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07 Jul 2023, 10:44
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C
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Difficulty:
75%
(hard)
Question Stats:
45% (02:12) correct
55%
(02:30)
wrong
based on 64
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If n is a positive integer, what is the value of n?
(1) For any positive integer y, the product of all integers from y to y+n is divisible by 35.
(2) n^2 - 8n + 7 < 0
(1) For any positive integer y, the product of all integers from y to y+n is divisible by 35.
(2) n^2 - 8n + 7 < 0
07 Jul 2023, 13:44
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OmkarSheth008
Statement 1
(1) For any positive integer y, the product of all integers from y to y+n is divisible by 35.
35 = 7 * 5
Hence, if we take any set of 7 consecutive integers, the product will be divisible by 35.
Therefore, the minimum value of 'n' is 6. Any value of n greater than or equal to 6 satisfies statement 1.
Hence, the statement alone is not sufficient to find a unique value of n. We can eliminate A and D.
Statement 2
(2) \(n^2 - 8n + 7 < 0\).
\(n^2 - 8n + 7 < 0\)
\((n-1)(n-7) < 0\)
------------ 1 ------------------------ 7 -------------
Any integer value between 1 and 7 satisfies the inequality. The statement alone is not sufficient to find a unique value of n. We can eliminate B.
Combined
From statement 1: \(n\geq6\)
From statement 2: \(1 < n < 7\)
The only common value = 6.
The statements combined help us identify a unique value of n.
Option C
07 Jul 2023, 17:27
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OmkarSheth008
Suppose y = 7x
7x . 7x+1 . ....7x+6 => 7(K ) form.
And more over 5 consecutive numbers will always be in the form of 5(K)
Hence this is divisible by 35.
So n can be any number greater than 6.
Hence statement 1 is not sufficient.
By factorizing statement 2, (n-1)(n-7)<0, n can be 2,3...6.
So even statement 2 is not self-sufficient.
But if we combine both 1 & 2, the Only possible value of n is 6.
IMO C
