A math problem: enriching uranium!

Mon, 17 Jan 2022 13:14:20 +0330

The uranium enriching process is a simple yet hard process. We want to reach 3.5% uranium and we only have got 0.7% uranium. To achieve this goal, we should feed the uranium to a centrifuge. For each 2 units of x% uranium which we feeds to the centrifuge, we gets 1 unit of (x-0.1)% and 1 unit of (x+0.1)% uranium. For example a unit of 0.6% and a unit of 0.8% for 2 units of 0.7%. 0.0% uranium will be discarded. Units are always positive integer and we cannot feed 1 unit of uranium, like to get two $ \frac{1}{2} $ unit.

So to reach 3.5% uranium, 3 questions get raised:

Will we be able to produce 3.5% uranium from $m$ units of 0.7% uranium?
For $m$ units of 0.7% uranium how many units of 3.5% uranium can we produce?
If with some units of 0.7% uranium, we cannot produce at least a unit of 3.5% uranium, what is the richest which we can produce?

You can solve these questions either by your math skills or your coding skills. Good luck :)

2^x+2^y=k

Sun, 16 Jan 2022 21:49:16 +0330

2x + 2y = k where x, y and k are positive integers and k is a constant. The question’s that for different values of k how many solution does our equation have? For any odd k there is no solution and for positive powers of 2 there is one. We want a general solution for this equation.

Solution by Susam Pal

We will take this fact for granted: Every positive integer can be uniquely expressed as the sum of distinct powers of 2. In other words, every positive integer has a unique binary representation.

Now consider the case where x = y. Let m = x = y. Then k = 2m + 2m = 2(m+1). Further, if k = 2(m+1) where m > 1, x = y = m is a solution. There cannot be a solution in which x != y because that would result in another binary representation of k apart from the existing representation of 2(m+1) which would contradict the uniqueness of the binary representation of k. Therefore there must be exactly 1 solution when k is a positive power of 2.

Now consider the case where x != y. Let x = m and y = n. Then k = 2m + 2n = 2n + 2m. Thus both (x = m, y = n) and (x = n, y = m) are solutions. There cannot be a solution in which x = y because that would lead to k = 2(x+1) which violates the uniqueness of the binary representation of k. Also, there cannot be another solution in which either x or y does not belong to {m, n} because that would too violate the uniqueness of the binary representation of k. We conclude that there must be exactly 2 solutions when k is a sum of two distinct positive powers of 2.

For any other positive integer k, no solutions are possible. We can show this by contradiction. Let k be such that it is neither a positive power of 2 nor a sum of two distinct positive powers of 2. If we assume that a solution exists for such k, then we have two positive integers x and y such that 2x + 2y = k. If x = y, it contradicts the fact that k is not a positive power of 2. If x != y, it contradicts the fact that k is a sum of two distinct positive powers of 2.

Thus, we can conclude that the equation has exactly 1 solution when k is a positive power of 2, 2 solutions when k is a sum of two distinct positive powers of 2, no solutions otherwise.

Math_puzzle on Farooq's

A math problem: enriching uranium!

2^x+2^y=k

Solution by Susam Pal