## Sunday, December 11, 2005

### There is an integer

There is an integer whose first digit is 3 having the property that if you take the 3 from the beginning of the number and place it at the end, you will have multiplied the original number by 3/2. What is that number?

Hint 1: it's not a small number.
Hint 2: The logic is hard, but the math is easy.

1. Hint 3: If A is the first number and B is the second, then 3A/2 = B. Therefore 3A/2 ends in 3, so A/2 must end in 1.

2. The number is 3,529,411,764,705,882, which when multiplied by 3/2 produces 5,294,117,647,058,823.

To see how to get these numbers, we'll call them A and B, respectively. The first step in determining A is to see that its last digit must be 2. That's because 3A/2 ends in 3, so A/2 must end in 1.

If A ends in 2, B must end in 23, because B is formed from A by putting the 3 at the end. But B = 3*(A/2), so A/2 must end in 41 (4 is the only digit whose product with 3 ends in 2). If A/2 ends in 41, A ends in 82.

Keep going, working right to left, until we get to a point where we encounter a 3, at which point the successive divisions end.

3. It was awesome

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