SOLUTION: Six teenagers get together to plan a party together. The product of their ages was 13071240. Find the sum of their ages?

Algebra.Com
Question 820201: Six teenagers get together to plan a party together. The product of their ages was 13071240. Find the sum of their ages?
Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
We try dividing 13071240 by 19, we get 687960.

We try dividing 687960 by 19 again, but we get a decimal fraction.

We try dividing 687960 by 18, and we get 38220.

We try dividing 38220 by 18 again, but we get a decimal fraction.

We try dividing 38220 by 17, but we get a decimal fraction.

We try dividing 38220 by 16, but we get a decimal fraction.

We try dividing 38220 by 15, and we get 2548.

No use to divide 2548 by 15 again, for no multiple of 15 ends in 8.

We try dividing 2548 by 14, and we get 182.

We try dividing 2548 by 14 again, and we get 13.

So their ages are 13, 14, 14, 15, 18, and 19  

The sum of their ages is 13+14+14+15+18+19 = 93 

Edwin

RELATED QUESTIONS

a mother and daughter decided to add their ages together. the sum of their ages is 76.... (answered by josgarithmetic)
the sum of the ages of two teenagers is 34. While the product of their ages is 289. How... (answered by Alan3354)
the ratio in the ages of A,B,C, six years before was 2:3:5 if the sum of their present... (answered by Fombitz)
The product of all the ages of the teenagers at a party was 2 971 987 200. The number of... (answered by addingup,ikleyn,greenestamps)
six year are the sum of the ages of two sisters was 22 and the product of their age was... (answered by ikleyn,greenestamps)
The sum of ages of two sisters is 22years.six years ago,the product of their ages was... (answered by Alan3354,josgarithmetic)
Three exteenagers find the product of their ages is 26,390. Find the sum of their... (answered by Alan3354)
A woman is now 4 times older than her daughter. Six years ago, the product of their ages... (answered by Merciful)
Could you please help me with this problem. Thank you in advance. Three... (answered by solver91311)