lheiannie07 wrote:How much greater is the square of the sum of three different positive integers than the sum of their squares?
(1) The sum of the products of all possible pairs of two different integers out of the original set of three is 61.
(2) The largest of the three integers, 7, is equal to the sum of the two smaller integers.
Is Option 1 sufficient? Why or why not?
OA A
Say the three different ineters are a, b and c.
The square of the sum of three different positive integers = (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca
The sum of their squares = a^2 + b^2 + c^2
(The square of the sum of three different positive integers) - (The sum of their squares) = [a^2 + b^2 + c^2 + 2ab + 2bc + 2ca] - [a^2 + b^2 + c^2]
= 2ab + 2bc + 2ca = 2(ab + bc + ca)
If we get the value of either ab + bc + ca or a, b and c, we get the answer.
Let's take each statement one by one.
(1) The sum of the products of all possible pairs of two different integers out of the original set of three is 61.
=> a^2 + b^2 + c^2 = 61
(The square of the sum of three different positive integers) - (The sum of their squares) = 2(ab + bc + ca) = 2*61 = 122. Sufficient
(2) The largest of the three integers, 7, is equal to the sum of the two smaller integers.
Case 1: We have the largest of the three integers = c = 7. Say the other two integers are a = 1 and b = 6.
Thus, 2(ab + bc + ca) = 2(ab + bc + ca) = 2(1*6 + 6*7 + 7*1) = 110
Case 2: We have the largest of the three integers = c = 7. Say the other two integers are a = 2 and b = 5.
Thus, 2(ab + bc + ca) = 2(ab + bc + ca) = 2(2*5 + 5*7 + 7*2) = 118.
No unqiue value. Insufficient.
For the sake of understandin better, we take another case.
Case 3: We have the largest of the three integers = c = 7. Say the other two integers are a = 3 and b = 4.
Thus, 2(ab + bc + ca) = 2(ab + bc + ca) = 2(3*4 + 4*7 + 7*3) = 122. No unqiue value.
The correct answer:
A
Hope this helps!
-Jay
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