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Posted - 07/08/2013 : 15:54:04 The Krane children were born 2 years apart. When Micheal, the youngest of the Six children was 1 year old, the sum of their ages was 1+3+5+7+9+11, or 36, a perfect square. How many years old was Micheal the next time the sum of their ages was a perfect square?
2 L A T E S T R E P L I E S (Newest First)
Posted - 07/22/2013 : 11:21:48 Right on Subhotosh Khan.
Letting M be Michael's age at any time, the sum of the six ages will next be a square when 36 + 6(M - 1) = N^2. The sum of the ages increases in increments of 6 making the square we seek an even square. The sum of the ages progresses as
Looks like Michael was 19 when the sum of the ages equaled 144 = 12^2, the next square after 36.
Another way of zeroing in on the answer is to simply find what value of M will create an even square. Those above 36 proceed as 64, 100, 144, 196, 256, etc. A few punches on the old calculator shows that M = 19 satisfies N^2 = 144.
Another way of attacking this one is as follows:
1--The first sum of ages is 36 at Michael's age 1. 2--The subsequent sums increase by 6 each year. 3--The sum sought can be defined as N^2 = 36 + 6(M - 1), M being Michael's attained age. 4--Written another way, N^2 - 36 = N^2 - n^2 = 6(M - 1) 5--N^2 - n^2 can be written as (N + n)(N - n) 6--Therefore, (N + n)(N - n) = 6(M - 1) 7--Then, (N + n) = (M - 1) and (N - n) = 6 8--Adding, 2N = M + 5 or N = (M + 5)/2 9--We can now write N^2 = (M - 1)^2/4 = 36 + 6(M - 1) 10--Expanding and simplifying, M^2 - 14M - 95 = 0 11--From the quadratic formula, M = [14+/-sqrt(196 + 380)]/2 12--Solving, M = 19.
Posted - 07/16/2013 : 18:35:01
quote:Originally posted by TchrWill
The Krane children were born 2 years apart. When Micheal, the youngest of the Six children was 1 year old, the sum of their ages was 1+3+5+7+9+11, or 36, a perfect square. How many years old was Micheal the next time the sum of their ages was a perfect square?
After 'n' years the sum of their ages would be
S = 36 + 6*n............The S must be divisible by 6