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Divisibility Problem

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by cans » Mon Sep 12, 2011 8:38 am
108,114,120,....750
6*18,..... 6*125..
Thus number = 125-18 + 1=108
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by neoshichhadva » Mon Sep 12, 2011 8:57 am
Thanks! Similarly, can you help me solve this:

How many 3 digit positive integers exist that when divided by 7 leave a remainder of 5?

a) 128
b) 129
c) 141
d) 142
e) 143
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by shankar.ashwin » Mon Sep 12, 2011 9:14 am
To leave a remainder of 5, the number should be of the form 7k+5.

The minimum value k can take is 14, which gives 7(14)+5=102 (3 digit number)
And the maximum value K can take is 142, 7(142)+5=999.

So k can take any value from 14 to 142 which is 129 values.
neoshichhadva wrote:Thanks! Similarly, can you help me solve this:

How many 3 digit positive integers exist that when divided by 7 leave a remainder of 5?

a) 128
b) 129
c) 141
d) 142
e) 143
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by pemdas » Mon Sep 12, 2011 9:19 am
nothing unusual, common case of divisibility p=i*q+r, where p will be three digit number (any), i is integer, q=7 and r is remainder

now by taking the lowest three digit number divisible by 7 with the remainder of 5 we set
100 + [+3]=14*7 +2 +[3] or 103=14*7 +5 {p=103, i=14, q=7, r=5}

the last three digit number will be 999 => 999= 142*7 +5, hence 142-14+1=129

b
neoshichhadva wrote:Thanks! Similarly, can you help me solve this:

How many 3 digit positive integers exist that when divided by 7 leave a remainder of 5?

a) 128
b) 129
c) 141
d) 142
e) 143
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by saketk » Mon Sep 12, 2011 9:54 am
neoshichhadva wrote:How many numbers between 103 and 750 are divisble by 6?

a) 107
b) 108
c) 113
d) 120
e) 125
The quickest way to solve these type of questions is to use Arithmetic Progression (I hope AP is allowed in GMAT) . I am assuming you are aware of AP.

If not then simple counting that may take some more time, will help.

By AP: -

First term (a) = 108
Last term divisible by 6 is = 750

you can calculate the number of terms using this AP formula:

Tn = a + (n-1)*d; where n = number of terms
d= common difference, in this case it is 6
Tn = last term

plug the values and you will have the answer.

n= [(750-108)/6]+1 = 108
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by navami » Mon Sep 12, 2011 9:55 am
108
This time no looking back!!!
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by cans » Mon Sep 12, 2011 9:56 am
How many 3 digit positive integers exist that when divided by 7 leave a remainder of 5?

a) 128
b) 129
c) 141
d) 142
e) 143
As pemdas and Shankar explained, its similar to divisibility by 6 question..
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by knight247 » Mon Sep 12, 2011 10:08 am
For this we need to first find the biggest and the smallest 3 digit number which when divided by 7 leaves a remainder of 5. 999 when divided by 7 leaves a remained of 5 so it is the biggest. Now when 100 is divided by 7 it leaves a remainder of 2. We have the form D=qd+R where D=Dividend q=Quotient d=Divisor and R=Remainder. When expressing 100 divided by 7 in this form we have
100=q7+2. The quotient q is irrelevant as we don't need that info. However we need a remainder of 5 so adding 3 on both sides we have 103=q7+5 so 103 is the smallest 3 digit number that leaves a remainder of 5. Now, the next 3 digit number that leaves a remainder of 5 is 110 and 117 etc. Do you see a pattern? They are all in increments of 7 starting from 103 until 999

Now, our own expert Mitch Hunt has condensed the resolution to such problems into a very simple formula.

Number of multiples withing a certain range=[(Biggest-Smallest)/Increment]+1

=[(999-103)/7]+1
=[896/7]+1
=128+1
=129 Hence [spoiler]B[/spoiler]
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by Abhishek009 » Mon Sep 12, 2011 10:19 am
neoshichhadva wrote:How many numbers between 103 and 750 are divisble by 6?

a) 107
b) 108
c) 113
d) 120
e) 125

Numbers divisible by 6 are multiples of 6...


Let's find out the numbers divisible by 6 :


6 , 12 , 18 , ................. 750



Now let's find the total no of terms divisible from 1 to 750


Terms divisible till 750 is = > 750/6 = 125



Now let's find the numbers divisible till 103


Terms divisible till 103 by 6 is => 103/6 = 17 numbers....




Now we know out of 125 terms divisible by 6 ( from 1 - 750) there are 17 ( from 1 to 103 ) terms which are divisible by 6 ....

So total no of terms divisible by 6 from 103 to 750 is :

125 - 17 = > 108
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by bijoyajj » Mon Sep 12, 2011 10:25 am
neoshichhadva wrote:How many numbers between 103 and 750 are divisble by 6?

a) 107
b) 108
c) 113
d) 120
e) 125
if the choices are not nearby numbers to each other, then just find the difference (750-103=647)and divide it by 6.. so nearly 108.. use this method only after looking at the choices.
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by GMATGuruNY » Mon Sep 12, 2011 1:24 pm
The solutions above look great. To get more practice with evenly spaced integers, check the following threads:

https://www.beatthegmat.com/for-any-posi ... 59401.html

https://www.beatthegmat.com/consecutive- ... 73388.html

https://www.beatthegmat.com/sum-of-even- ... 85000.html

https://www.beatthegmat.com/depreciation-t88719.html

https://www.beatthegmat.com/how-many-int ... 88553.html

https://www.beatthegmat.com/tricky-one-t22481.html
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