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# M25-02

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 01:22
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45% (medium)

Question Stats:

58% (00:35) correct 42% (00:51) wrong based on 467 sessions

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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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Math Expert
Joined: 02 Sep 2009
Posts: 46297

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16 Sep 2014, 01:22
2
8
Official Solution:

How many integers are divisible by 3 between $$10!$$ and $$10! + 20$$ inclusive?

A. 6
B. 7
C. 8
D. 9
E. 10

Since 10! itself is a multiple of 3 (10!=2*3*...*10), then the question boils down to: how many integers from some multiple of 3 to that multiple of 3 + 20, inclusive are divisible by 3?

Or: how many integers are divisible by 3 from 0 to 20, inclusive?

# of multiples of $$x$$ in the range $$= \frac{\text{Last multiple of x in the range - First multiple of x in the range}}{x}+1$$.

So, $$\frac{18-0}{3}+1=7$$.

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Joined: 13 Jul 2014
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01 Oct 2014, 02:24
Bunuel wrote:
Official Solution:

How many integers are divisible by 3 between $$10!$$ and $$10! + 20$$ inclusive?

A. 6
B. 7
C. 8
D. 9
E. 10

Since 10! itself is a multiple of 3 (10!=2*3*...*10), then the question boils down to: how many integers from some multiple of 3 to that multiple of 3 + 20, inclusive are divisible by 3?

Or: how many integers are divisible by 3 from 0 to 20, inclusive?

# of multiples of $$x$$ in the range $$= \frac{\text{Last multiple of x in the range - First multiple of x in the range}}{x}+1$$.

So, $$\frac{18-0}{3}+1=7$$.

Hi Bunnel, if there isn't the word "inclusive" do we still +1?
Math Expert
Joined: 02 Sep 2009
Posts: 46297

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02 Oct 2014, 02:48
safwanst wrote:
Bunuel wrote:
Official Solution:

How many integers are divisible by 3 between $$10!$$ and $$10! + 20$$ inclusive?

A. 6
B. 7
C. 8
D. 9
E. 10

Since 10! itself is a multiple of 3 (10!=2*3*...*10), then the question boils down to: how many integers from some multiple of 3 to that multiple of 3 + 20, inclusive are divisible by 3?

Or: how many integers are divisible by 3 from 0 to 20, inclusive?

# of multiples of $$x$$ in the range $$= \frac{\text{Last multiple of x in the range - First multiple of x in the range}}{x}+1$$.

So, $$\frac{18-0}{3}+1=7$$.

Hi Bunnel, if there isn't the word "inclusive" do we still +1?

# of multiples of $$x$$ in the range $$= \frac{\text{Last multiple of x IN the range - First multiple of x IN the range}}{x}+1$$.

So, if it were: how many integers are divisible by 3 from 0 to 20, NOT inclusive, then the answer would be (18 - 3)/3 +1 = 6.

Check here for more: how-many-multiples-of-4-are-there-between-12-and-94862.html#p730075
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Joined: 10 May 2014
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26 Dec 2014, 13:05
Hi Bunuel,

I solved this problem in the following way.

I didn´t pay attention to the actual range 10! to 10! + 20. I only knew that there were 21 positive integers.

I think that in any range of consecutive positive integers, 1 out of 3 of those consecutive integers will be divisible by 3 and the other 2 will not.

This can be explained algebraically in the following way:
Suppose x is any positive integer.
A range of 3 integers can be described as x + (x + 1) + (x + 2). This yelds 3x + 3, or 3(x + 1). Therefore this expression is divisble by 3.

If you have 21 integers, you have to do this 7 times.

Does this make sense?
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Math Expert
Joined: 02 Sep 2009
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27 Dec 2014, 03:54
minwoswoh wrote:
Hi Bunuel,

I solved this problem in the following way.

I didn´t pay attention to the actual range 10! to 10! + 20. I only knew that there were 21 positive integers.

I think that in any range of consecutive positive integers, 1 out of 3 of those consecutive integers will be divisible by 3 and the other 2 will not.

This can be explained algebraically in the following way:
Suppose x is any positive integer.
A range of 3 integers can be described as x + (x + 1) + (x + 2). This yelds 3x + 3, or 3(x + 1). Therefore this expression is divisble by 3.

If you have 21 integers, you have to do this 7 times.

Does this make sense?

Yes. As in my solution: the question boils down to "how many integers from some multiple of 3 to that multiple of 3 + 20, inclusive are divisible by 3?"
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28 Dec 2014, 21:33
Bunuel wrote:
minwoswoh wrote:
Hi Bunuel,

I solved this problem in the following way.

