How many numbers between 103 and 750 are divisble by 6?
a) 107
b) 108
c) 113
d) 120
e) 125
Divisibility Problem
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108,114,120,....750
6*18,..... 6*125..
Thus number = 125-18 + 1=108
6*18,..... 6*125..
Thus number = 125-18 + 1=108
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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
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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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.
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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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
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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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.neoshichhadva wrote:How many numbers between 103 and 750 are divisble by 6?
a) 107
b) 108
c) 113
d) 120
e) 125
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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As pemdas and Shankar explained, its similar to divisibility by 6 question..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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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]
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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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
Abhishek
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.neoshichhadva wrote:How many numbers between 103 and 750 are divisble by 6?
a) 107
b) 108
c) 113
d) 120
e) 125
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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
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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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
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As a tutor, I don't simply teach you how I would approach problems.
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For more information, please email me (Mitch Hunt) at [email protected].
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