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24 Jun 2020, 11:27
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A
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Difficulty:
75%
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Question Stats:
55% (02:00) correct
45%
(01:53)
wrong
based on 370
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If x is a positive integer and x^2 + 3 divided by 32 leaves a reminder of 7. Which of the following must be an integer?
I. x/2
II. x/4
III. x/6
A. I only
B. II only
C. I and II only
D. I and III only
E. I, II and III
I. x/2
II. x/4
III. x/6
A. I only
B. II only
C. I and II only
D. I and III only
E. I, II and III
24 Jun 2020, 11:38
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When \((x^2+3)\) is divided by 32 let the quotient be 'a'
So, \(x^2+3 = 32a+7\)
or, \(x=\sqrt{32a+4}\)
or, \(x=2\sqrt{8a+1}\)
From this we can conclude that x MUST be divisible by 2.
It CAN be divisible by 6 as well (for e.g., if a=10, x=2*9=18) but it is not a MUST.
Answer: A
So, \(x^2+3 = 32a+7\)
or, \(x=\sqrt{32a+4}\)
or, \(x=2\sqrt{8a+1}\)
From this we can conclude that x MUST be divisible by 2.
It CAN be divisible by 6 as well (for e.g., if a=10, x=2*9=18) but it is not a MUST.
Answer: A
General Discussion
24 Jun 2020, 11:47
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Bunuel
We know that,
Dividend = Divisor * Quotient + Remainder
Therefore, according to this question:
\(x^2\)+ 3 = 32 * Q + 7
Upon simplifying -
\(x^2\) = 32 * Q + 4
Taking 4 common -
\(x^2\) = 4 (8 * Q + 1)
x = \(\sqrt{4 (8 * Q + 1)}\) = 2 \(\sqrt{(8 * Q + 1)}\)
x is definitely divisible by 2 and definitely not divisible by 4. (one can take further examples to verify this)
Now, we are given that x has to be an integer, let's substitute integer values of Q to determine which ones map to integral values of x and whether those values are divisible by 3. (to check whether x can be divisble by 6)
When Q = 1, 8*1 + 1 = 9 (square of 3) divisible by 3!
When Q = 2, 8*2 + 1 = 17 (square root of 17 is not integral, thus, x cannot be integral when Q = 2, this value can be disregarded)
When Q = 3, 8*3 + 1 = 25 (square of 5) not divisible by 3.
Thus, x is only divisble by 2, since the questions says which one "MUST" be an integer and we determined a condition where x is not divisble by 3, thus, cannot be divisble by 6.
Answer - (A)
