U, V, C are 3 distinct digits..values can vary from 1 to 9.

CU is a concatenated number

Solve for U,V, C from the following relationship:

$U^V$ X $V^U $ = $CU $

This will give some basis to upcoming Unique Pan digital Fraction problems.


2 Answers 2


U = 2, V = 3, C = 7: 23 × 32 = 72.
U and V can't be too big, if U is 2 the product becomes greater than 100 even for V = 4. On the other hand, if U or V = 1, the equation becomes U = CU or V = CU, single digit on the left and two digits on the right, which is also impossible.
If we try 2 and 3, which are not too big and not too small, we get the product 72.

  • 1
    $\begingroup$ ninjaed by you... :(, nvm, have an upvote! $\endgroup$ May 18, 2019 at 4:09
  • $\begingroup$ Essentially, the number set can be constructed from U, V..This info will be helpful for the next puzzle to be posted on a unique set . $\endgroup$
    – Uvc
    May 18, 2019 at 9:30

If $U=1, 1*V=CU$ (not possible)
If $V=1, U*1=CU$ (not possible)
When $U=2$,

When $V=1, 2^1*1^2=2$ (rejected)
When $V=2, U=V$ (rejected)
When $V=3, 2^3*3^2=72$ (possible)

When U=3,

When $V=1, 3^1*1^3=3$ (rejected)
When $V=2, 3^2*2^3=72$ (rejected)
When $V=3, U=V$ (rejected)

When U=4+,

$V=1$ (rejected)

Final Answer:

When $U=2, V=3,$ and $C=7, 2^3*3^2=72$

  • $\begingroup$ ninja-ed by @Mariia, yet just wanna post this as a more complete explanation $\endgroup$ May 18, 2019 at 4:09

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.