Use pure logic and deductive reasoning only. Document detailed steps in arriving at final solution.

You can use calculator minimally.

$ Given$:

Utilizing These two unique set of Pan digital fractional relations, solve for all the digits. You can try solving first without helping hand from this U V C equation

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

CU is a concatenated number

$U^V x V^U$ = $CU$

$Unique $Set of $Pan digital $ $fractional $ and $Decimal Relationships:$

All the letters represent distinct digits(1 to 9).

All the words in both equations on either side represent concatenated Numbers




$U = 2$ $V = 3$ $C=7$ Reasoning
$u^v \times v^u = cu$ means that the multiplication must result in a 2 digit number.
This means u and v are small digits and there are not so many cases. We can try them out.
None of them can be 1 as the result it would be a 1 digit number.
u=2, v=3 results in 8x9 = 72. Good
u=2, v=4 results in 16x16>100. Stop incrementing v
u=3, v=2 results in 9x8 = 72. Not good.
Rest of the cases go over 100.

From the second division:

I actually transformed it into a multiplication and concluded that a can be 0 or 5 otherwise we don't get a n integer from ๐‘‚๐ฟ๐ด๐บ๐ธ๐‘‰๐ต. But A cannot be 0 since the first division contains one factor that starts with A. SO A=5.


From the same second division, then decimals are now o75. This, multiplied by 72 must be divisible with 1000 in order to get an integer. The only possible values for o are 3 and 8, but 3 is already taken. So o=8.
From the same equation I got that b=1 since a six digit number (ignore decimals) multiplied by 72 we get a 7 digit number. if b>=2 we get an 8 digit number.
Taking decimals again we get 875*72 = 63000. This means that 2*L+3 = 1 (mod 10). So 2*l = 8 (mod 10). So L can be 9 or 4.
but bugvel.oca = 12g3el.875 > 120000. And 120000*72 = 864000 so ๐‘‚๐ฟ๐ด๐บ๐ธ๐‘‰๐ต > 864000. So L>6. So L=9.

Now we are left with...

2 letters g,e and 2 digits. 4,6. We could probably calculate them but there are only 2 combinations. So let's do the math.
g=4, e = 6: 124369,875ร—72 = 8954631. Valid.
g=6 e=4: 126349,875ร—72 = 9097191. Not valid.

Now we got all:

B=1, U=2, V=3, G=4, A=5, E=6, C=7, O=8, L=9.

  • $\begingroup$ Excellent deductive quick work...would you be able to solve it without Uvc relation? $\endgroup$ – Uvc May 24 '19 at 15:24
  • $\begingroup$ Not sure. I didn't even tiuch one of the divisions. I didnt even check it because i am lazy. Maybe, maybe not $\endgroup$ – Marius May 24 '19 at 16:48

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