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Which of the following is two more than the square of an odd integer?

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Which of the following is two more than the square of an odd integer?  [#permalink]

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New post 02 Sep 2015, 22:54
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

60% (02:20) correct 40% (02:32) wrong based on 157 sessions

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Re: Which of the following is two more than the square of an odd integer?  [#permalink]

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New post 06 Sep 2015, 05:42
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6
Bunuel wrote:
Which of the following is two more than the square of an odd integer?

(A) 14,173
(B) 14,361
(C) 14,643
(D) 14,737
(E) 14,981


Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

From the size of the answer choices, we should recognize that there must be some trick – a way to find the answer quickly within 2 minutes, since theoretically, every GMAT problem must be solvable that way.

Let’s start from first principles. We can represent a general odd integer as 2k + 1, where k is an integer. So the square of an odd integer can be written this way:

(2k + 1)^2

= 4k^2 + 4k + 1

Since the first two terms (4k^2 and 4k) contain 4 as a factor, we can see that they are both multiples of 4, and thus their sum is also a multiple of 4. This means that the square of any odd integer is 1 more than a multiple of 4. Thus, the number we are looking for (“two more than the square…”) is 3 more than a multiple of 4.

To quickly determine whether a number is a multiple of 4, we can just examine the last 2 digits. The reason is that 100 is divisible by 4, so only the tens and the ones digit matter. The hundreds digit, the thousands digit, etc. will never affect divisibility by 4.

Looking at the answer choices and cutting off everything but the last 2 digits, we are left with 73, 61, 43, 37, and 81. Of these, only 43 is 3 more than a multiple of 4 (40). Thus, C is the right answer. (Incidentally, 14,643 is 2 + 121^2.)

The correct answer is (C).
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Re: Which of the following is two more than the square of an odd integer?  [#permalink]

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New post 03 Sep 2015, 08:02
3
3
Bunuel wrote:
Which of the following is two more than the square of an odd integer?

(A) 14,173
(B) 14,361
(C) 14,643
(D) 14,737
(E) 14,981


Kudos for a correct solution.


one of the given options should be a square if 2 is subtracted from it.
the options would be
(A) 14,171
(B) 14,359
(C) 14,641
(D) 14,735
(E) 14,979

We can solve this question with many approaches. Here are the 2 easiest methods.

Method 1:-
If a square has odd unit's digit, then it's ten's digit will always be even.
Only 14,641 satisfies this condition.

Method 2:-
If you know about palindromic numbers,
then you can clearly see that 14,641 is a palindromic square.

Answer:- C
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Re: Which of the following is two more than the square of an odd integer?  [#permalink]

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New post 03 Sep 2015, 12:00
1
Solution:

First sub 2 from all numbers:
14171
14359
14641
14735
14979

These are closer to 14400 which is 120^2.
S0, try 121^2 and you will get 14641 as answer.

Option C

I know this is crude method and hope that Bunuel will provide a better one.
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Re: Which of the following is two more than the square of an odd integer?  [#permalink]

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New post 03 Sep 2015, 13:44
kunal555 wrote:
Bunuel wrote:
Which of the following is two more than the square of an odd integer?

(A) 14,173
(B) 14,361
(C) 14,643
(D) 14,737
(E) 14,981


Kudos for a correct solution.


one of the given options should be a square if 2 is subtracted from it.
the options would be
(A) 14,171
(B) 14,359
(C) 14,641
(D) 14,735
(E) 14,979

We can solve this question with many approaches. Here are the 2 easiest methods.

Method 1:-
If a square has odd unit's digit, then it's ten's digit will always be even.
Only 14,641 satisfies this condition.


Method 2:-
If you know about palindromic numbers,
then you can clearly see that 14,641 is a palindromic square.

Answer:- C


Could someone give some background on why the above highlighted is true... I didn't know what :( :(
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Which of the following is two more than the square of an odd integer?  [#permalink]

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New post 03 Sep 2015, 15:58
3
DropBear wrote:
kunal555 wrote:
Bunuel wrote:
Which of the following is two more than the square of an odd integer?

