If n is a positive integer and r is the remainder when (n – 1)(n...
(1) n is not divisible by 2. Insufficient on its own, but this statement says that n=odd--> n−1 and n+1 are consecutive even integers --> (n−1)(n+1) must be divisible by 8 (as both multiples are even and one of them will be divisible by 4. From consecutive even integers one is divisible by 4: (2, 4); (4, 6); (6, 8); (8, 10); (10, 12), ...).
(2) n is not divisible by 3. Insufficient on its own, but form this statement either n−1 or n+1 must be divisible by 3 (as n−1, n, and n+1 are consecutive integers, so one of them must be divisible by 3, we are told that it's not n, hence either n−1 or n+1).
(1)+(2) From (1) (n−1)(n+1) is divisible by 8, from (2) it's also divisible by 3, therefore it must be divisible by 8∗3=24, which means that remainder upon division (n−1)(n+1) by 24 will be 0. Sufficient.
View all questions of this test
If n is a positive integer and r is the remainder when (n – 1)(n...
We can use the division algorithm to write:
n = dq + r
where d is a positive integer and 0 ≤ r < q.="" />
Rearranging this equation, we get:
r = n - dq
We know that n is a multiple of 13, so we can write:
n = 13k
where k is a positive integer.
Substituting this into the equation above, we get:
r = 13k - dq
Since r is less than q, we can write:
r = q - s
where s is a positive integer.
Substituting this into the equation above, we get:
q - s = 13k - dq
Simplifying, we get:
dq + s = 13k - q
Since d and q are positive integers, we know that dq + s is greater than q. Therefore, we can write:
dq + s = q + 13m
where m is a positive integer.
Substituting this into the equation above, we get:
q + 13m = 13k - q
Simplifying, we get:
2q + 13m = 13k
This shows that 2q is a multiple of 13. Since 13 is a prime number, this means that q must be a multiple of 13.
Therefore, the remainder r when n is divided by q must be less than q, and it must also be a multiple of 13. The only multiple of 13 that is less than q is 0. Therefore, the remainder r must be 0.
If n is a positive integer and r is the remainder when (n – 1)(n...
Correct answer will be option C
how?
statement-1 put any odd value of N in the equation, you will get different remainder.out
statement-2 put any value of N other than a number multiple of 3 in the equation.remainder will be different.
combining both we can say N is not divisible by either 2 or 3 that means any value of N=6K is not possible.
But N can be 6K+1 or 6K-1
so (N-1)(N+1)=6k*(6k+2)=12k(3k+1)
now see if k is odd 3k+1 will be even ,that implies
12*odd*even= 12*even ,that implies that even number is a multiple of 2 ,so 12*2 =24 will be the minimum value of the equation, which when divided by 24 will leave 0 remainder
and same you can prove for N=6K-1.
so answer is C