If y + 1 is the square of an integer, which of the following could NOT

Question

If y + 1 is the square of an integer, which of the following could NOT be y?

(A) 3,024
(B) 3,135
(C) 3,246
(D) 3,363
(E) 3,480

ArunSharma12 · Accepted Answer

Given: y+1 =  x^2

lets add 1 to each of the options and observer which ones can be a square of a number:
3025
3136
3247
3364
3481

useful properties to remember :
square of a number can not end in: 2,3,7 or 8
square of a number always end in: 00,1,4,6,9,25

looking at numbers, we can safely mark 3247 as the number.
Ans: C

Kinshook · Answer

y + 1 = n^2
y = n^2 - 1

(A) 3,024 = 3025 - 1 = 55^2 - 1
(B) 3,135 = 3136 - 1 = 56^1 - 1
(C) 3,246 = 3247 - 1 ; 3247 is not a perfect square
(D) 3,363 = 3364 - 1 = 58^2 - 1
(E) 3,480 = 3481 - 1 = 59^2 - 1

IMO C

Archit3110 · Answer

square of number will give 1,4,9,6,5,0
for given values of y only option C 3246 +1 ; 3247 is not square of number
option C is correct

Basshead · Answer

A square of a number can ONLY end in  1, 4, 9, 6, 5, 0.  It can not end in 2, 3, 7, or 8.

Lets take a look at the choices:

(A) 3,024; 3,024 + 1 = 3,025 
(B) 3,135; 3,125 + 1 = 3,136
(C) 3,246; 3,246 + 1 = 2,24 7 . Not possible.
(D) 3,363; 3,363 + 1 = 3,364
(E) 3,480; 3,480 + 1 = 3,481

Answer is C.

Sabby · Answer

Hi,

You can solve the question through the concept of 1, 4, 9, 6, 5, 0 being the only possible units for square numbers; otherwise, you can notice the pattern in the last two digits of each answer choices :
y+1= x^2
therefore y=x^2-1 ==> y= (x-1)(x+1)
This tells us that the two last digits will be odd consecutive integers or even consecutive integers.

(A) 3,024 => 24 = 4*6
(B) 3,135 => 35= 5*7
(C) 3,246 => Doesn't work, this can't be y
(D) 3,363=> 63 = 7*9
(E) 3,480 => 80= 8*10

Answer C)

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If y + 1 is the square of an integer, which of the following could NOT

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If y + 1 is the square of an integer, which of the following could NOT

Prep Toolkit

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My Rewards