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# M01 Question 18

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M01 Question 18 [#permalink]  01 Jun 2009, 15:51
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How many of the three-digit numbers are divisible by 7?

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

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D

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Re: M01 Question 18 [#permalink]  01 Jun 2009, 16:45
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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:
$$\frac{994-105}{7}+1=128$$
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3 Digit Numbers Divisible by 7 (and others) [#permalink]  03 Aug 2009, 23:02
Hi All,

Here’s a question that came up in M01 on the gmatclub tests.

How many of the 3 digit numbers are divisible by 7?

Solution (provided by gmatclub)

Take the first and last multiples of 7 in the range.

They are 105 and 994.

(994 – 105) / 7 + 1 = 128

Does this technique hold true for all numbers in a range? I’ve tried it with a couple of other examples and it seems to work
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Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]  04 Aug 2009, 02:17
Aztec wrote:
Hi All,

Here’s a question that came up in M01 on the gmatclub tests.

How many of the 3 digit numbers are divisible by 7?

Solution (provided by gmatclub)

Take the first and last multiples of 7 in the range.

They are 105 and 994.

(994 – 105) / 7 + 1 = 128

Does this technique hold true for all numbers in a range? I’ve tried it with a couple of other examples and it seems to work

yes, it applies to all the numbers in the range..
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Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]  10 Aug 2009, 00:00
It should work every time on all the numbers.
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Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]  10 Aug 2009, 01:46
Just to mention,
if it says "excluding" then the formula becomes, (max-min)/divisibility - 1.
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Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]  10 Aug 2009, 01:54
Economist wrote:
Just to mention,
if it says "excluding" then the formula becomes, (max-min)/divisibility - 1.

Great now we have all the rules in one page , thanks
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Re: M01 Question 18 [#permalink]  01 Oct 2009, 20:55
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
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Re: M01 Question 18 [#permalink]  04 Jan 2010, 20:02
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7*15=105
7*142=994

then 142-15+1 = 128
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Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]  10 Jan 2010, 16:25
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.
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Re: M01 Question 18 [#permalink]  20 Sep 2010, 04:52
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105 is the first 3 digit number divisible by 7
994 is the last 3 digit number divisible by 7

(994 - 105)/7 + 1 = 128
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Re: M01 Question 18 [#permalink]  20 Sep 2010, 20:29
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
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Re: M01 Question 18 [#permalink]  20 Sep 2010, 21:55
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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

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 $$\frac{994-105}{7}+1=128$$.

Hope it helps.
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Re: M01 Question 18 [#permalink]  25 Sep 2010, 04:09
The range of 3 digit numbers that are divisible by 7 are 105 to 994 inclusive. If we consider the set of these numbers as in Arithmetic Progression, {105, 112, 119 ... 994}, we can find the number of values in the set using the formula; Last term = First term + (n-1)(Common Difference), which implies, 994 = 105 + (n-1)(7). Solving n = 128
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Re: M01 Question 18 [#permalink]  27 Sep 2010, 03:54
Hi Bunuel nice explanation.
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Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]  28 Dec 2010, 22:23
Can someone solve the same problem with some addtiional details
How many of the 3 digit numbers are divisible by 7?
Nos divisble by 7 = 128
Nos divisible by 5 = 180
Can we find nos divisible by 35 on the basis of above info ( not the traditional formula used for this problem , but using set theory )
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Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]  28 Dec 2010, 22:43
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.
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Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]  28 Dec 2010, 23:59
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
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Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]  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 $$\frac{994-105}{7}+1=128$$.

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]  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 $$\frac{994-105}{7}+1=128$$.

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] 02 Jan 2011, 00:13

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# M01 Question 18

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