How many integers are divisible by 3 between 10! and 10!+20 inclusive?
6
7
8
9
10
I dint understand the question whts the meaning?
How many integers are divisible by 3 between 10! and 10!+20
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Hi shibsriz,
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!
The same applies to 4! (4! = 1x2x3x4). We can factor out a 3 and get 3(1x2x4); so this means that 3 divides evenly into 4! In this same way, we know that 3 divides evenly into 5!, 6!, 7!, etc. We now know that 3 divides evenly into 10!.
Does 3 divide into 3! + 1? No, because you CAN'T factor out a 3.
Does 3 divide into 3! + 2? No, because you CAN'T factor out a 3.
Does 3 divide into 3! + 3? YES, because you CAN factor out a 3. You'd have 3(1x2 + 1).
This same rule applies to the range of values between 10! and 10! + 20
3 will divide evenly into:
10!
10! + 3
10! + 6
10! + 9
10! + 12
10! + 15
10! + 18
Final Answer: B
GMAT assassins aren't born, they're made,
Rich
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!
The same applies to 4! (4! = 1x2x3x4). We can factor out a 3 and get 3(1x2x4); so this means that 3 divides evenly into 4! In this same way, we know that 3 divides evenly into 5!, 6!, 7!, etc. We now know that 3 divides evenly into 10!.
Does 3 divide into 3! + 1? No, because you CAN'T factor out a 3.
Does 3 divide into 3! + 2? No, because you CAN'T factor out a 3.
Does 3 divide into 3! + 3? YES, because you CAN factor out a 3. You'd have 3(1x2 + 1).
This same rule applies to the range of values between 10! and 10! + 20
3 will divide evenly into:
10!
10! + 3
10! + 6
10! + 9
10! + 12
10! + 15
10! + 18
Final Answer: B
GMAT assassins aren't born, they're made,
Rich
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The intent of the problem seems to be as follows:
(multiple of k) + (multiple of k) = (multiple of k).
(multiple of k) + (non-multiple of k) = (non-multiple of k).
Since 10! = 10*9*8*7*6*5*4*3*2, 10! is a multiple of 3.
To yield subsequent multiples of 3, we must add to 10! consecutive multiples of 3
Options between 10! and 10! + 20, inclusive:
10!
10! + 3
10! + 6
10! + 9
10! + 12
10! + 15
10! + 18.
Total options = 7.
The correct answer is B.
Two number property rules:How many integers between 10! and 10!+20, inclusive, are divisible by 3?
6
7
8
9
10
(multiple of k) + (multiple of k) = (multiple of k).
(multiple of k) + (non-multiple of k) = (non-multiple of k).
Since 10! = 10*9*8*7*6*5*4*3*2, 10! is a multiple of 3.
To yield subsequent multiples of 3, we must add to 10! consecutive multiples of 3
Options between 10! and 10! + 20, inclusive:
10!
10! + 3
10! + 6
10! + 9
10! + 12
10! + 15
10! + 18.
Total options = 7.
The correct answer is B.
Last edited by GMATGuruNY on Fri Sep 05, 2014 11:36 pm, edited 1 time in total.
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There's a nice rule that says: If M is divisible by k, and N is divisible by k, then (M + N) is divisible by k.[email protected] wrote:How many integers are divisible by 3 between 10! and 10!+20 inclusive?
A) 6
B) 7
C) 8
D) 9
E) 10
First, since 10! = (10)(9)(8)..(3)(2)(1), we know that 10! is divisible by 3.
So, by the above rule, we know that 10! + 3 is divisible by 3
And 10! + 6 is divisible by 3
10! + 9 is divisible by 3
10! + 12 is divisible by 3
10! + 15 is divisible by 3
10! + 18 is divisible by 3
So, there are 7 integers from 10! to 10! + 20 inclusive that are divisible by 3.
Answer: B
Cheers,
Brent
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HI Guru/Brent,GMATGuruNY wrote:The intent of the problem seems to be as follows:
Two number property rules:How many integers between 10! and 10!+20, inclusive, are divisible by 3?
A)6
B)7
C)8
D)9
E)10
(multiple of k) + (multiple of k) = (multiple of k).
(multiple of k) + (non-multiple of k) = (non-multiple of k).
Since 10! = 10*9*8*7*6*5*4*3*2, 10! is a multiple of 3.
To yield subsequent multiples of 3, we must add to 10! consecutive multiples of 3?
Options between 10! and 10! + 20, inclusive:
10!
10! + 3
10! + 6
10! + 9
10! + 12
10! + 15
10! + 18.
Total options = 7.
The correct answer is B.
If the question is How many integers between 10! and 10!+20, are divisible by 3?
A)6
B)7
C)8
D)9
E)10
What would be the ans?
Are you including 10!?
Thanks
Nandish
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Since 3 is a factor of 10!, 10! is divisible by 3. From this point, we can keep adding 3 to 10! to obtain integers that are divisible by 3, so the integers between 10! and 10!+20 inclusive that are divisible by 3 are:[email protected] wrote:How many integers are divisible by 3 between 10! and 10!+20 inclusive?
6
7
8
9
10
10!, 10! + 3, 10! + 6, 10! + 9, 10! + 12, 10! + 15 and 10! + 18
We see that there are 7 such integers.
Answer: B
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