# Find the values of U, V, C based on the given relationship…useful for upcoming puzzles

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.

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.

• ninjaed by you... :(, nvm, have an upvote! – Omega Krypton May 18 at 4:09
• 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 . – Uvc May 18 at 9:30

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)

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