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the children's ages combined is three times the age of mike, who is three years younger than Shawn. The cube [^3] of Shawn's age, plus the square [^2] of Mikes age plus Laura's age totals a multiple of 100. IF ALL THE AGES ARE IN YEARS, HOW OLD IS LAURA?
(also tell the ages of the other children anyway)
showing work would help u get closer to the ten points but make sure its readable!

2007-03-21 12:38:34 · 3 answers · asked by XxVaneSsaXx214 3 in Education & Reference Trivia

3 answers

Here are the equations:

M + S + L = 3M, or S + L = 2M or S = 2M - L
M + 3 = S
S^3 + M^2 + L is a multiple of 100 (this isn't that useful yet)

Using the valule for S (eq 1) in eq 2 we get
M + 3 = 2M - L or 3 = M - L or 3+ M = L.

So now we know that Mike is three years younger than Shawn, but also three years older than Laura.

We also know that Shawn and Laura's age must be odd. Why?
Since they are all 3 years apart, there are two possible patterns -
Odd Even Odd or Even Odd Even.
Applying the last equation to the first combination, we will end up with an odd + even + odd which adds to an even.
But if we apply it to the second combination, we will end up with an even + odd + even, which is odd and cannot end up being a multiple of 100.

The numbers that fit this pattern are 5, 8 and 11. Laura is 11 years old.

Algebraically, this is (M+3)^3 + M + (M - 3), or
(m^3 + 9m^2 + 18M + 27) + 2M -3, or
m^3 + 9m^2 + 20M + 24

2007-03-22 05:19:04 · answer #1 · answered by Anonymous · 0 0

S = Shawns age M = Mikes etc. 3m = s + l + m. s3 + m2 +l = 100. If you add 1 to s or l, the other one goes up 1. l = about 1.52 years s = about 4.48 years and m = about 3 years.

2007-03-23 09:00:11 · answer #2 · answered by Anonymous · 0 0

2 years..... I am pretty goood in math or it is 3

2007-03-24 20:54:15 · answer #3 · answered by JOjo 2 · 0 0

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