If \(n=20!+17\), then \(n\) is divisible by which of the

[swerve]

If \(n=20!+17\), then \(n\) is divisible by which of the following?

I. 15
II. 17
III. 19

A. None
B. I only
C. II only
D. I and III
E. II and III

The OA C

Source: Official Guide

[joao_cardoso123]

It's only divisible by 17 because "20!" is divisible by all the three of them (15, 17, 19) so if you add 17 to "20!" it is not divisible anymore by 15 nor by 19

[Scott@TargetTestPrep]

If \(n=20!+17\), then \(n\) is divisible by which of the following?

I. 15
II. 17
III. 19

A. None
B. I only
C. II only
D. I and III
E. II and III

The OA C

Source: Official Guide

Since 20! is a multiple of 17 and 17 is a multiple of 17, we know that 20! + 17 is divisible by 17. On the other hand, since 20! is a multiple of 15 but 17 is not, then 20! + 15 is not divisible by 15. Likewise, 20! + 17 is not divisible by 19, either.

Answer: C

[[email protected]]

Hi All,

We're told that N = (20! + 17). We're asked which of the following is N divisible by. This question is ultimately about "factoring" and why numbers divide evenly into other numbers. I'm going to start with a simple example and work up to the details in this prompt. You probably know that 3 divides evenly into 3! (3! = 1x2x3). We can factor out a 3 and get 3(2); mathematically, this means that 3 divides evenly into 3!

Does 3 divide into 3! + 1? Does 3 divide into 7? No, because you CAN'T factor out a 3.
Does 3 divide into 3! + 2? Does 3 divide into 8? No, because you CAN'T factor out a 3.
Does 3 divide into 3! + 3? Does 3 divide into 9? YES, because you CAN factor out a 3. You'd have 3(1x2 + 1).

This same rule applies to this much larger value: 20! + 17

We can factor out a 17, which would give us (17)(20x19x18x16x15....x3x2x1 + 1). This tells us that N is absolutely divisibly by 17. That extra "+1" in the calculation means that NONE of the other numbers from 1 to 20 will divide into 20! + 17 though (since they cannot be factored out). Thus, the only one of the 3 Roman Numerals that divides evenly in is 17.

Final Answer: C

GMAT assassins aren't born, they're made,
Rich

[Brent@GMATPrepNow]

If \(n=20!+17\), then \(n\) is divisible by which of the following?

I. 15
II. 17
III. 19

A. None
B. I only
C. II only
D. I and III
E. II and III

The OA C

Source: Official Guide

Answer choice I: is 20! + 17 divisible by 15?
20! + 17 = (20)(19)(18)(17)(16)(15)(other stuff) + 15 + 2
= (15)(some number + 1) + 2
So, if we divide (15)(some number + 1) + 2 by 15, the remainder will be 2
So, 20! + 17 is NOT divisible by 15


Answer choice II: is 20! + 17 divisible by 17?
20! + 17 = (20)(19)(18)(17)(other stuff) + 17
= (17)(some number + 1)
If we divide (17)(some number + 1) by 17, the remainder will be 0
So, 20! + 17 IS divisible by 17


Answer choice III: is 20! + 17 divisible by 19?
20! + 17 = (20)(19)(other stuff) + 17
= (19)(some number) + 17
If we divide (19)(some number) + 17 by 19, the remainder will be 17
So, 20! + 17 is NOT divisible by 19

Answer: C

Cheers,
Brent