Below, you’ll find the answers to last week’s puzzles. For those who didn’t get a chance, if you would like to take a crack at the Week 3 challenge, just click here.

**1.** (2/8+6)*4 = 25. The easiest way to come to this answer is through trial and error. But that would be pretty frustrating if you don’t take a few things into account.

We know that adding, subtracting, or multiplying two even numbers will give you a number that is also even. In order to get to 25, an odd number, therefore, we’re going to have to use division somehow. So perhaps the next thing you might ask yourself is how you can get to 25 through division. The most obvious options are 100/4 and 50/2. That would mean you would just need to get to 100 by using 2, 6, and 8, or to 50 by using 4, 6, and 8. But unfortunately, no matter how long and hard you think about it, you’re not going to be able to do that.

So what else can you do? You can remember the properties of elementary operators. If you know that multiplication and division can operate over addition and subtraction, then you might be able to, in a sense, use a number more than once without running afoul of the rules. The other thing you can keep in mind is that you can use the divisor to create a fraction, as long as we’re left with that whole number, 25, in the end.

If you divide 2 by 8 and then multiply the result by 4, you get 1. Then you’re left with 6. So how can you use the numbers 1 and 6 to get 25? Remember that distributive property! You’ve already multiplied 2/8 by 4 — but you can also multiply 6 by 4 if you allow it to operate over addition. 24+1=25.

**2.** There are four perfect squares whose digits are all the same (0, 1, 4, and 9). They’re all a single digits. How can we show that? We can start with the fact that any square has to end in 0, 1, 4, 5, 6, or 9. Then, since we’re looking to see if any numbers with more then one digit fulfill our requirement, let’s look at the tens digit. The possible endings are 11, 44, 55, 66, or 99. We can rule out 55 immediately, because all squares ending in 5 also end in 25. Squares that end in 1, 4, and 9 have even digits in the tens place, while squares that end in 6 have an odd number in the tens. That means we can eliminate 1, 9, and 6. Four is still an option, though. We know that 44 is not a square, we can move on to the hundreds spot. But it turns out that any square ending in 44 has an odd number in the hundreds place. So we’re left with our original four options: 0, 1, 4, and 9.

**3.** 9376 is the square root of a number that ends in 9676. The easiest way to approach this problem is also to rule out the various options. We know that the number will end in 0, 1, 5, or 6. How do we know that? If the last digit is 0, its square will end in 0. If it’s 1 or 9, its square ends in 1. If it’s 2 or 8, its square ends in 4. It’s 3 or 7, its square ends in 9. If it’s 4 or 6, it ends in 6, and if it’s 5, it ends in 5 (actually 25). Since the digits of the square root have to match the the last four digits of the square, we can rule out 4 and 9, and we’re left with 0, 1, 5, and 6.

Then we have to look at the tens place. Here, the easiest thing is to use a little brute force. Think of the number as x0, x1, x5, or x6. Then we can use trial and error, plugging in each integer in the x slot, squaring the two-digit number, and seeing if you end up with a number that ends in the original two-digit number. (So, for instance, try squaring 01, 11, 21, 31, 41, 51, etc.) What works for x0? Nothing! Great, we can rule it out. For 1, only 01 works. 25 works for x5. For x6, we have 76.

Then we move onto the hundreds spot and do the same thing. Nothing works for x01. For x25, 625 is still a candidate. For x76, we’ve got 376.

Then, finally, you do the same thing with the thousands. There isn’t a four digit number ending in 625 whose square ends with those same digits. But there is one for x376. Since I used ascending numbers for x, it’s the last one I tried: 9376.