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How many 3 digit positive integers exist that when divided

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How many 3 digit positive integers exist that when divided [#permalink] New post 29 Aug 2011, 23:56
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How many 3 digit positive integers exist that when divided by 7 leave a remainder of 5?

A. 128
B. 142
C. 143
D. 141
E. 129
[Reveal] Spoiler: OA
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Re: divisibility by 7 [#permalink] New post 30 Aug 2011, 02:29
First three digit number which when divided by 7 leaves remainder 5 is 103. here's how.
14*7=98
98+5=103.
next number will be (15*7+5=110). next will be (16*7+5=117)
Sequence 103,110,117,124,......992,999
Last term=first term + (n-1)common difference
999=103+(n-1)7
999-103=(n-1)7
896/7=n-1
128+1=n
n=129.

OA E

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Re: divisibility by 7 [#permalink] New post 01 Sep 2011, 19:08
Very Triky Question.

Minimum three digit number is 100 and maximum three digit number is 999.
The first three digit number that leaves remainder 5 when divided by 7 is 103.
14 * 7 = 98 +5 = 103
The second three digit number that leaves remainder 5 when divided by 7 is 110.
15 * 7 = 105 +5 =110
The third three digit number that leaves remainder 5 when divided by 7 is 117
and so on

The last three digit number that leaves remainder 5 when divided by 7 is 999
142 * 7 = 994 + 5 = 999

Therefore, we identify the sequence
103,110,117.....999

use the formula of last term
Last term = first term + (n - 1) * common difference

you will get the answer 129 that is definitely E.
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Re: divisibility by 7 [#permalink] New post 01 Sep 2011, 21:09
1000/7 = 142 is the result with 6 as remainder
100/7= 14 with 2 as remainder

so 15*7=105 is the first three digit # which is divisible by 7 and 994 is the last three digit #. So the total three digit # divisible by 7 = 142-14-1 = 129

and as 994+ 5 =999 is a three digit #

so the # of three digit # which after divided by 7 leave a remainder of 5 are 129
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How many 3 digit positive integers exist that when divided [#permalink] New post 22 Aug 2013, 10:56
How many 3 digit positive integers exist that when divided by 7 leave a remainder of 5?

A. 128
B. 142
C. 143
D. 141
E. 129

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Re: How many 3 digit positive integers exist that when divided [#permalink] New post 22 Aug 2013, 11:00
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Re: How many 3 digit positive integers exist that when divided [#permalink] New post 22 Aug 2013, 11:14
Used sheer calculations.

3 digit no. divided by 7 leaves remainder 5
xyz=7a+5
Started with a =13, xyz=096
for a=14, xyz=105
therefore 1st value of a is 14.
now for highest value of a, started with a =140 => xyz=985
for a=141 => xyz=992
for a=142 => xyz=999
therefore highest value of a is 142
that makes the no. of values for which xyz remains 3 digit no. are - 142-14+1 = 129

What is the simplest way to solve this?
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Re: divisibility by 7 [#permalink] New post 14 Oct 2013, 16:48
jamifahad wrote:
First three digit number which when divided by 7 leaves remainder 5 is 103. here's how.
14*7=98
98+5=103.
next number will be (15*7+5=110). next will be (16*7+5=117)
Sequence 103,110,117,124,......992,999
Last term=first term + (n-1)common difference
999=103+(n-1)7
999-103=(n-1)7
896/7=n-1
128+1=n
n=129.

OA E


Can someone explain what this means, and where I can read up on this concept? I have never run into it
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Re: divisibility by 7 [#permalink] New post 15 Oct 2013, 08:55
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AccipiterQ wrote:
jamifahad wrote:
First three digit number which when divided by 7 leaves remainder 5 is 103. here's how.
14*7=98
98+5=103.
next number will be (15*7+5=110). next will be (16*7+5=117)
Sequence 103,110,117,124,......992,999
Last term=first term + (n-1)common difference
999=103+(n-1)7
999-103=(n-1)7
896/7=n-1
128+1=n
n=129.

OA E


Can someone explain what this means, and where I can read up on this concept? I have never run into it


For arithmetic progression if the first term is a_1 and the common difference of successive members is d, then the n_{th} term of the sequence is given by: a_ n=a_1+d(n-1).

For more check here: math-number-theory-88376.html

Hope it helps.

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Re: divisibility by 7 [#permalink] New post 15 Oct 2013, 12:22
Bunuel wrote:
AccipiterQ wrote:
jamifahad wrote:
First three digit number which when divided by 7 leaves remainder 5 is 103. here's how.
14*7=98
98+5=103.
next number will be (15*7+5=110). next will be (16*7+5=117)
Sequence 103,110,117,124,......992,999
Last term=first term + (n-1)common difference
999=103+(n-1)7
999-103=(n-1)7
896/7=n-1
128+1=n
n=129.

OA E


Can someone explain what this means, and where I can read up on this concept? I have never run into it


For arithmetic progression if the first term is a_1 and the common difference of successive members is d, then the n_{th} term of the sequence is given by: a_ n=a_1+d(n-1).

For more check here: math-number-theory-88376.html

Hope it helps.



Thank you, are there other problems like this one to practice on?
Re: divisibility by 7   [#permalink] 15 Oct 2013, 12:22
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