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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 _________________

My dad once said to me: Son, nothing succeeds like success.

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

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

Re: How many 3 digit positive integers exist that when divided [#permalink]

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22 Aug 2013, 12:14

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

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

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: \(a_ n=a_1+d(n-1)\).

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: \(a_ n=a_1+d(n-1)\).

Re: How many 3 digit positive integers exist that when divided [#permalink]

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18 Jun 2016, 02:33

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