If the difference between the squares of the first and last terms of a

Question

If the difference between the squares of the first and last terms of a sequence of consecutive odd positive integers is 720,
and the first and last terms share the same ones unit, what is the sum of the first and last terms?

A. 64
B. 68
C. 72
D. 76
E. 80

ashudhall · Accepted Answer

Answer should be C.
Since the number share the same unit digit, difference between the terms would be in multiple of 10.
Let no be x & y => x^2-y^2=720
(x+y)(x-y)=720
(X+y )*10n = 720
X+y=72/n
Since n is an integer and there is no factor of 72 present in the list, answer should be 72.

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Bunuel · Answer

Say the first term is y and the last term is y (notice y > x).

y^2 - x^2 = 720 ;

(y - x)(y + x) = 2^4*3^2*5 .

Since both x and y are odd then both y - x and y + x are even. 
Since the units digits of x and y are the same then the units digit of y - x is 0.

So, y - x is at least 5*2 = 10 and y + x is at most 3^2*2^3 = 72. In this case y = 41 and x = 31. These two numbers satisfy all the conditions and since a PS question cannot have two correct answer then y + x = 72.

Answer: C.

Alternatively for the last step: y - x is at least 5*2 and y + x is at most 3^2*2^3 = 72. From this: 
y - x cannot be 5*2*3 = 30 and x = 3*2^3 = 24 because in this case y - x < y + x, which cannot happen. 
y - x cannot be 5*2^2 = 20 and x = 3^2*2^2 = 36 because in this case in this case both x and y turn to be even, not odd.

Those were only two other cases, so y - x = 5*2 = 10 and y + x = 3^2*2^3 = 72

Hope it's clear.

gracie · Answer

if the first and last terms share the same units digit, 
then the range of the sequence will be a multiple of 10
because all the answer choices have only two digits, assume the range is 10
let x=the first term; y=the last term
(y+x)(y-x)=720
y-x=range=10
y+x=720/10=72
C

bumpbot · Answer

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