# Math Help - Help the poor math grads !!!!

1. ## Help the poor math grads !!!!

Two MIT math grads bump into each other at Fairway on the upper west side. They haven't seen each other in over 20 years.

The first grad says to the second
: "how have you been?"

Second
: "great! i got married and i have three daughters now"
First
: "really? How old are they?"

Second
: "well, the product of their ages is 72, and the sum of their ages is the same as the number on that building over there.."

First
: "right, ok.. oh wait.. hmm, i still don't know"

Second
: "oh sorry, the oldest one just started to play the piano"
F
irst: "wonderful! My oldest is the same age!"
How old are the daughters?

2. Originally Posted by MathLearner
Two MIT math grads bump into each other at Fairway on the upper west side. They haven't seen each other in over 20 years.
The first grad says to the second
: "how have you been?"
Second
: "great! i got married and i have three daughters now"
First
: "really? How old are they?"
Second
: "well, the product of their ages is 72, and the sum of their ages is the same as the number on that building over there.."
First
: "right, ok.. oh wait.. hmm, i still don't know"
Second
: "oh sorry, the oldest one just started to play the piano"
F
irst: "wonderful! My oldest is the same age!"
How old are the daughters?
Haha, wow. i'd like to see the solution to this. there must be some hidden message i am not seeing

3. Yay !
I love it =) =)

Since they haven't seen each other for 20 years and that it seems he wasn't married yet (thus no children), the age of the children is < 20.

$\boxed{72=2^33^2}$
S:sum of the ages

\begin{aligned} 72 &=1*8*9 \leftarrow S=18\\
&=1*4*18 \leftarrow S=23 \\
&=1*6*12 \leftarrow S=19 \\
&=2*2*18 \leftarrow S=22 \\
&=2*3*12 \leftarrow S=17 \\
&=2*4*9 \leftarrow S=15 \end{aligned}

\begin{aligned} 72&=2*6*6 \leftarrow S={\color{red}14} \\
&=3*3*8 \leftarrow S={\color{red}14} \\
&=3*4*6 \leftarrow S=13 \end{aligned}

Since the guy says he still can not know what the age are, you have to retrieve the same sum for two occurrences.

That is to say it's either {2,6,6} either {3,3,8}.

But then, he says that the eldest one blah blah blah

---> 3,3 & 8

4. Originally Posted by Moo
Yay !
I love it =) =)

Since they haven't seen each other for 20 years and that it seems he wasn't married yet (thus no children), the age of the children is < 20.

$\boxed{72=2^33^2}$
S:sum of the ages

\begin{aligned} 72 &=1*8*9 \leftarrow S=18\\
&=1*6*12 \leftarrow S=19 \\
&=2*2*18 \leftarrow S=22 \\
&=2*3*12 \leftarrow S=17 \\
&=2*4*9 \leftarrow S=15 \end{aligned}

\begin{aligned} 72&=2*6*6 \leftarrow S={\color{red}14} \\
&=3*3*8 \leftarrow S={\color{red}14} \\
&=3*4*6 \leftarrow S=13 \end{aligned}

Since the guy says he still can not know what the age are, you have to retrieve the same sum for two occurrences.

That is to say it's either {2,6,6} either {3,3,8}.

But then, he says that the eldest one blah blah blah

---> 3,3 & 8
You're so smart!

5. Originally Posted by Moo
Yay !
I love it =) =)

...

Since the guy says he still can not know what the age are, you have to retrieve the same sum for two occurrences.

That is to say it's either {2,6,6} either {3,3,8}.

But then, he says that the eldest one blah blah blah

---> 3,3 & 8
Nice work. What a neat problem!

6. Originally Posted by Moo
Yay !
I love it =) =)

Since they haven't seen each other for 20 years and that it seems he wasn't married yet (thus no children), the age of the children is < 20.

$\boxed{72=2^33^2}$
S:sum of the ages

\begin{aligned} 72 &=1*8*9 \leftarrow S=18\\
&=1*6*12 \leftarrow S=19 \\
&=2*2*18 \leftarrow S=22 \\
&=2*3*12 \leftarrow S=17 \\
&=2*4*9 \leftarrow S=15 \end{aligned}

\begin{aligned} 72&=2*6*6 \leftarrow S={\color{red}14} \\
&=3*3*8 \leftarrow S={\color{red}14} \\
&=3*4*6 \leftarrow S=13 \end{aligned}

Since the guy says he still can not know what the age are, you have to retrieve the same sum for two occurrences.

That is to say it's either {2,6,6} either {3,3,8}.

But then, he says that the eldest one blah blah blah

---> 3,3 & 8
Isn't this all assuming that permarital sex is impossible?

7. Originally Posted by Mathstud28
Isn't this all assuming that permarital sex is impossible?
Actually, that initial information seems irrelevant (and good thing too, since it says they haven't seen each other for at least 20 years):

The only natural numbers > 20 that divide 72 are 72, 36, and 24. So if one child is greater than 20 years old, we would have one of the following sums:

$72 + 1 + 1 = 74$

$36 + 1 + 2 = 39$

$24 + 1 + 3 = 28$

Since none of these sums are equal, and since none of them are equal to any other sum of the factors, the only case that works is the one Moo found.

8. Originally Posted by Mathstud28
Isn't this all assuming that permarital sex is impossible?
Its MIT.