If n is a positive integer, is n^3 + 4n^2 - 5n divisible by 8 ?
(1) n = 4b + 1, where b is a positive integer.
(2) n^2 - n is divisible by 24.
OA D
Source: Princeton Review
If n is a positive integer, is n^3 + 4n^2 - 5n divisible by
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We have n^3 + 4n^2 - 5nBTGmoderatorDC wrote:If n is a positive integer, is n^3 + 4n^2 - 5n divisible by 8 ?
(1) n = 4b + 1, where b is a positive integer.
(2) n^2 - n is divisible by 24.
OA D
Source: Princeton Review
=> n(n^2 + 4n - 5) = n(n^2 + 5n - n - 5) = n[n(n + 5) - 1(n + 5)] = n(n - 1)(n + 5)
Thus, we have to find out whether n(n - 1)(n + 5) is divisible by 8.
Let's take each statement one by one.
(1) n = 4b + 1, where b is a positive integer.
Plugging-in the value of n = 4b + 1 in n(n - 1)(n + 5), we get (4b + 1)(4b + 1 - 1)(4b + 1 + 5) = (4b + 1)(4b)(4b + 6) = 8b(4b + 1)(2b + 3)
So, we see that n^3 + 4n^2 - 5n is a factor of 8; thus, it is divisible by 8. Sufficient.
(2) n^2 - n is divisible by 24.
=> n(n - 1) is divisible by 24.
Since n^3 + 4n^2 - 5n = n(n - 1)(n + 5), it is a factor of n(n - 1). Since n(n - 1) is divisible by 24, it is obviously be divisible by 8. Thus, n^3 + 4n^2 - 5n is divisible by 8. Sufficient.
The correct answer: D
Hope this helps!
-Jay
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$$n^3+4n^2-5n$$
$$n\left(n^2+4n-5\right)$$
$$\frac{\left[n\left(n-1\right)\left(n+5\right)\right]}{8}??$$
Statement 1=> n = 4b + 1; where b is a positive integer.
$$n\left(n-1\right)\left(n+5\right)=\left(4b+1\right)\left(4b+1-1\right)\left(4b+1+5\right)$$
$$=\left(4b+1\right)\left(4b\right)\left(4b+6\right)$$
$$=\left(4b+1\right)\left(16b^2+24b\right)$$
$$Therefore;\ n\left(n-1\right)\left(n+5\right)=\frac{\left[\left(4b+1\right)\left(8b\right)\left(2b+3\right)\right]}{8}$$
It is divisible by 8. Thus, statement 1 is SUFFICIENT.
Statement 2=>
$$n^2-n\ is\ divisible\ by\ 24.$$
If n(n-1) is divisible by 24, it will also be divisible by 8. Thus, n(n-1)(n+5) is also divisible by 8. Statement 2 is also SUFFICIENT.
Therefore, each statement alone are SUFFICIENT
Answer = Option D
$$n\left(n^2+4n-5\right)$$
$$\frac{\left[n\left(n-1\right)\left(n+5\right)\right]}{8}??$$
Statement 1=> n = 4b + 1; where b is a positive integer.
$$n\left(n-1\right)\left(n+5\right)=\left(4b+1\right)\left(4b+1-1\right)\left(4b+1+5\right)$$
$$=\left(4b+1\right)\left(4b\right)\left(4b+6\right)$$
$$=\left(4b+1\right)\left(16b^2+24b\right)$$
$$Therefore;\ n\left(n-1\right)\left(n+5\right)=\frac{\left[\left(4b+1\right)\left(8b\right)\left(2b+3\right)\right]}{8}$$
It is divisible by 8. Thus, statement 1 is SUFFICIENT.
Statement 2=>
$$n^2-n\ is\ divisible\ by\ 24.$$
If n(n-1) is divisible by 24, it will also be divisible by 8. Thus, n(n-1)(n+5) is also divisible by 8. Statement 2 is also SUFFICIENT.
Therefore, each statement alone are SUFFICIENT
Answer = Option D