Question

A prime number is an integer greater than 1 with exactly two different positive factors, 1 and the number itself. There are three children in a family. Each of their ages is a prime number. The sum of their ages is 41 and at least two of the children have ages that differ by 16. Determine all possibilities for the ages of the children.

Answer 1

Answer:

Hence,

a=19,b=19,c=3

a=11,b=7,c=23

Step-by-step explanation:

$Let\ their\ ages\ be\ a,b\ and\ c\ respectively, where\ a,b,c\ are\ primes\\Hence,\\a+b+c=41\\b-c=16\\Listing\ out\ all\ the\ primes\ from\ 0\ to\ 41\ we\ get:\\2,3,5,7,11,13,17,19,23,29,31,37,\\Hence,\\Lets\ seperate\ out\ the\ primes\ who\ differ\ by\ 16 :\\3\ and\ 19, 7\ and\ 23\ ,$

Hence,

$Lets\ consider\ this\ set\ as\ b,c:\\Hence,\\To\ find\ a \\a=41-(b+c)\\Hence, By\ doing\ this\ indiviually\ we\ get:$

$a=41-(19+3)\\a=41-22\\a=19\\Hence, a=19,b=19,c=3$

$a=41-(23+7)\\a=41-30\\a=11\\Hence, a=11,b=7,c=23$