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16 Sep 2014, 01:22



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Re: M2502 [#permalink]
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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, \(\frac{180}{3}+1=7\).
Answer: B Hi Bunnel, if there isn't the word "inclusive" do we still +1?



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Re: M2502 [#permalink]
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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, \(\frac{180}{3}+1=7\).
Answer: B 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: howmanymultiplesof4aretherebetween12and94862.html#p730075
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Re: M2502 [#permalink]
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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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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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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 121, 222, 323, 424, etc...
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Re: M2502 [#permalink]
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10 Jul 2015, 10:03
Hello Bunnel,
I have a query about this question.
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.
Please help.
Thanks



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Re: M2502 [#permalink]
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11 Jul 2015, 02:49
vikasbansal227 wrote: Hello Bunnel,
I have a query about this question.
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.
Please help.
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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Re: M2502 [#permalink]
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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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Re: M2502 [#permalink]
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29 Aug 2015, 04:12
Bunuel wrote: vikasbansal227 wrote: Hello Bunnel,
I have a query about this question.
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.
Please help.
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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Re: M2502 [#permalink]
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29 Aug 2015, 05:56
reto wrote: Bunuel wrote: vikasbansal227 wrote: Hello Bunnel,
I have a query about this question.
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.
Please help.
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? Engr2012Your question is not clear. The question specifically says "divisible by 3 between 10! and 10!+20 inclusive"
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Re: M2502 [#permalink]
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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!+2010!' leaving only 20, so multiples from 0 to 20 inclusive.



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Re: M2502 [#permalink]
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11 Jan 2016, 13:57
The factorial makes it seem more confusing than the question actually is:
(10!+2010!) = 20 ( to prove it just think x+20x = 20 or 3!+13! = 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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Re: M2502 [#permalink]
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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, \(\frac{180}{3}+1=7\).
Answer: B 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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Re: M2502 [#permalink]
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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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Re: M2502 [#permalink]
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13 Mar 2016, 10:40
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: numberpropertiestipsandhints174996.html
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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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Re: M2502 [#permalink]
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23 Aug 2017, 05:11
Bunuel is this method fine? First multiple= 10! .... Last Multiple= 10!+ 18 n= (10!+1810!)/3 + 1 n= 18/3 +1 n= 7
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