Knowing your A,B and C's?
Given: C C × A A = A B B A
and A + B = C , A = B + 1
then what is the value of C = ?
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2 solutions
Thank you.
The first equation can be written as ( 1 0 C + C ) × ( 1 0 A + A ) = 1 0 0 1 A + 1 1 0 B
Substituting from the other two equations A = B + 1 and C = 2 B + 1 into this will produce an equation in B with solution B = 4
So that A = 5 and C = 9
Thank you for the nice solution.
If we know, that these numbers are written in the decimal system, we can get the same unique solution without knowing, that A = B + 1 and C = 2B + 1.
1 1 C × 1 1 A = 1 0 0 1 A + 1 1 0 B
1 1 C × 1 1 A = 1 1 ( 9 1 A + 1 0 B )
1 1 A C = 9 1 A + 1 0 B
A × ( 1 1 C − 9 1 ) = 1 0 B
Now, since A and B are positive, (11C - 91) has to be positive as well. This only happens, when C = 9.
(Which gives us A = 5 (since both sides 10| ) and B = 4. Indeed, 55 × 99 = 5445.)
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I would post your comment as a solution too.
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Posted it as a solution, as you suggested.
We can get the same unique (decimal) solution without knowing, that A = B + 1 and C = 2B + 1.
C C × A A = A B B A
( 1 0 C + C ) × ( 1 0 A + A ) = 1 0 0 0 A + 1 0 0 B + 1 0 B + A
1 1 C × 1 1 A = 1 0 0 1 A + 1 1 0 B
1 1 C × 1 1 A = 1 1 ( 9 1 A + 1 0 B )
1 1 A C = 9 1 A + 1 0 B
A × ( 1 1 C − 9 1 ) = 1 0 B
Now, since A and B are positive, (11C - 91) has to be positive as well. This only happens, when C = 9.
(Which gives us A = 5 (since both sides 10| ) and B = 4. Indeed, 55 × 99 = 5445 (and A = B + 1 and C = 2B + 1 are also true).)