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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Sign up to join this communityU = 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.
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 $C=7, 2^3*3^2=72$