3
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Given:

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.

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6
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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.

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  • 1
    $\begingroup$ ninjaed by you... :(, nvm, have an upvote! $\endgroup$ – Omega Krypton May 18 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 at 9:30
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If $U=1, 1*V=CU$ (not possible)
If $V=1, U*1=CU$ (not possible)
When $U=2$,

$2^V*V^2<=92$
$V<=3$
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,

$3^V*V^3<=93$
$V<=3$
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+,

$4^V*V^4<=94$
$V=1$ (rejected)

Final Answer:

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

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

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