**1. Sally is thinking of a 6-digit number. The sum of the digits is 43. And only two of the following three statements about the number are true: (1) it's a square number. (2) it's a cube number, and (3) the number is under 500000. What number was Sally thinking of?**

**Answer:** 499849.

The sum of the digits of Sally's number is 43. 43 is 7 (mod 9), which means when you divide 43 by 9, you get 7 as a remainder.

That means Sally's number is 7 (mod 9). That's because the remainder (mod 9) doesn't change when you replace a number by the sum of its digits. Try a few small examples and you'll see that's true.

There are no cubes that are equivalent to 7 (mod 9). To prove this, just try all nine possibilities, (mod 9):

0^{3} = 0 (mod 9)

1^{3} = 1 (mod 9)

2^{3} = 8 (mod 9)

3^{3} = 0 (mod 9)

4^{3} = 1 (mod 9)

5^{3} = 8 (mod 9)

6^{3} = 0 (mod 9)

7^{3} = 1 (mod 9)

8^{3} = 8 (mod 9)

So for this to be true, Sally's number must be a square between 99999 and 500000 that is 7 (mod 9). The only numbers whose squares are 7 (mod 9) are numbers that are equivalent to 4 or 5 (mod 9) -- again, try all nine possibilities to convince yourself this is true. Also, the only numbers whose squares are between 99999 and 500000 are between 316 and 708. There are 88 numbers that are equivalent to 4 or 5 (mod 9) between 316 and 708. With a spreadsheet program (like Excel) you can check them all pretty quickly to see which one has a square whose digits sum to 43.

**2. This is a variation on a classic "guess the numbers" puzzle: **

I think of 2 single-digit numbers from 1 to 9. I tell Peter Griffin their product and Lois Griffin their sum.

1. Peter: "I don't know the numbers."

2. Lois: "I don't know the numbers."

3. Peter: "I don't know the numbers."

4. Lois: "I don't know the numbers."

5. Peter: "I don't know the numbers."

6. Lois: "I don't know the numbers."

7. Peter: "I don't know the numbers."

8. Lois: "I don't know the numbers."

9. Peter: "Now I know the numbers."

What are the numbers?

**Answer: **2 and 8

Statement 1 by Peter tells you (and Lois) that the product must be 4, 6, 8, 9, 12, 16, 18, 24, or 36, because these are the products that can be made two different ways. The possibilities are:

2 * 2 = 4 (sum 4)

1 * 4 = 4 (sum 5)

2 * 3 = 6 (sum 5)

1 * 6 = 6 (sum 7)

2 * 4 = 8 (sum 6)

1 * 8 = 8 (sum 9)

3 * 3 = 9 (sum 6)

1 * 9 = 9 (sum 10)

3 * 4 = 12 (sum 7)

2 * 6 = 12 (sum 8)

2 * 8 = 16 (sum 10)

4 * 4 = 16 (sum 8)

3 * 6 = 18 (sum 9)

2 * 9 = 18 (sum 11)

4 * 6 = 24 (sum 10)

3 * 8 = 24 (sum 11)

6 * 6 = 36 (sum 12)

4 * 9 = 36 (sum 13)

Statement 2 by Lois tells you the sum can't be 4, 12, or 13, because of all the different ways to make one of Peter's products, these three sums would give away the numbers -- 2+2=4, 6+9=12, and 4+9=13.

1 * 4 = 4 (sum 5)

2 * 3 = 6 (sum 5)

1 * 6 = 6 (sum 7)

2 * 4 = 8 (sum 6)

1 * 8 = 8 (sum 9)

3 * 3 = 9 (sum 6)

1 * 9 = 9 (sum 10)

3 * 4 = 12 (sum 7)

2 * 6 = 12 (sum 8)

2 * 8 = 16 (sum 10)

4 * 4 = 16 (sum 8)

3 * 6 = 18 (sum 9)

2 * 9 = 18 (sum 11)

4 * 6 = 24 (sum 10)

3 * 8 = 24 (sum 11)

Statement 3 by Peter tells you the product can't be 4, because if it were, then Peter would know the numbers are 1 and 4, since statement 2 by Lois eliminated 2 and 2.

