How many integers are divisible by 3 between 10! and 10!+20 inclusive?
A)6
B)7
C)8
D)9
E)10
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10! and 10!+20
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j_shreyans
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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.
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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.How many integers are divisible by 3 between 10! and 10!+20 inclusive?
A) 6
B) 7
C) 8
D) 9
E) 10
Conversely, If M is divisible by k, and Q is NOT divisible by k, then (M + Q) is NOT divisible by k.
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
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Hi j_shreyans,
The GMAT Quant section oftentimes tests you on math rules/concepts that you know, but in ways that you're not used to thinking about. This question is built on a division rule that you probably know: factoring.
Here's a simple example of the concept:
Is 10 + 8 divisible by 2?
Yes; both 10 and 8 are divisible by 2, so adding them together gives us a bigger number that's divisible by 2.
This can be proven by factoring:
(10 + 8) = 2(5 + 4).
With these ideas in mind, you can now tackle the prompt:
10! = (10)(9)(8)(7)(6)(5)(4)(3)(2)(1) so we know that this is divisible by lots of numbers. There are 21 integers between 10! and 10!+20 that we have to consider. Which of these integers is divisibly by 3?
We already know that 10! is divisible by 3 (since there's a "3" that can be factored out). Now, what other numbers, when added to 10! are ALSO divisible by 3?
10!
10! + 3
10! + 6
10! + 9
10! + 12
10! + 15
10! + 18
Total integers = 7
Final Answer: B
GMAT assassins aren't born, they're made,
Rich
The GMAT Quant section oftentimes tests you on math rules/concepts that you know, but in ways that you're not used to thinking about. This question is built on a division rule that you probably know: factoring.
Here's a simple example of the concept:
Is 10 + 8 divisible by 2?
Yes; both 10 and 8 are divisible by 2, so adding them together gives us a bigger number that's divisible by 2.
This can be proven by factoring:
(10 + 8) = 2(5 + 4).
With these ideas in mind, you can now tackle the prompt:
10! = (10)(9)(8)(7)(6)(5)(4)(3)(2)(1) so we know that this is divisible by lots of numbers. There are 21 integers between 10! and 10!+20 that we have to consider. Which of these integers is divisibly by 3?
We already know that 10! is divisible by 3 (since there's a "3" that can be factored out). Now, what other numbers, when added to 10! are ALSO divisible by 3?
10!
10! + 3
10! + 6
10! + 9
10! + 12
10! + 15
10! + 18
Total integers = 7
Final Answer: B
GMAT assassins aren't born, they're made,
Rich
