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Re: M01 Question 18 [#permalink]
01 Jun 2009, 16:45

4

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Expert's post

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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: \(\frac{994-105}{7}+1=128\) _________________

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

Re: M01 Question 18 [#permalink]
04 Feb 2011, 10:46

1

This post received KUDOS

Expert's post

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 _________________

Q: How many 3 digits numbers are divisible by 7 OR How many multiples of 7 are there between 100 and 999, inclusive?

Choose the first number greater than equal to 100 that is divisible by 7. Remainder of 100/7 is 2. 100-2+7=105 "105" is the first number greater than equal to 100 that is divisible by 7.

Choose the first number less than equal to 999 that is divisible by 7. Remainder of 999/7 is 5. 999-5=994 "994" is the first number less than equal to 999 that is divisible by 7.

\(n=\frac{994-105}{7}+1=127+1=128\)

Ans: "D"

************************ This rules hold good for all numbers.

Find the numbers divisible by 3 between 1 and 18, inclusive.

First Number greater than equal to 1 that is divisible by 3 = 3 First Number less than equal to 18 that is divisible by 3 = 18

\(Number of multiples = \frac{18-3}{3}+1=5+1=6\)

Validate: Count of {3,6,9,12,15,18} = 6

Try it with any number and you should get the count. _________________

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.

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 _________________

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 ) _________________

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 _________________