Algebra Problems, Math Problems

Problem of the Day: 1/3/13

Six years ago, Sally was half Tommy’s age. Two years ago, Sally was three-fourths of Tommy’s age. How old are Sally and Tommy now?

Solution to yesterday’s problem:
First, we should count the 1xN rectangles, 10 of them.

Now, the thing with perfect squares is that if their roots are prime (2, 3, 5, etc.), then that root becomes its only prime factor. For instance, the only prime factorization of 9 is 3×3. So if the root is prime, we don’t have to look for any other rectangles associated with it, because there are no other factor pairs.

So we don’t have to look at 4, 9, 25, and 49 because their roots are prime. However, we do have to look at 16, 36, 64, 81, and 100.

16: The factors of 16 are 1 and 16, 2 and 8, and 4 and 4. The only one of these that has unequal dimensions and hasn’t already been counted is 2×8. 11 so far!

36: Factors are 1, 36; 2, 18; 3, 12; 4, 9; 6, 6. We’ll add 2×18, 3×12, and 4×9 to our list to make 14 total.

64: Factors are 1, 64; 2, 32; 4, 16; 8, 8. We’ll add 2×32 and 4×16 to make 16 total.

81: Factors are 1, 81; 3, 27; 9, 9. Adding 3×27, we have 17 total.

100: Factors are 1, 100; 2, 50; 4, 25; 5, 20; 10, 10. Adding 2×50, 4×25, and 5×20, we have 20 total. So there are 20 such rectangles.

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