Are true meaning must be true??

I would say III only.

N IS EVEN N+4 IS ALSO EVEN THEREFORE MUST BE DIVISIBLE BY 2

n^2 is not divisible by 600 since n^2 will be missing a 2


n^2+19 will not be divisible by 19 since n^2 is not

i think C.
coz...divic=nd by n^2 , 2 remains in Denom.
in 3...
n +4 = even
n = even -even =even
so...true
1 is wrong , 3 is true
no need to cal 2.as only option left with this is C

Brent Hanneson wrote:If n = 2×3×5×7×11×13×17, then which of the following statements are true?
I. n^2 is divisible by 600
II. n+19 is divisible by 19
III. (n+4)/2 is even

A) I only
B) II only
C) III only
D) I and III
E) None of the above

I - 600 = 2^3 * 5^2 * 3

therefore n^2 is not divisible by 600, coz in denominator we would be left with one 2.

Eliminate - A and D.

II - n+19 is divisible by 19
Only way this could be possible is when n is a multiple of 19 i.e. prime number
Eliminate - B


III
(n+4)/2 = even

Since we have all prime numbers here, we can do a test with only 2 prime numbers

let n=2*3 = 6

(6+4)/2 = 5

hence eliminate C

Therefore, E.

OA?

Be careful. The question refers to n+19, not n^2+19

Thanks that was a typo.

If n was divisible by 19 then n+19 is divisible by 19

If n^2 is divisible by 19 then n^2+19 is divisible by 19


In the problem above neither n nor n^2 is divisible by 19 (If n was divisible by 19 then n^2 is also divisible by 19)

Rushing does not help! sure to make mistakes like I did :roll:

2*3*5*7*11*13*17 + 4 /2

2 (3*5*7*11*13*17 + 2) /2

= 1 * (odd+even)

= 1 * odd
=odd

Hence E)

Good solution PC and a good problem Brent!

Very good problem Brent.
it touched multiple concepts.

I got E as well. but could not get it under 2 mins.
was stuck on II for a long time.

Thanks