I didn´t pay attention to the actual range 10! to 10! + 20. I only knew that there were 21 positive integers.

I think that in any range of consecutive positive integers, 1 out of 3 of those consecutive integers will be divisible by 3 and the other 2 will not.

This can be explained algebraically in the following way:
Suppose x is any positive integer.
A range of 3 integers can be described as x + (x + 1) + (x + 2). This yelds 3x + 3, or 3(x + 1). Therefore this expression is divisble by 3.

If you have 21 integers, you have to do this 7 times.

Does this make sense?

Yes. As in my solution: the question boils down to "how many integers from some multiple of 3 to that multiple of 3 + 20, inclusive are divisible by 3?"

Hi Bunuel,

Thanks for your answer. I understand your point, but I thought you were saying that the first integer of the 21 must be divisible by 3. Kind of like in a sequence a, b, c, a, b, c… in which a is divisible by 3 but b and c are not.

I was saying that if you have 21 consecutive positive integers, it really doesn´t matter if the first, second, or third is the one that is divisible by 3. If you have 21 consecutive positive integers, you will have 7 integers that are divisible by 3 anyway.

For instance, you have 7 integers divisble by 3 in all ranges 1-21, 2-22, 3-23, 4-24, etc...
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10 Jul 2015, 10:03
Hello Bunnel,

While attempting this question in test mode, I understood the idea about number of multiples in 20.

In solution you have mentioned that 10! contains one multiple 1*2*3.....but actually there are three multiples of 3 in 10! i.e. .....3*6*9....

So I picked D i.e. 09....could you please help me with this problem. I tried every possible alternate but still arriving at same solution i.e. 09.

Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 46297

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11 Jul 2015, 02:49
4
vikasbansal227 wrote:
Hello Bunnel,

While attempting this question in test mode, I understood the idea about number of multiples in 20.

In solution you have mentioned that 10! contains one multiple 1*2*3.....but actually there are three multiples of 3 in 10! i.e. .....3*6*9....

So I picked D i.e. 09....could you please help me with this problem. I tried every possible alternate but still arriving at same solution i.e. 09.

Thanks

The solution does NOT say that 10! contains one multiple of 3, it says that 10! IS a multiple of 3, it does not matter what is the power of 3 in 10!.

Multiples of 3 between $$10!$$ and $$10! + 20$$ inclusive are:

$$10!$$;
$$10! + 3$$;
$$10! + 6$$;
$$10! + 9$$;
$$10! + 12$$;
$$10! + 15$$;
$$10! + 18$$.

Total of 7.
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Joined: 25 Nov 2014
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18 Jul 2015, 03:51
Bunuel wrote:
minwoswoh wrote:
Hi Bunuel,

I solved this problem in the following way.

I didn´t pay attention to the actual range 10! to 10! + 20. I only knew that there were 21 positive integers.

I think that in any range of consecutive positive integers, 1 out of 3 of those consecutive integers will be divisible by 3 and the other 2 will not.

This can be explained algebraically in the following way:
Suppose x is any positive integer.
A range of 3 integers can be described as x + (x + 1) + (x + 2). This yelds 3x + 3, or 3(x + 1). Therefore this expression is divisble by 3.

If you have 21 integers, you have to do this 7 times.

Does this make sense?

Yes. As in my solution: the question boils down to "how many integers from some multiple of 3 to that multiple of 3 + 20, inclusive are divisible by 3?"

yes, but a certain thing to determine is whether the corner values of the range are divisible or not...
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29 Aug 2015, 04:12
Bunuel wrote:
vikasbansal227 wrote:
Hello Bunnel,

While attempting this question in test mode, I understood the idea about number of multiples in 20.

In solution you have mentioned that 10! contains one multiple 1*2*3.....but actually there are three multiples of 3 in 10! i.e. .....3*6*9....

So I picked D i.e. 09....could you please help me with this problem. I tried every possible alternate but still arriving at same solution i.e. 09.

Thanks

The solution does NOT say that 10! contains one multiple of 3, it says that 10! IS a multiple of 3, it does not matter what is the power of 3 in 10!.

Multiples of 3 between $$10!$$ and $$10! + 20$$ inclusive are:

$$10!$$;
$$10! + 3$$;
$$10! + 6$$;
$$10! + 9$$;
$$10! + 12$$;
$$10! + 15$$;
$$10! + 18$$.

