U, V, C are three distinct digits ( 0 to 9 ).

UVVVV and CVVVV.U are concatenated numbers.

Dot “.” Stands for decimal.



Find U, V , C

  • 2
    $\begingroup$ hmm inspiration for a mathematical puzzle with t, o and m or t, 0 and m maybe... :-) $\endgroup$
    – tom
    Commented May 19, 2019 at 16:31

3 Answers 3


Finding $C$

  1. If dividing an integer by $C$ gives a fraction with exactly one digit after the decimal point (note that $U=0$ doesn't work), then $C$ must be non-coprime with $10$, i.e. it must be one of $2,4,5,6,8$.

  2. If $C\geq45$, then the right-hand side is more than $40,000$, and after multiplying by $C$ it won't be a 5-digit number any more. So we must have $C=2$.

Finding $U$ and $V$

  1. Since $C=2$, the division by $C$ must give $U=5$.

  2. Since $UVVVV$ divided by $2$ is not an integer, $V$ must be odd. Trying the possibilities in turn shows that $V=9$ is the only one which works.


$U=5,V=9,C=2$. The equation is $59999/2=29999.5$.

  • 1
    $\begingroup$ C only has to be even (or 5) for part 1. $\endgroup$
    – JMP
    Commented May 19, 2019 at 15:58
  • $\begingroup$ To be precise, the condition is that the prime factors of C contain 2 and/or 5. This makes the following numbers valid: 2,4,5,6,8. Point 2 can still get the right value with the same method. $\endgroup$
    – Leo
    Commented May 20, 2019 at 1:42
  • $\begingroup$ @JonMarkPerry and Leo: Oops! Thanks for the tip; I modified my answer just slightly to take these possibilities into account. $\endgroup$ Commented May 20, 2019 at 7:53
  • $\begingroup$ you can use $\gcd(C,10)\gt1$ for 'non-coprime' $\endgroup$
    – JMP
    Commented May 20, 2019 at 8:03



$C=1,2,3$ due to RHS being $\sim C^2$ in magnitude, which must be five digits. $C=1$ means $U=0$ which is impossible, and $C=3$ means $(C\times .U) \pmod 1 \equiv 0$ which is also impossible. Therefore $C=2, U=5$.

and then:

We now have$\frac{5VVVV}{2}=2VVVV.5$ which leads to $5000+\frac{VVVV}{2}=VVVV.5$ by cancelling $20,000$ from each side. So $10000+VVVV=2VVVV+1$ and then $VVVV=9999$, so $V=9$.


Ok so others have got there before me, but I have a slightly different way of approaching I think...

C cannot be 1,3,7,9,0 because.. 1 would not give a decimal point, 3 7 9 would give more than one decimal point or no decimal points e.g. 2/3 = 0.66666etc., and we cannot conveniently divide by 0 - thus C must be 2,4,5,6,8


C = U/C more or less because when the divide the first digit, U, by C we need to get C in the first column... Now for C=2 we could have U=4, but we can't have .4 at the end of the answer we need .5 so U must be 5 if C=2 - this works because 5VVVV/2 = 2VVVV.5 is possible if V is odd... but if we try C=4,5,6,8 we cannot find a single digit that will fit e.g. 8VVVV/4 = 2VVVV... and we need 4 at the front... thus for C>2 we cannot get the first digit to work - thus C=2 and U=5 (unless, of course, C=3 and U=9, but then we would not get the single digit after the decimal point at the end of number... )


By inspection V=9 works if C=2 and U=5 and V=1,3,5,7 do not work (of course even V does not give .5 at the end) so we have $$59999/2=29999.5$$


this works for any number of 9s in the middle... 5/2 = 2.5; 59/2 = 29.5;....; 59999999999/2 = 29999999999.5 etc.

  • $\begingroup$ $\frac{33}{6}=5.5$ $\endgroup$
    – JMP
    Commented May 19, 2019 at 16:32
  • $\begingroup$ @JonMarkPerry - oh rats... now I understand your comment above better... to the first answer -- answer edited to correct this $\endgroup$
    – tom
    Commented May 19, 2019 at 16:33
  • $\begingroup$ @tom..sure..generating all infinite number of non-prime palindromes $\endgroup$
    – Uvc
    Commented May 19, 2019 at 16:43

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