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M01 Question 18

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Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink] New post 29 Dec 2010, 00:40
Expert's post
How many of the three-digit numbers are divisible by 7?
(A) 105
(B) 106
(C) 127
(D) 128
(E) 142

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

For example: how many multiples of 5 are there between -7 and 35, not inclusive?

Last multiple of 5 IN the range is 30;
First multiple of 5 IN the range is -5;

\frac{30-(-5)}{5}+1=8.

OR:
How many multiples of 7 are there between -28 and -1, not inclusive?
Last multiple of 7 IN the range is -7;
First multiple of 7 IN the range is -21;

\frac{-7-(-21)}{7}+1=3.

Back to the original question:

Last 3-digit multiple of 7 is 994;
First 3-digit multiple of 7 is 105;

So # of 3-digt multiples of 7 is \frac{994-105}{7}+1=128.

Answer: D.

gettinit wrote:
this strategy should work for most of these types of problems. I have never seen an exclusive scenario.


The above approach will work for both inclusive and exclusive cases, which is also.
hirendhanak wrote:
gettinit wrote:
this strategy should work for most of these types of problems. I have never seen an exclusive scenario.

Also to figure out those 3 digit numbers divisible by 35 I would just use the same formula.


Thanks for your reply, i understand that i can find the nos with the same formula

But i wish to understand whethr the same is possible with set theory.. because the q could be asked as a DS question in some different format also


What do you mean by the red part?

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Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink] New post 02 Jan 2011, 00:13
Bunuel wrote:
How many of the three-digit numbers are divisible by 7?
(A) 105
(B) 106
(C) 127
(D) 128
(E) 142

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

For example: how many multiples of 5 are there between -7 and 35, not inclusive?

Last multiple of 5 IN the range is 30;
First multiple of 5 IN the range is -5;

\frac{30-(-5)}{5}+1=8.

OR:
How many multiples of 7 are there between -28 and -1, not inclusive?
Last multiple of 7 IN the range is -7;
First multiple of 7 IN the range is -21;

\frac{-7-(-21)}{7}+1=3.

Back to the original question:

Last 3-digit multiple of 7 is 994;
First 3-digit multiple of 7 is 105;

So # of 3-digt multiples of 7 is \frac{994-105}{7}+1=128.

Answer: D.


Hi Bunuel,

I'm a little confused. The general formula above, does it give us the number of integers inclusive of the upper and lower limits or non inclusive of the limits?

For example: what is the number of integers between 1 and 5, inclusive? The answer is 5, Which is also substantiated by the formula above.

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Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink] New post 02 Jan 2011, 01:38
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gmatpapa wrote:
Hi Bunuel,

I'm a little confused. The general formula above, does it give us the number of integers inclusive of the upper and lower limits or non inclusive of the limits?

For example: what is the number of integers between 1 and 5, inclusive? The answer is 5, Which is also substantiated by the formula above.


You should read the solution carefully: it doesn't matter whether the the range is inclusive or exclusive, the formula gives you the correct answer in any case.

How many multiples of 5 are there between 5 and 35, not inclusive?

Last multiple of 5 IN the range is 30 (not 35 but 30 as 35 is not IN the range);
First multiple of 5 IN the range is 10 (not 5 but 10 as 5 is not IN the range);

\frac{30-10}{5}+1=5: 10, 15, 20, 25 and 30.


How many multiples of 5 are there between 5 and 35, inclusive?

Last multiple of 5 IN the range is 35;
First multiple of 5 IN the range is 5;

\frac{35-5}{5}+1=7: 5, 10, 15, 20, 25, 30 and 35.

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Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink] New post 02 Jan 2011, 10:17
Expert's post
A quick demonstration of the formula and why it is intuitive:

How many numbers are there from 3 to 7 (inclusive)?

3, 4, 5, 6, 7 i.e. 5 numbers
When we do (7 - 3 = 4), we remove 3 from the list.
I I I I I I I : 7 sticks
[strike]I I I[/strike] I I I I : (7 - 3 = 4) sticks
The third stick has been removed but we need it to be included too. So we add 1 to the result 4 + 1 = 5 sticks

These 5 sticks are: I I I I I

Now, how many numbers are there from 3 to 7 (exclusive)?

