Question 1178360: Ann, Ben and Cathy have their birthdays today. The sum of their ages is 23. The product of their ages is 113 more than the product of their ages on their birthday last year. What is the sum of the squares of their ages?
Found 2 solutions by ikleyn, mananth:
Answer by ikleyn(52786)   (Show Source):
You can put this solution on YOUR website! .
Ann, Ben and Cathy have their birthdays today. The sum of their ages is 23.
The product of their ages is 113 more than the product of their ages on their birthday last year.
What is the sum of the squares of their ages?
~~~~~~~~~~~~~~~
Simple calculations show that
abc - (a-1)*(b-1)*(c-1) = ab + ac + bc - (a + b + c) + 1
Substituting here (a + b + c) = 23, we get this equality
113 = ab + ac + bc - 23 + 1,
or
ab + ac + bc = 113 + 23 - 1 = 135.
Now, from the general formula
(a + b + c)^2 = a^2 + b + c^2 + 2(ab + ac + bc)
we have
a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + ac + bc) = 23^2 - 2*135 = 259.
ANSWER. The sum of squares of their ages is 259.
Solved.
////////////
@Manantrh successfully re-wrote my solution, by adding nothing new to it . . . .
O, no --- . . . by adding his name to it (!)
Probably, he (or she) wanted to say that he (or she) was totally agree with my solution
and confirmed my calculations by repeating them again . . .
Dear @Mananth, thank you for supporting and popularizing my solution (!)
Answer by mananth(16946)  (Show Source):
You can put this solution on YOUR website! x+y+z=23
xyz=113+(x-1)(y-1)(z-1)
xyz = 113 +( xy-x-y+1)(z-1)
xyz= 113+ xyz-xy-xz+x-yz+y+z-1 (xyz) cancels off
xy+yz+zx =113+22
xy+yz+zx = 135
2(xy+yz+zx) =270
x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)
= 23^2-270 -259
the sum of the squares of their ages =259
|
|
|
