Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 59720

Which of the following is two more than square of an odd integer?
[#permalink]
Show Tags
15 Nov 2019, 00:40
Question Stats:
65% (02:14) correct 35% (02:34) wrong based on 54 sessions
HideShow timer Statistics
Competition Mode Question Which of the following is two more than square of an odd integer? A. 14,173 B. 14,361 C. 14,643 D. 14,737 E. 14,981 Are You Up For the Challenge: 700 Level Questions
Official Answer and Stats are available only to registered users. Register/ Login.
_________________



GMAT Club Legend
Joined: 18 Aug 2017
Posts: 5479
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)

Re: Which of the following is two more than square of an odd integer?
[#permalink]
Show Tags
15 Nov 2019, 01:06
for odd integer ; 1,3,5,7,9 ; the square gives ; 1,9,5,9,1 as the unit digit and odd unit digit + 2 ; odd unit digit A. 14,173 2 ; 14171 ; not divisible by 3 or 9 or 11NO B. 14,361 ; 14359 ; not divisible by 7 C. 14,643 ; 14641 ; divisible by 11 and square of 121 ; 14641 ; yes sufficient D. 14,737 ; 14735 factorising does not give odd integer E. 14,981 ; not divisible by 7
IMO C ; 14,643
Which of the following is two more than square of an odd integer?
A. 14,173 B. 14,361 C. 14,643 D. 14,737 E. 14,981



VP
Joined: 20 Jul 2017
Posts: 1147
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)

Re: Which of the following is two more than square of an odd integer?
[#permalink]
Show Tags
15 Nov 2019, 01:13
Any odd number can be assumed as "2n + 1" for any non negative integer n Square of odd number = (2n + 1)^2 = 4n^2 + 4n + 1 2 more than square of an odd number = 4n^2 + 4n + 3 = 4(n^2 + n) + 3 = 4p + 3, for any positive integer p > The correct option (2 more than square of an odd number) when divided by 4 should leave a remainder 3
Rule for divisible by 4 > Last 2 digits should be divisible by 4
A. 14,173 : 73 = 4*18 + 1 > NO B. 14,361 : 61 = 4*15 + 1 > NO C. 14,643 : 43 = 4*10 + 3 > YES (Remainder = 3) D. 14,737 : 37 = 4*9 + 1 > NO E. 14,981 : 81 = 4*20 + 1 > NO
IMO Option C



Director
Joined: 24 Nov 2016
Posts: 963
Location: United States

Re: Which of the following is two more than square of an odd integer?
[#permalink]
Show Tags
15 Nov 2019, 04:49
Quote: Which of the following is two more than square of an odd integer?
A. 14,173 B. 14,361 C. 14,643 D. 14,737 E. 14,981 A. 14,1732=14171 B. 14,3612=14359 C. 14,6432=14641 D. 14,7372=14735: f(14735)=5*2947…2947≠f(5)…not perf.square E. 14,9812=14979: f(14979)=3*4993…4993≠f(3)…not perf.square 100*100=10,000 200*200=40,000 150*150=22,500 120*120=14,400 x is an odd close to 120 \(x=119:x^2=120*120120119=14400239=14161≠(A)\) \(x=121:x^2=120*120+120+121=14400+241=14641=(C)\) \(x=123:x^2=121*121+(2)121+(2)123=14641+2(244)=15129>14979;invalid\)



Senior Manager
Joined: 07 Mar 2019
Posts: 456
Location: India
WE: Sales (Energy and Utilities)

Re: Which of the following is two more than square of an odd integer?
[#permalink]
Show Tags
15 Nov 2019, 08:15
Which of the following is two more than square of an odd integer? A. 14,173 B. 14,361 C. 14,643 D. 14,737 E. 14,981 So, let 'x' be the integer. \(x^2 + 2 = 14173 , x^2 = 14169\) \(x^2 + 2 = 14361 , x^2 = 14359\) \(x^2 + 2 = 14643 , x^2 = 14641\) \(x^2 + 2 = 14737 , x^2 = 14735\) \(x^2 + 2 = 14981 , x^2 = 14979\) A closer observation reveals that 120^2 = 14400. Thus A and B are straight forward eliminated. Out of all the options left only C looks like a square of an integer because of its similarity to its approach to get square of 11 or 111. As \(11^2 = 121\) and \(111^2 = 12321\) Hence \(121^2 = 14641\) Answer C.
_________________
Ephemeral Epiphany..!
GMATPREP1 590(Q48,V23) March 6, 2019 GMATPREP2 610(Q44,V29) June 10, 2019 GMATPREPSoft1 680(Q48,V35) June 26, 2019



Senior Manager
Joined: 01 Mar 2019
Posts: 338
Location: India
Concentration: Strategy, Social Entrepreneurship
GPA: 4

Which of the following is two more than square of an odd integer?
[#permalink]
Show Tags
Updated on: 18 Nov 2019, 09:45
A. 14,173 B. 14,361 C. 14,643 D. 14,737 E. 14,981
Since all the options are around....14,000.... nearest to 120^2= 14400
119^2= 14161 A is eliminated
Next odd number is 121^2= 14641....after adding 2= 14643.....option C
OA:C
Posted from my mobile device
Originally posted by madgmat2019 on 15 Nov 2019, 09:14.
Last edited by madgmat2019 on 18 Nov 2019, 09:45, edited 1 time in total.