4, 5, 6 i.e. 3 numbers

When we do (7 - 3 = 4), we still have 7 in the list.
I I I I I I I : 7 sticks
[strike]I I I[/strike] I I I I : (7 - 3 = 4) sticks
The seventh stick is still there which has to be removed. So we subtract 1 from the result 4 - 1 = 3 sticks

These 3 sticks are: I I I

or you can think of the exclusive case as (inclusive case - 2)
You say in inclusive case: (Last term - First term + 1): Both end terms are included here.
Then you subtract 2 to remove both the end terms and you get: (Last term - First term + 1) - 2 = Last term - First term - 1

I have also explained this in the post: http://gmatquant.blogspot.com/2010/11/four-prongs.html
Check out the subtraction part.

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Re: M01 Question 18 [#permalink] New post 04 Feb 2011, 10:27
Bunuel wrote:
jayasimhaperformer wrote:
a 7*15=105
b 106/7=15.1
c 127/7=18.1
d 128/7=18.12 aprox
e 142/7=20.--- aprox
so a is right & a straight ans as its div by 7*15 :P


Hi, and welcome to Gmat Club.

It seems that you misinterpreted the question.

The question is "How many of the three-digit numbers are divisible by 7?" It's not necessary the answer itself to be divisible by 7

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

For example: how many multiples of 5 are there between -7 and 35, not inclusive?

Last multiple of 5 IN the range is 30;
First multiple of 5 IN the range is -5;

\frac{30-(-5)}{5}+1=8.

OR:
How many multiples of 7 are there between -28 and -1, not inclusive?
Last multiple of 7 IN the range is -7;
First multiple of 7 IN the range is -21;

\frac{-7-(-21)}{7}+1=3.

Back to the original question:

Last 3-digit multiple of 7 is 994;
First 3-digit multiple of 7 is 105;

So # of 3-digt multiples of 7 is \frac{994-105}{7}+1=128.

Answer: D.

Hope it helps.


Hello Bunuel, nice explanation as usual :)
is there any shortcut to find last and first multiples? for example how to find 994 ad 105 of 7?
or we have to use the long method of division?
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Re: M01 Question 18 [#permalink] New post 04 Feb 2011, 11:03
Bunuel wrote:
tinki wrote:
Hello Bunuel, nice explanation as usual :)
is there any shortcut to find last and first multiples? for example how to find 994 ad 105 of 7?
or we have to use the long method of division?
thx 4 response


Well, it depends multiple of which number you want to find and the range you are looking in. Common sense, and divisibility rules should help you in this but sometimes 'trial and error' is also a good method. Check for example divisibility rules here: math-number-theory-88376.html



i did not know about 7 and 11 . nowhere seen the rule before.
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Re: M01 Question 18 [#permalink] New post 22 Sep 2011, 05:27
hockeyplaya182 wrote:
1) Take all numbers between 1 and 999 divisible by 7: 999/7=142
2) Subtract the number that are not three digits: 99/7=14

142-14=128


Great explanation. Thanks.
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Re: M01 Question 18 [#permalink] New post 22 Sep 2011, 09:07
Here is a simple Solution :

The count of numbers<=999 divisible by seven = rounded[999/7] = 142

The number of 2 digit numbers divesible by 7 = rounded[99/7] = 14

The number of 3 digit numbers divisible by 7 = 142 - 14 = 128
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Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink] New post 22 Sep 2011, 09:30
james12345 wrote:
I'm kept my lame questions to a minimum. I see why plus 1 work for inclusive. I don't see for exclusive why minus 1 would work. If both of the end number "excluded" are actually multiples of the number in question it changes it. but that should be the case for inclusive as well. in other words, how many multiples of 7 between 8 and 22 exclusive means 9 to 21 which means the answer is 2. 21-9 is 12 which divided by 7 is 1 so plus 1 = 2. however 7 to 28 exclusive means 8-27 = 19 which is 2 7s which is actually correct.