(A) 14,173
(B) 14,361
(C) 14,643
(D) 14,737
(E) 14,981


Kudos for a correct solution.


one of the given options should be a square if 2 is subtracted from it.
the options would be
(A) 14,171
(B) 14,359
(C) 14,641
(D) 14,735
(E) 14,979

We can solve this question with many approaches. Here are the 2 easiest methods.

Method 1:-
If a square has odd unit's digit, then it's ten's digit will always be even.
Only 14,641 satisfies this condition.


Method 2:-
If you know about palindromic numbers,
then you can clearly see that 14,641 is a palindromic square.

Answer:- C


Could someone give some background on why the above highlighted is true... I didn't know what :( :(


Please find attached for the proof. I like doing on paper than typing. That's why i have taken a pic and attached it.

Hope I was helpful.
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IMG_0198.JPG
IMG_0198.JPG [ 1.86 MiB | Viewed 1898 times ]

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Re: Which of the following is two more than the square of an odd integer?  [#permalink]

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New post 05 Sep 2015, 11:14
1
Bunuel wrote:
Which of the following is two more than the square of an odd integer?

(A) 14,173
(B) 14,361
(C) 14,643
(D) 14,737
(E) 14,981


Kudos for a correct solution.


If 2 is subtracted from the numbers in the option, we will get a square of an odd integer.

The numbers are closer to square of 120, 120 x120 = 14400

try 121, you get 14641.

option C = 14643 - 2 = 14641
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Which of the following is two more than the square of an odd integer?  [#permalink]

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New post 06 Mar 2016, 20:12
Bunuel wrote:
Which of the following is two more than the square of an odd integer?

(A) 14,173
(B) 14,361
(C) 14,643
(D) 14,737
(E) 14,981


Kudos for a correct solution.


wow..now this is smth I had no idea how to approach..
my first calculations = 100x100 = 10,000
so def need smth 12X*120X, otherwise I would not get 14XXX
and def smth less than 130x130 =because we get 16,900
so only options are: 121, 123, 125, 127, 129.
so started with 121*121 => 14,641 - which is 2 less than C..
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Which of the following is two more than the square of an odd integer?  [#permalink]

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New post Updated on: 30 Oct 2016, 10:46
1
Summary:
All perfect squares end in 1, 4, 5, 6, 9 or 00 (i.e. Even number of zeros). Therefore, a number that ends in 2, 3, 7 or 8 is not a perfect square.
Digital roots are 1, 4, 7 or 9. No number can be a perfect square unless its digital root is 1, 4, 7, or 9. You might already be familiar with computing digital roots. (To find digital root of a number, add all its digits. If this sum is more than 9, add the digits of this sum. The single digit obtained at the end is the digital root of the number.)
If unit digit ends in 5, ten’s digit is always 2.
If it ends in 6, ten’s digit is always odd (1, 3, 5, 7, and 9) otherwise it is always even. That is if it ends in 1, 4, and 9 the ten’s digit is always even (2, 4, 6, 8, 0).
Total numbers of prime factors of a perfect square are always odd

Source: burningmath blogspot com_2013_09_how-to-check-if-number-is-perfect-square

Originally posted by manishtank1988 on 25 Aug 2016, 11:39.
Last edited by manishtank1988 on 30 Oct 2016, 10:46, edited 1 time in total.
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Which of the following is two more than the square of an odd integer?  [#permalink]

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New post 30 Oct 2016, 08:44
According to the question we have:
(2x+1)^2=y-2
4x^2+4x+3=y => 4(x^2+x)+3=y
So y is 3 more than a multiple of 4. Now check our options for divisibility by 4:
1) 73-3=70 No.
2) 61-3=58 No.
3) 43-3=40 Yes.
4) 37-3=34 No.
5) 81-3=78 No.
We have choice C.
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Re: Which of the following is two more than the square of an odd integer?  [#permalink]

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