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16 Jan 2020, 01:07
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Difficulty:
95%
(hard)
Question Stats:
36% (02:08) correct
64%
(02:10)
wrong
based on 116
sessions
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A prime number P greater than 100 leaves a remainder Q when divided by 28. How many values of Q exist?
A. 5
B. 9
C. 12
D. 15
E. 28
A. 5
B. 9
C. 12
D. 15
E. 28
16 Jan 2020, 02:53
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DisciplinedPrep
So \(P=28x+Q\)...
As Q is a remainder, it has to be <28.
Now if Q shares a common factor with 28, P will also be multiple of that factor and, therefore, will not be a prime number..
multiples of 2 till 28 --- 14 in number
multiples of 7 till 28 --- 4 in number
multiple of both 2 and 7 -- 2 in number ( 14 and 28).. these are common to both 14 and 4.
Thus remaining possibilities of Q = 28-14-4+2=12
C
General Discussion
09 Sep 2025, 10:16
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chetan2u, I calculated it differently.
P=28X+Q, where 0<Q<28.
1.Q can not be equal to 0 as then, P=28X and will have 2,7 as factors. Hence, P will not be a prime number.
2. Also, since the remainder is always smaller than the divisor, then Q<28.
Since, Q can not share any common factor with 28. It must not have 2 and 7 as prime factors.
So, total possible values of Q is 27. (1,2,3...27)
numbers with 2 as factor-13 (2,4,6...26)
numbers with 7 as factor-3 (7,14,21)
numbers with both 2 and 7 as factors-1 (14)
So, total number of possible values of Q = 27-(13+3-1)= 12 (Ans)
P=28X+Q, where 0<Q<28.
1.Q can not be equal to 0 as then, P=28X and will have 2,7 as factors. Hence, P will not be a prime number.
2. Also, since the remainder is always smaller than the divisor, then Q<28.
Since, Q can not share any common factor with 28. It must not have 2 and 7 as prime factors.
So, total possible values of Q is 27. (1,2,3...27)
numbers with 2 as factor-13 (2,4,6...26)
numbers with 7 as factor-3 (7,14,21)
numbers with both 2 and 7 as factors-1 (14)
So, total number of possible values of Q = 27-(13+3-1)= 12 (Ans)
chetan2u
