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18 May 2020, 01:57
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C
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Difficulty:
95%
(hard)
Question Stats:
31% (02:26) correct
69%
(02:34)
wrong
based on 162
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How many pairs of integer (a, b) are possible such that a^2 – b^2 = 288?
A. 6
B. 12
C. 24
D. 32
E. 48
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18 May 2020, 02:20
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\(a^2-b^2 = 288\)
\((a+b)(a-b)= 288\)
\(288=2^5*3^2\)
We'll get integral solutions od (a,b) when (a+b) and (a-b) are even. (both can't be odd as their product is even)
Number of ways to write 288 as a product of 2 positive even numbers
\(2x*2y = 288\)
\(x*y=2^3*3^2\)
total unordered solutions of (x,y)=\(\frac{(3+1)*(2+1)}{2}=6\)
Hence positive integral solutions of \(a^2-b^2 = 288\) is equal to 6
Since we require integral solutions, (a,b) can be (++), (-+), (+-) and (--).
integral solutions of \(a^2-b^2 = 288\) is equal 6*4 = 24
\((a+b)(a-b)= 288\)
\(288=2^5*3^2\)
We'll get integral solutions od (a,b) when (a+b) and (a-b) are even. (both can't be odd as their product is even)
Number of ways to write 288 as a product of 2 positive even numbers
\(2x*2y = 288\)
\(x*y=2^3*3^2\)
total unordered solutions of (x,y)=\(\frac{(3+1)*(2+1)}{2}=6\)
Hence positive integral solutions of \(a^2-b^2 = 288\) is equal to 6
Since we require integral solutions, (a,b) can be (++), (-+), (+-) and (--).
integral solutions of \(a^2-b^2 = 288\) is equal 6*4 = 24
18 May 2020, 04:23
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288 = 2^5 * 3^2.
Thus, there are (5+1) * (2+1) = 18 factors of 288 listed as following products:
1) 1 * 288,
2) 2 * 144,
3) 3 * 96,
4) 4 * 72,
5) 6 * 48,
6) 8 * 36,
7) 9 * 32,
8) 12 * 24, and
9) 16 * 18
a^2 – b^2 = (a-b).(a+b)
So, if both a and b are integers, (a+b) and (a–b) must be both odd or both even. Among the abovementioned lists, 1 * 288, 3 * 96, and 9 * 32 don't satisfy such condition, but the remaining 6 product sets do:
1) 2 * 144,
2) 4 * 72,
3) 6 * 48,
4) 8 * 36,
5) 12 * 24, and
6) 16 * 18
Now, take an example of 16*18=(a-b).(a+b). There are 4 possible combination of (a, b) for 16*18 alone:
I. a=17, b=1
II. a=-17, b=-1
III. a=-17, b=1
IV. a=17, b=-1
So, for each of the remaining 6 product sets, we will have 4 possible combination. THUS, total number of (a, b) that will satisfy this equation = 6 * 4 = 24.
FINAL ANSWER IS (C)
Posted from my mobile device
Thus, there are (5+1) * (2+1) = 18 factors of 288 listed as following products:
1) 1 * 288,
2) 2 * 144,
3) 3 * 96,
4) 4 * 72,
5) 6 * 48,
6) 8 * 36,
7) 9 * 32,
8) 12 * 24, and
9) 16 * 18
a^2 – b^2 = (a-b).(a+b)
So, if both a and b are integers, (a+b) and (a–b) must be both odd or both even. Among the abovementioned lists, 1 * 288, 3 * 96, and 9 * 32 don't satisfy such condition, but the remaining 6 product sets do:
1) 2 * 144,
2) 4 * 72,
3) 6 * 48,
4) 8 * 36,
5) 12 * 24, and
6) 16 * 18
Now, take an example of 16*18=(a-b).(a+b). There are 4 possible combination of (a, b) for 16*18 alone:
I. a=17, b=1
II. a=-17, b=-1
III. a=-17, b=1
IV. a=17, b=-1
So, for each of the remaining 6 product sets, we will have 4 possible combination. THUS, total number of (a, b) that will satisfy this equation = 6 * 4 = 24.
FINAL ANSWER IS (C)
Posted from my mobile device
