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

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Bunuel
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If y + 1 is the square of an integer, which of the following could NOT [#permalink] New post  17 Mar 2020, 07:21
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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
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Re: If y + 1 is the square of an integer, which of the following could NOT [#permalink] New post  17 Mar 2020, 07:31
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Bunuel wrote:
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


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
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Re: If y + 1 is the square of an integer, which of the following could NOT [#permalink] New post  17 Mar 2020, 07:26
Bunuel wrote:
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


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
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Re: If y + 1 is the square of an integer, which of the following could NOT [#permalink] New post  16 Oct 2020, 10:59
Bunuel wrote:
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


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
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Basshead
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Re: If y + 1 is the square of an integer, which of the following could NOT [#permalink] New post  16 Oct 2020, 11:48
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 = 2,247. Not possible.
(D) 3,363; 3,363 + 1 = 3,364
(E) 3,480; 3,480 + 1 = 3,481

Answer is C.
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Re: If y + 1 is the square of an integer, which of the following could NOT [#permalink] New post  26 Feb 2022, 23:32
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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Re: If y + 1 is the square of an integer, which of the following could NOT [#permalink] New post  31 May 2023, 12:33
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Re: If y + 1 is the square of an integer, which of the following could NOT [#permalink]
31 May 2023, 12:33
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