Puzzling Weekend! UVc is back again with Lots of Magic... Enjoy the Show!

Given: LOTS is being multiplied by F to yield MAGIC.

Digit 0 to 9 are all represented by ten different letters in this multiplication Alphametic.

LOTS, MAGIC represent two concatenated Numbers.

$ LOTS $ X $F$ = $MAGIC$


1) F represents 3,

2). M, G, C are consecutive, but may not be in order.

3). A, I are also consecutive.

4). Also, LOTS contain 3 consecutive digits not necessarily in order. One of the consecutive digits adds upto rest of the two consecutive digits.

Figure them out and Enjoy the Show!!!

  • $\begingroup$ the rest instead of test? $\endgroup$ May 18, 2019 at 15:11
  • $\begingroup$ Thx for catching it. $\endgroup$
    – Uvc
    May 18, 2019 at 15:12
  • $\begingroup$ welcome:) according to (4), the 3 digits must be 1,2,3. yet F=3. possible error? $\endgroup$ May 18, 2019 at 15:12
  • $\begingroup$ F is the multiplying digit...I wanted to bring it down to show it better..LOTS is being multiplied by single digit F to yield MAGIC. $\endgroup$
    – Uvc
    May 18, 2019 at 15:15
  • $\begingroup$ i mean will one of LOTS repeat with F? $\endgroup$ May 18, 2019 at 15:16

2 Answers 2




$1$ or $2$

Since $F=3$

Even if $L = 9$, $M$ would also be $2$. Carry is not possible since $M$ would then be 3 ($M=F$)


$3$ (since only then there would be carry)





$F=3$. $M=1,2$, and so $MGC=\{0,1,2\}$, but $C\ne0$. As $M+G+C=3$ and our answer is divisible by $3$, $3|(A+I)$, so $A+I=4+5, 7+8$. From clue 4, $A+I=7+8$, and $LOTS=\{4,5,6,9\}$. $S=4$ as $3\times4=2$, and no other value produces an output in the range required, so $C=2, M=1, G=0$. The rest I guessed.

  • $\begingroup$ Note that last hint that the sum of two consecutive digits in LOTS must add up to the third, but $4+5\ne 6$. $\endgroup$ May 18, 2019 at 15:36
  • $\begingroup$ @TheSimpliFire; I think he means $4+5=9$. $\endgroup$
    – JMP
    May 18, 2019 at 15:37
  • $\begingroup$ How did you deduce M = 1,2 ? $\endgroup$ May 18, 2019 at 15:50
  • $\begingroup$ @NoLand'sMan; LOTS is at most 9876, which would make M=2, or it could be 1. I assumed it was not zero, because leading zeroes are frowned on. $\endgroup$
    – JMP
    May 18, 2019 at 15:54
  • $\begingroup$ Note \mid for $\mid$. $\endgroup$ May 18, 2019 at 15:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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