Amalgum23
the age of father mother and three childrens be f,m,a,b,c. Moreover a>b>c.
given that : F=3*(a+b) and F= b2+c2+11 .
The two statements standalone are not sufficient. F+m+a+b+c =105 and 40<M<F.
combining both we get, we find F is a multiple of 3 and greater than 40. List of 3k multiples which is greater than 40 are : 42, 45, 48, 51, 54, 57...... but there is a constraint of mother greater than 40. Extending beyond 40 we cannot accommodate the three children. If father is 57, the sum of the two children’s become 19. In worst case 57+41+19 exceeds 105. So, at exactly 48, we get 48+41+16 =105. We need to accomdate the third child. Now, we are left with two cases 45 = 3*15 and 42= 3*14. Using F=b2+c2+11. Let’s take sum of squares = 45-11 = 34. Can 34 be represented as sum of squares. 9+25. So, both combined is sufficient.
Amalgum23
the age of father mother and three childrens be f,m,a,b,c. Moreover a>b>c.
given that : F=3*(a+b) and F= b2+c2+11 .
The two statements standalone are not sufficient. F+m+a+b+c =105 and 40<M<F.
combining both we get, we find F is a multiple of 3 and greater than 40. List of 3k multiples which is greater than 40 are : 42, 45, 48, 51, 54, 57...... but there is a constraint of mother greater than 40. Extending beyond 40 we cannot accommodate the three children. If father is 57, the sum of the two children’s become 19. In worst case 57+41+19 exceeds 105. So, at exactly 48, we get 48+41+16 =105. We need to accomdate the third child. Now, we are left with two cases 45 = 3*15 and 42= 3*14. Using F=b2+c2+11. Let’s take sum of squares = 45-11 = 34. Can 34 be represented as sum of squares. 9+25.
So both statements together is sufficient. Nice question. Thanks for sharing