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so we have
\[a \cdot b \cdot c=7429\]

we know prime numbers are odd for the most part

and we know 7429 is not even so none of the primes are 2

two of her children are teenagers
so they could be 13,17, or 19

I would probably try to write the prime factorization of 7429

can you do that?

wait what are prime numbers? :$

prime numbers are integers that are only divisible by themselves and 1.

the key here is just to write 7429 in prime factorization
that is all there is to the problem

like for example
25=5*5 or 5^2
31 is prime
25 was not
umm...
2 is the only even prime

I dont even know what factorization means >.< I missed few days of school.

one tip... given the last digit is 9... I'd divide the original number by 17

example:
the prime factorization of 12=2*2*3

thanks <3

then see if the other 2 primes in the teens divides evenly

I don't know , I will try to figure it out thanks though

well the key to help was the teenager part

yeah , teens

"so they could be 13,17, or 19"
exactly two of these numbers will be factors

then you have to find the other factor by divide 7429 by the product of two teenagers

what is 7429 divided by 13 ?

2476.3

so you only have to find one last factor

\[\frac{7429}{17 \cdot 19}=?\]

So far I have done , 17/7429 = 437 /23 = 19

that leaves me with 1 , if i divide it within 19 itself

is this what you meant to say:
\[\frac{7429}{17 \cdot 19}=\frac{7429}{323}=23\]

which means 17*19*23=7429

kinda

oh I think I get what you are saying but those fractions aren't equal

\[\frac{7429}{17}=437 \\ \frac{437}{23}=19\]

first fraction gives 7429=17(437)
then second fraction says 437=23(19)

so replace 437 in 7429=17(437) with 23(19)
so 7429=17(23*19)

and you did it girl
good job!

Oh my gosh thank you so much <3

and 17,19,23 are all prime because they are only divisible by themselves or 1