In a family of five members (father, mother and three children), the product of the (integral) ages of all members equals $1998$. Knowing that the father is $10$ years older than the mother, determine the ages of all members.
Problem
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Tags: system of equations, algebra
SilverDragon246
04.06.2021 06:36
$1998 = 2 * 3^3 * 37$ Conveniently $37 = 27 + 10$. So you can have the father be $37$, the mother be $27$, and the three children be $1$, $1$, and $2$ in some order.
natmath
04.06.2021 06:52
Should these problems be in MSM?
ilikemath40
09.06.2021 05:36
natmath wrote: Should these problems be in MSM? Is it too easy or too hard?
SigmaPiE
09.06.2021 05:55
ilikemath40 wrote: natmath wrote: Should these problems be in MSM? Is it too easy or too hard? I assume that this thread was previously in a different forum, most likely either HSM or College Math. MSM is the "lowest" admin-run site you get to. Therefore, it probably was too easy for whichever forum it was originally posted in.
kante314
09.06.2021 16:38
$1998=2 \cdot 3^{3} \cdot 37$
$37$ is the father, $27$ is the mother. The children are $2$, $1$, and $1$
asdf334
09.06.2021 17:04
natmath wrote: Should these problems be in MSM? idk, this problem seems painfully easy
kante314
09.06.2021 17:07
asdf334 wrote: natmath wrote: Should these problems be in MSM? idk, this problem seems painfully easy It says grade 8 so yeah
Apple321
09.06.2021 17:37
kante314 wrote: asdf334 wrote: natmath wrote: Should these problems be in MSM? idk, this problem seems painfully easy It says grade 8 so yeah grade 8 problem 1 lol