At the annual High school Parents' Dance contest, the ages of the winners were two smallest consecutive integers whose squares differ by more than 100. What is the least possible sum of the ages of the two winners?
Kind of tricky.
Thanks
2006-11-02 13:29:19 · 7 answers · asked by babigurl34 2 in Science & Mathematics ➔ Mathematics
y^2 - x^2 = 100
y = x + 1
(x + 1)^2 - x^2 = 100
x^2 + 2x + 1 - x^2 = 100
2x + 1 = 100
2x = 99
x = 49.5
so, 50 and 51
2006-11-02 13:37:12 · answer #1 · answered by Anonymous · 0⤊ 0⤋
First let x = the age of the fist person
So, x+2 = the age of the second person
Now the hard part, (x+2)^2 - x^2 > 100
(x+2)(x+2) - x^2 > 100
x^2 +4x +4 - x^2 > 100
4 x + 4 >100
4 x > 96
x > 24
So, x + 2 = 26
Lets see if it works
26^2 - 24^2
676 - 576
100
So to get the exact answer, we would need the next smallest consecutive intergers. Which would be 25 and 27
27^2 - 25 ^ 2
729 - 625
104
So, your answer is 25 and 27
2006-11-02 21:38:11 · answer #2 · answered by danjlil_43515 4 · 0⤊ 0⤋
Its 50 or to 51
2006-11-02 21:38:54 · answer #3 · answered by John S 1 · 0⤊ 0⤋
The integers must be consecutive.
50 and 51 are the lowest of such numbers.
The sum is 101.
2006-11-03 05:28:44 · answer #4 · answered by Akilesh - Internet Undertaker 7 · 0⤊ 0⤋
if their squares can differ by 100 then its 10 and 11 so 21, if it has to be larger than 100 then its 11 and 12 so 23
2006-11-02 21:33:11 · answer #5 · answered by J 3 · 0⤊ 0⤋
50 & 51
2006-11-02 21:41:42 · answer #6 · answered by george g 5 · 0⤊ 0⤋
50 and 51 2500 and 2601
2006-11-02 21:34:52 · answer #7 · answered by wheeldave2 2 · 1⤊ 0⤋