Given a simple weighting scale with two pans, we are given a weight T and some other weights which are the powers of a specific number a, our goal is to balance these pans using the given weights. More formally we need to satisfy this equation,
T + (some power of a) = (some other powers of a)
Remember that we are given exactly one weight corresponding to one power.
T = 11 and a = 4 Then scale can be balanced as, 11 + 4 + 1 = 16
We can see that in this problem our first goal is to represent T in terms of powers of a so when we write T with the base of a, if representation has only 1s and 0s then we know that weight corresponding to 1s can make the T. For example,
If T is 10 and a is 3 then representing 10 on base of 3 we get 101 i.e. using 0th power of 3 and 2nd power of 3 (1 + 9) we can get 10
Now if all digits of base representation are not 1 or 0, then we can’t balance the scale, except the case when in base representation some digit is (a – 1) because in that case, we can transfer corresponding power to T’s side and increase the left number in base representation by 1. For example,
If T is 7 and a is 3 then representing 7 on base 3 we get 021. Now while looping over representation from right to left we get 2 (which is 3 - 1), in this case, we can transfer corresponding power(3^1) to T’s side and increase the left number by 1 i.e., T’s side 7 + 3 = 10 and representation 101 (9 + 1) which has only 1s and 0s now, so scale can be balanced.
So now algorithm for solving this problem can be represented as follows, represent the T in base of a, if all digits of base representation are 1s or 0s, scale can be balanced otherwise loop from right side of the representation if any number is (a – 1) then increase left side number by 1, keep doing this and ignore 1, 0, (a – 1) and a cases in representation. If complete base representation is processed scale can be balanced otherwise not.
Balance is possible
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