It makes me very nervous to apply formulas without being crystal clear as to why they're right. Maybe I've misunderstood part of the premise of the question.



The key point is that in a simple subtraction, we are acting inclusive in one side and exclusive in the other side. For example, when we say 10-6=4, we have not considered 6 itself (exclusive in one side) but we have considered 10 (inclusive in the other side). This is the rule. So, if you figure that out, you see that when GMAT says exclusive, we have to just exclude the side which was not originally excluded. Since one side is not excluded in a normal subtraction, we just subtract 1, not 2. On the other hand, when GMAT says inclusive, we have already included one side, so we just add 1, not 2, to include the other side.

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Re: M01 Question 18 [#permalink] New post 24 Sep 2012, 04:18
Ans is 128.

14*7 = 98 next will be a 3 digit factor..

and 994 /7 = 142 next will be a 4 digit..

so 142 -14 = 128 :)

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Re: M01 Question 18 [#permalink] New post 24 Sep 2012, 04:22
Total 3 digit nos 100 to 999
among them total digits div.by 7 ={(last digit div.by 7 Minus 1st digit div by 7)/7 }+1
{(994-105)/7}+1 = 128

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Re: M01 Question 18 [#permalink] New post 24 Sep 2012, 07:47
T
How do we find out the largest 3 digit number that is divisible by 7 ?
Well one way is trial and error, but that is not too efficient.
Rather , let's divide 999 by 7 to find the remainder. Subtract the remainder from 999 and voila !

For eg : When 999 is divided by , the remainder is 5. Subtract from 999, the answer is 994.

If m/n leaves "r" as the remainder, then m-r is divisible by n.

What is the logic behind this ? Well when you divided a number by another divisor, the remainder represents the piece that is not divisible by the divisor. Remove
the offending remainder from the quotient and the new quotient will be divisible by the divisor

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Re: M01 Question 18 [#permalink] New post 27 Feb 2013, 02:54
bb wrote:
Is something not clear in the Official Explanation?



Approach One: Divide all of the three-digit numbers 999 - 100 + 1 = 900 (Don't forget to add 1 to get the number of all the 3-digit numbers) by 7, which is 128.57, and then round it off to 128.

Approach Two:
Find the first and the last multiples of 7 WITHIN the range (which is between 105 and 994). Find the positive difference, divide by 7, and add 1:
\frac{994-105}{7}+1=128


Hi,
I don't understand how and why approach 1 works. Please help explain why it works and provide some examples.

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Re: M01 Question 18 [#permalink] New post 23 Sep 2013, 07:34
bb wrote:
Is something not clear in the Official Explanation?

Approach One: Divide all of the three-digit numbers 999 - 100 + 1 = 900 (Don't forget to add 1 to get the number of all the 3-digit numbers) by 7, which is 128.57, and then round it off to 128.

Approach Two:
Find the first and the last multiples of 7 WITHIN the range (which is between 105 and 994). Find the positive difference, divide by 7, and add 1:
\frac{994-105}{7}+1=128


Be careful using approach one because it can lead to wrong answers. It is best used as an approximation because it will be within 1 of the correct answer.

How many two-digit numbers that start with 3 are divisible by 7: 39 - 30 + 1 = 10. Divide by 7 to get 1.43. Round down to 1 (Correct because we have 35 only.)
How many two-digit numbers that start with 4 are divisible by 7: 49 - 40 + 1 = 10. Divide by 7 to get 1.43. Round down to 1 (Incorrect because we have 42 and 49.)
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Re: M01 Question 18 [#permalink] New post 23 Sep 2013, 19:05
How many of the three-digit numbers are divisible by 7?

(A) 105
(B) 106
(C) 127
(D) 128
(E) 142


Firstly,find the first 3 digit number that is divisible by 7 ie 105 and the last 3 digit number that is divisible by 7 is 994.

so, series become like this:

105,112,.....................994.

l = a + (n-1) d

994 = 105+(n-1)7
994-105+7=7n
n=896/7=128

so, right answer is

128.
Re: M01 Question 18   [#permalink] 23 Sep 2013, 19:05
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