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Re: Which of the following is two more than square of an odd integer?
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15 Nov 2019, 01:06
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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?
Re: Which of the following is two more than square of an odd integer?
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15 Nov 2019, 01:13
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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
Re: Which of the following is two more than square of an odd integer?
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15 Nov 2019, 04:49
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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. 14,173-2=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
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.
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Ephemeral Epiphany..!
GMATPREP1 590(Q48,V23) March 6, 2019 GMATPREP2 610(Q44,V29) June 10, 2019 GMATPREPSoft1 680(Q48,V35) June 26, 2019
Re: Which of the following is two more than square of an odd integer?
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15 Nov 2019, 12:30
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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.
Re: Which of the following is two more than square of an odd integer?
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15 Nov 2019, 22:37
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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.
Re: Which of the following is two more than square of an odd integer?
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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.
Re: Which of the following is two more than square of an odd integer?
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16 Nov 2019, 20:14
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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 \(121^2\).
Re: Which of the following is two more than square of an odd integer?
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16 Nov 2019, 21:21
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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
close
68% (02:25) correct 
