Re: Which of the following is two more than square of an odd integer?
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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: 14,173-2 = 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.