Base a and b
A number can be represented as 123 in base a . The same number can also be represented as 146 in base b . Find the minimum value of a + b .
The answer is 15.
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2 solutions
It's really neat that both sides of the equation were two away from perfect squares. Nice solution!
Notice that from ( a + 1 ) 2 = ( b + 2 ) 2 , you cannot tell that ( a + 1 ) = ( b + 2 )
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This is a good point, but since neither a nor b can be negative numbers, the second solution (a+1=-b-2) is trivial.
We must solve a 2 + 2 a + 3 = b 2 + 4 b + 6 . A neat trick is to write s for the sum of a and b , and d for their difference. a = 2 s + d , b = 2 s − d . By doing so, I know that the squares s 2 and d 2 will cancel on both sides. 4 1 ( s + d ) 2 + ( s + d ) + 3 = 4 1 ( s − d ) 2 + 2 ( s − d ) + 6 . s d − s + 3 d − 3 = 0 . ( s + 3 ) ( d − 1 ) = 0 . Thus, either the sum s should be − 3 (which is impossible, since no negative bases are allowed), OR the difference d should be equal to 1. That gives a = b + 1 .
But b ≥ 7 because a digit "6" is used. Thus we choose a + b = 8 + 7 = 1 5 .
123 in base a = 146 in base b
a 2 + 2a + 3= b 2 +4b +6
( a + 1 ) 2 +2 = ( b + 2 ) 2 +2
(a+1)^2 = (b+2)^2
As a and b are positive, a+1=b+2
a-1=b
The minimum value of b is 7
a=8
8+7= 15