SVP
Joined: 03 Jun 2019
Posts: 1884
Location: India

Re: Which of the following is two more than square of an odd integer?
[#permalink]
Show Tags
15 Nov 2019, 09:43
Asked: Which of the following is two more than square of an odd integer? N = (2k+1)^2 +2 = 4k^2 + 4k + 1 +2 = 4k(k+1) + 3 N3 = 4k(k+1)
A. 14,173 14173  3 = 14170 = 4* 3542 + 2; Not divisible by 4 B. 14,361 14361 3 = 14358 = 4* 3589 + 2; Not divisible by 4 C. 14,643 14643  3 = 14640 = 4* 3660 = 4* 60*61; k=60; D. 14,737 E. 14,981
IMO C



Manager
Joined: 31 Oct 2015
Posts: 94

Re: Which of the following is two more than square of an odd integer?
[#permalink]
Show Tags
15 Nov 2019, 12:30
I figured that all values were close to 14400 which is square of 120. I tried the square of the next odd number and added 2, found that it is option C.
C is the answer.
Although I look forward to an logical answer



Director
Joined: 18 May 2019
Posts: 550

Re: Which of the following is two more than square of an odd integer?
[#permalink]
Show Tags
15 Nov 2019, 22:37
We are to determine which of the answer choices is two more than the square of an odd number.
The first step is to subtract 2 from an answer choice. After doing that find out if the unit digit is either of the following: 1, 9, or 5. If it is then you can proceed to step two, which is to determine if the tens digit is even. These two steps are enough to solve this question.
Step1: Subtract 2 from all the answer choices and check if the unit digits are 1, 5, or 9. As shown below, all five answer choices have met this criterion. A. 14,173: 14,1732 = 14,171 B. 14,361: 14,3612 = 14,359 C. 14,643: 14,6432 = 14,641 D. 14,737: 14,7372 = 14,735 E. 14,981: 14,9812 = 14,979
Step 2: Check the tens digit to see if it is an even number. Any perfect square ending in 1, 5, or 9 always has a tens digit which is even. For example 81, (9^2), has a tens digit is 8 which is even; 121, (11^2), has a tens digit is 2 an even number; 625, (25^2), has a tens digit of 2 which is even; 49, (7^2), has a tens digit of 4 which is even; 729, (27^2), has a tens digit of 2 which is even.
subjecting the difference between the answer choices and 2 to step 2 above, we realize only option C has a tens digit of 4 which is even. Hence, option C must be our number.
The answer is, therefore, option C.



Intern
Joined: 01 Jul 2016
Posts: 11

Re: Which of the following is two more than square of an odd integer?
[#permalink]
Show Tags
16 Nov 2019, 04:57
First substract 2 from all choices. Then we need to see which one is closer to the square number. Per the digits, it seems it is somewhere closer to square of 120. I.e. 14400, then if we try 121^2 , we get 14,641.
Posted from my mobile device



Intern
Joined: 10 May 2016
Posts: 3

Re: Which of the following is two more than square of an odd integer?
[#permalink]
Show Tags
16 Nov 2019, 12:23
C. 14,643 because of 14,6432=14,641=121*121.



Intern
Joined: 10 May 2016
Posts: 3

Re: Which of the following is two more than square of an odd integer?
[#permalink]
Show Tags
16 Nov 2019, 12:24
C. 14,643 because of 14,6432=14,641=121*121.



Manager
Joined: 11 Mar 2018
Posts: 158

Re: Which of the following is two more than square of an odd integer?
[#permalink]
Show Tags
16 Nov 2019, 20:14
After seeing the options, a Square that will fit in the series will be ending with 2 zeroes to have 10 from 100 and beginning from 144, which is square of 12.
Hence perfect square number among this list would be \(14400 = 120^2\)  (1)
As we have options ending with 3, 1 and 7, we can have only numbers ending with 3 or 7 as the correct option, because 3 and 7 are never the end digits of a square.
Now, We are left with only three options  14,173 OR 14,643 OR 14,737  (2)
From (1) and (2), we can deduce that numbers can be either \(119^2\) OR \(121^2\),
From the looks of it seems, that 2 options can be checked by calculating \(121^2\).
After doing so, \(121^2\) = 14,641
As per question, \(121^2\) + 2 = 14,643
Hence IMO C



Intern
Joined: 14 Nov 2013
Posts: 22
Location: India
Concentration: Finance, Sustainability
WE: Corporate Finance (NonProfit and Government)

Re: Which of the following is two more than square of an odd integer?
[#permalink]
Show Tags
16 Nov 2019, 21:21
The answer is 14643
Method i used
All options are in 14000 range the closest perfect square we all know is 14400 which is 120^2
Question asks for a number which is 2 greater than the square of an odd integer
since we know that 120^2 is 14400 we can try the two closest odd numbers to 120 119^2 & 121^2
121^2 we can check first because it has 1s & 2s (easy multiplication )
we get 14641 we can add 2 to it
we get 14643
which is option C raw method.. probably a more technical method maybe there.. but aint doing a phd on the gmat... so be chill keep it simple do the question in a minute



Intern
Joined: 25 Mar 2013
Posts: 7

Re: Which of the following is two more than square of an odd integer?
[#permalink]
Show Tags
16 Nov 2019, 21:34
C
All the options given are around 14000
12^2 = 144 13^2 = 169
Implies 120^2 = 14400 and 130^2 = 16900
Since all the option are closer to 120^2,I checked 121^2 + 2 which is available in the options.
Posted from my mobile device




Re: Which of the following is two more than square of an odd integer?
[#permalink]
16 Nov 2019, 21:34






