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

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

A. 14,173-2=14171
B. 14,361-2=14359
C. 14,643-2=14641
D. 14,737-2=14735: f(14735)=5*2947…2947≠f(5)…not perf.square
E. 14,981-2=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

\(x=119:x^2=120*120-120-119=14400-239=14161≠(A)\)
\(x=121:x^2=120*120+120+121=14400+241=14641=(C)\)
\(x=123:x^2=121*121+(2)121+(2)123=14641+2(244)=15129>14979;invalid\)

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.
_________________

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

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
N-3 = 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
_________________

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

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: 14,173-2 = 14,171
B. 14,361: 14,361-2 = 14,359
C. 14,643: 14,643-2 = 14,641
D. 14,737: 14,737-2 = 14,735
E. 14,981: 14,981-2 = 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.

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

C. 14,643 because of 14,643-2=14,641=121*121.

C. 14,643 because of 14,643-2=14,641=121*121.

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
_________________

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

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.


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