2 * 3 = 6 (sum 5)

1 * 6 = 6 (sum 7)

2 * 4 = 8 (sum 6)

1 * 8 = 8 (sum 9)

3 * 3 = 9 (sum 6)

1 * 9 = 9 (sum 10)

3 * 4 = 12 (sum 7)

2 * 6 = 12 (sum 8)

2 * 8 = 16 (sum 10)

4 * 4 = 16 (sum 8)

3 * 6 = 18 (sum 9)

2 * 9 = 18 (sum 11)

4 * 6 = 24 (sum 10)

3 * 8 = 24 (sum 11)

Statement 4 by Lois tells you the sum can't be 5, because if it were, then Lois would know the numbers are 2 and 3, since statement 3 by Peter eliminated 1 and 4.

1 * 6 = 6 (sum 7)

2 * 4 = 8 (sum 6)

1 * 8 = 8 (sum 9)

3 * 3 = 9 (sum 6)

1 * 9 = 9 (sum 10)

3 * 4 = 12 (sum 7)

2 * 6 = 12 (sum 8)

2 * 8 = 16 (sum 10)

4 * 4 = 16 (sum 8)

3 * 6 = 18 (sum 9)

2 * 9 = 18 (sum 11)

4 * 6 = 24 (sum 10)

3 * 8 = 24 (sum 11)

Statement 5 by Peter tells you the product can't be 6, because if it were, then Peter would know the numbers are 1 and 6, since statement 4 by Lois eliminated 2 and 3.

2 * 4 = 8 (sum 6)

1 * 8 = 8 (sum 9)

3 * 3 = 9 (sum 6)

1 * 9 = 9 (sum 10)

3 * 4 = 12 (sum 7)

2 * 6 = 12 (sum 8)

2 * 8 = 16 (sum 10)

4 * 4 = 16 (sum 8)

3 * 6 = 18 (sum 9)

2 * 9 = 18 (sum 11)

4 * 6 = 24 (sum 10)

3 * 8 = 24 (sum 11)

Statement 6 by Lois tells you the sum can't be 7, because if it were, then Lois would know the numbers are 3 and 4, since statement 5 by Peter eliminated 1 and 6.

2 * 4 = 8 (sum 6)

1 * 8 = 8 (sum 9)

3 * 3 = 9 (sum 6)

1 * 9 = 9 (sum 10)

2 * 6 = 12 (sum 8)

2 * 8 = 16 (sum 10)

4 * 4 = 16 (sum 8)

3 * 6 = 18 (sum 9)

2 * 9 = 18 (sum 11)

4 * 6 = 24 (sum 10)

3 * 8 = 24 (sum 11)

Statement 7 by Peter tells you the product can't be 12, because if it were, then Peter would know the numbers are 2 and 6, since statement 6 by Lois eliminated 3 and 4.

2 * 4 = 8 (sum 6)

1 * 8 = 8 (sum 9)

3 * 3 = 9 (sum 6)

1 * 9 = 9 (sum 10)

2 * 8 = 16 (sum 10)

4 * 4 = 16 (sum 8)

3 * 6 = 18 (sum 9)

2 * 9 = 18 (sum 11)

4 * 6 = 24 (sum 10)

3 * 8 = 24 (sum 11)

Statement 8 by Lois tells you the sum can't be 8, because if it were, then Lois would know the numbers are 4 and 4, since statement 7 by Peter eliminated 2 and 6.

2 * 4 = 8 (sum 6)

1 * 8 = 8 (sum 9)

3 * 3 = 9 (sum 6)

1 * 9 = 9 (sum 10)

2 * 8 = 16 (sum 10)

3 * 6 = 18 (sum 9)

2 * 9 = 18 (sum 11)

4 * 6 = 24 (sum 10)

3 * 8 = 24 (sum 11)

Now Peter knows the numbers, according to his statement 9. The only way he could know the numbers is if their product is 16, so the numbers must be 2 and 8.

**3. Find three three-digit square numbers that together use each of the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 exactly once.**

361, 529, and 784.

The 3-digit squares without repeated digits are 169, 196, 256, 289, 324, 361, 529, 576, 625, 729, 784, 841, and 961.

Of these, 324 and 361 are the only squares that contain "3", so one of them must be used.

If 324 is used, then squares containing "2" or "4" are eliminated, leaving 169, 196, 576, and 961. But in that case, the "8" isn't used at all, so 324 is a bad choice.

If 361 is used, then squares containing "1" or "6" are eliminated, leaving 289, 529, 729, and 784. Of those, only 784 contains a "4," and only 529 contains a "5," so the three squares are 361, 529, and 784.