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Using ab = 351 - a - b , we can substitute and narrow the range such that

300 ≤ ab ≤ 348

Let a = 12 such that

12 b = 351 - 12 - b 13 b = 339 b = 26.08

Thus, 12 is not one of the values of a or b . Continuing this process, we find a = 15 and b = 21. Thus, | a - b | = |15 - 21| = 6.

February Challenge: How many such pairs of real numbers exist that the sum, the product, and the quotient of these two numbers are all equal?

SOLUTION: Let x , y be real numbers such that x + y = xy = x / y . This means that x + y = x, x + y = x / y, and xy = x / y .

Solving xy = x / y, y 2 = 1, so y = ±1.

If y = 1, then x + 1 = x , so 1 = 0. There are no real solutions to this equation.

If y = - 1, then x - 1 = - x . Thus, x = 1 / 2.

Therefore, ((1 / 2, - 1) is the only real solution and there is only one pair of real numbers that satisfy the equa tions.

March Challenge: We note that n ! = 1 × 2 × 3 × … × ( n - 1) × n . If n ! = 2 15 × 3 6 × 5 3 × 7 2 × 11 × 13, then n = ?

SOLUTION: Given the definition of a factorial, we know that n ! = 215 × 36 × 53 × 72 × 11 × 13 can be rewrit ten as the product of consecutive integers as follows:

n ! = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5 × 7 × 7 × 11 × 13 n ! = 2 × 3 × (2 × 2) × 5 × (2 × 3) × 7 × (2 × 2 × 2) × (3 × 3) × (2 × 5) × 11 × (2 × 2 × 3) × 13 × (2 × 7) × (3 × 5) × (2 × 2 × 2 × 2) n ! = 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10 × 11 × 12 × 13 × 14 × 15 × 16

n is the largest integer. Therefore, n = 16.

Virginia Mathematics Teacher vol. 46, no. 2

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