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Wendy from China asks, "three digit number. When Written in base eight, the number still has three digits, but theY are eXactlY the reVerse of its base ten representation. What is the number?"

If I understand you correctly, you're looking for a base ten number with three digits, such that if you reverse the digits, you get the same number written as a base eight numeral?

In other words, you're looking for digits x,y, and z, such that:

xyz_{ten }= zyx_{eight}.

Well, let's explore this question and see what happens. First, since xyz is a base ten number, we know that x, y, and z must all be less than 10. On the other hand, zyx is a base eight number, so we can conclude that they are actually less than 8.

So we have three digits between zero and seven inclusive. So let's suppose there is a number that works. We can write the following equation:

100x + 10y + z = 64z + 8y + x

2y = 63z - 99x

2y = 9(7z - 11x)

Now, you might be wondering why I moved the z and x terms to the right and then factored out a 9. It wasn't a completely random choice. This equation is what we call a Diophantine equation (an equation in which all the solutions must be integers). In Diophantine equations, we often are looking for factors and multiples, so looking for something we can factor out is often helpful. Since I spotted that both 63 and 99 are divisible by 9, I thought it would be helpful to arrange it like this. So what does that do for us?

Well, since x and z are both integers, we can conclude that the right-hand side of the equation is a multiple of 9. This means that the left-hand side must also be a multiple of 9. But wait a minute! How can 2y be a multiple of 9? Only if y is a multiple of 9. Our first instinct would be to say that y must equal 9, but remember that it has to be less than eight. So what could it be? Only one possibility: y = 0.

But if y equals zero, that means that 7z - 11x must also be zero, in order for the two sides of the equation to be zero.

7z = 11x, or x = (7/11)z.

Since x is an integer, that means (7/11)z is also an integer. But that can only happen if z is a multiple of 11. How many integers between 0 and 7 inclusive are multiples of 11? Only one! z = 0. But this means x = 0 too!

So in the end, the only answer is a very uninteresting one: the number is 000.

000_{ten} = 000_{eight}.

I'd be hard presed to say that 000 is a three-digit number, so I'd prefer to say that there's no solution.

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