Three ex-teenagers find that the product of their ages is 17,710. What is the sum of their ages?

17,710 = 2 * 5 * 7 * 11 * 23, all primes.


Since there are three people, these numbers need to be put into three groups. Two groups will have 2 numbers, and 1 group will have 1 number. The only number which fits ';ex-teenager'; is 23.





The remaining numbers are 2, 5, 7, and 11


Consider the 2. It needs to be in a group with one other number.


2*5 = 10, doesn't fit ';ex-teenager';


2*7 = 14, doesn't fit ';ex-teenager';


2*11 = 22, this fits.


That leaves 5 and 7 for the other group. 5*7 = 35.





The ages are 23, 22, and 35. Sum is 80.Three ex-teenagers find that the product of their ages is 17,710. What is the sum of their ages?
i have solved using trial and error, hence may be wrong





product = 17710


= 2*5* 7*11*23 (finding factors)


= (2*11)(5*7)*23( arranging such that there are three factors and each factor is more than nineteen).


=22*35*23.


hence sum = 22+35+23=80(ans)





NOTE: if this is a question from calculus, we have to use the concept of maxima and minima and solveThree ex-teenagers find that the product of their ages is 17,710. What is the sum of their ages?
Prime factors of 17710 are 2,5,7,11,23


The only way of grouping these to make three numbers each %26gt;= 20


(2,11) (5,7) 23





The ages are 22,35,23
17710 = 2 x 5 x 7 x 11 x 23. Since all possible ages must be %26gt;19, this makes the ages 23, 22 and 35 for a sum of 80
more information please?