| Shame on you for blowing off formulas, especially when some were used mentally.
On paper, the ages of the children last year were a, b, and c.
abc = 224 and involves partitioning of 2*2*2*2*2*7 into the three factors.
Adding 1 to each of last year's ages
(a+1)(b+1)(c+1)=
abc + ab + ac + bc + a + b + c +1 =
224 + ab + ac + bc + a + b + c +1 =
225 + ab + ac + bc + a + b + c =
360.
ab + ac + bc + a + b + c = 135.
Because there is a single odd factor, ab + ac + bc must be even, a + b + c must be odd, and the odd factor must be one of the latter three addends. Let it be a that = 7, then bc must = 2^5 = 32.
7b + 7c + 32 + 7 + b + c = 135.
8b + 8c = 96.
b + c = 12.
It is clear that the correct partition of five factors all 2's is (2*2)*(2*2*2).
Last year's ages must be 4, 7, and 8; this year's 5, 8, and 9. |