Total of 7.

So the word "between" excludes the 10! from counting the multiples? Very tricky to understand... What if the question asked, "How many integers are divisible by 3 from and inclusive 10! and 10!+20 inclusive?

It then would be different, correct? Engr2012
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Posts: 46297

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29 Aug 2015, 05:56
1
reto wrote:
Bunuel wrote:
vikasbansal227 wrote:
Hello Bunnel,

While attempting this question in test mode, I understood the idea about number of multiples in 20.

In solution you have mentioned that 10! contains one multiple 1*2*3.....but actually there are three multiples of 3 in 10! i.e. .....3*6*9....

So I picked D i.e. 09....could you please help me with this problem. I tried every possible alternate but still arriving at same solution i.e. 09.

Thanks

The solution does NOT say that 10! contains one multiple of 3, it says that 10! IS a multiple of 3, it does not matter what is the power of 3 in 10!.

Multiples of 3 between $$10!$$ and $$10! + 20$$ inclusive are:

$$10!$$;
$$10! + 3$$;
$$10! + 6$$;
$$10! + 9$$;
$$10! + 12$$;
$$10! + 15$$;
$$10! + 18$$.

Total of 7.

So the word "between" excludes the 10! from counting the multiples? Very tricky to understand... What if the question asked, "How many integers are divisible by 3 from and inclusive 10! and 10!+20 inclusive?

It then would be different, correct? Engr2012

Your question is not clear.

The question specifically says "divisible by 3 between 10! and 10!+20 inclusive"
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Joined: 17 Nov 2014
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01 Nov 2015, 11:17
Hi Bunnel,

Wanted to check if we could solve it the way given below:

Between 10! and 10!+20 inclusive implies '10!+20-10!' leaving only 20, so multiples from 0 to 20 inclusive.
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11 Jan 2016, 13:57
The factorial makes it seem more confusing than the question actually is:

(10!+20-10!) = 20 ( to prove it just think x+20-x = 20 or 3!+1-3! = 1)

How many multiples of 3 are in 20 including 0?
(20/3) + 1 = 7 integer multiples of 3.

Ans. B
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Joined: 05 Nov 2012
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05 Feb 2016, 10:42
Bunuel wrote:
Official Solution:

How many integers are divisible by 3 between $$10!$$ and $$10! + 20$$ inclusive?

A. 6
B. 7
C. 8
D. 9
E. 10

Since 10! itself is a multiple of 3 (10!=2*3*...*10), then the question boils down to: how many integers from some multiple of 3 to that multiple of 3 + 20, inclusive are divisible by 3?

Or: how many integers are divisible by 3 from 0 to 20, inclusive?

# of multiples of $$x$$ in the range $$= \frac{\text{Last multiple of x in the range - First multiple of x in the range}}{x}+1$$.

So, $$\frac{18-0}{3}+1=7$$.

Hi Bunuel,

The question here says between 10! and 10 + 20! and that is why I did not count 10!. Shouldn't the question say from and then say inclusive?
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13 Mar 2016, 10:37
Hello,
I am struggling to understand why are we counting the "0" as multiple of 3. the range is from 0 to 20, but 3:0 is undefined, I thought the answer was 6.
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Joined: 02 Sep 2009
Posts: 46297

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13 Mar 2016, 10:40
2
Avigano wrote:
Hello,
I am struggling to understand why are we counting the "0" as multiple of 3. the range is from 0 to 20, but 3:0 is undefined, I thought the answer was 6.

0 is not a divisor of any integer, but a multiple of every integer. So, 0 is divisible by 3: 0/3 = 0 = integer.

ZERO:

1. 0 is an integer.

2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.

Check more here: number-properties-tips-and-hints-174996.html
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14 Oct 2016, 01:05
Hello ,

The range is between 10 ! and 10 ! +20 and factorial starts from 1 i.e 10 ! = 1*.....10 , so the first multiple of 3 is 3 ...why 0??? even though the range is from 1 to 20 ...?

Bunuel , can you please explain !
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23 Aug 2017, 05:11
Bunuel is this method fine?

First multiple= 10!
....
Last Multiple= 10!+ 18

n= (10!+18-10!)/3 + 1
n= 18/3 +1
n= 7
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23 Aug 2017, 05:13
Shiv2016 wrote:
Bunuel is this method fine?

First multiple= 10!
....
Last Multiple= 10!+ 18

n= (10!+18-10!)/3 + 1
n= 18/3 +1
n= 7

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Yes, that's correct.
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