GmatKiss wrote: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 II
E.II and III

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

GmatKiss wrote: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 II
E.II and III

Solution:

We are given that n = 20! + 17 and need to know whether n is divisible by 15, 17, and/or 19. To determine this, we rewrite the given expression for n using each answer choice.

Thus, we have:

Does (20! + 17)/15 = integer?

Does (20! + 17)/17 = integer?

Does (20! + 17)/19 = integer?

We now use the distributive property of division over addition to determine which of these expressions is/are equal to an integer.

The distributive property of division over addition tells us that (a + c)/b = a/b + c/b. We apply this rule as follows:

I.

Does (20! + 17)/15 = integer?

Does 20!/15 + 17/15 = integer?

Although 20! is divisible by 15, 17 is NOT, and thus (20! + 17)/15 IS NOT an integer.

We can eliminate answer choices B and D.

II.

Does (20! + 17)/17 = integer?

Does 20!/17 + 17/17 = integer?

Both 20! and 17 are divisible by 17, and thus (20! + 17)/17 IS an integer.

We can eliminate answer choice A.

III.

Does (20! + 17)/19 = integer?

Does 20!/19 + 17/19 = integer?

Although 20! is divisible by 19, 17 is NOT, so (20! + 17)/19 IS NOT an integer.

We can eliminate answer choice E.

Thus, II is the only correct statement.

Answer:C

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