2 = 8
4×3 = 12
4×4 = 16
4×5 = 20
4×6 = 24
4×7 = 28
4×8 = 32
4×9 = 36

So we need D = 0 or 5 for the resulting units digit to be 0. But A = 0 is already taken, so D = 5. But now we have the largest possible product is 09875×4 = 39500, but if D = 5 we need the result to be at least 50000. So we cannot have either case, and A = 0 does not work.

Therefore A = 2. Again using the times table for 4, we need D to be 3 or 8 so that D×4 ends in a 2. But if A = 2, the number will at least be 20000×4 = 80000. So clearly D = 8 is the only possibility.

The resulting number EDCBA = 8DCB2 is a multiple of 4, so its last 2 digits must be a multiple of 4. Thus B2 is a multiple of 4.

4×0 = 0
4×1 = 4
4×2 = 8
4×3 = 12
4×4 = 16
4×5 = 20
4×6 = 24
4×7 = 28
4×8 = 32
4×9 = 36

From the times table, it could be B2 = 12 or 32. But if B = 3, then we would have 23000×4 = 92000 which is too large since E = 8 and the result is in the 80000s. So we must have B = 1.

Now we have 21CD8 times 4 equals 8DC12. So we must have D8×4 be a number with the last two digits 12. We can check the possibilities:

4×08 = 32
4×18 = 72
4×28 = 112
4×38 = 152
4×48 = 192
4×58 = 232
4×68 = 272
4×78 = 312
4×88 = 352
4×98 = 392

We either have 4×28 = 112 or 4×78 = 312 so D = 2 or 7. But A = 2, so we must have D = 7.

We are almost done. We can solve for C with some algebra.

21C78 × 4 = 87C12
(21078 + 100C)4 = 87012 + 100C
84312 + 400C = 87012 + 100C
300C = 2700
C = 9

To summarize, A = 2, B = 1, C = 9, D = 7, E = 8, meaning 21978 × 4 = 87912, and that’s the unique solution!

References

web2.0calc
https://web2.0calc.com/questions/if-abcde-x-4-edcba-what-is-the-number-abcde-as-usual

AGF Tutoring
https://www.agftutoring.com/abcde-4-edcba/

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