How many 3 digit positive integers exist that when divided

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

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?

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?

ouch...I ignored the 103 number...
my approach differ from others...
i started with finding the multiples of 5...
lowest one 105 or 15*7
greatest one 994 or 142*7
142-15 = 127.
for each multiple of 7, we can add 5, and get a number that when divided by 7, will yield a remainder of 5.
since 994 = 999-5, 999 works too.
127+1 = 128.
now..if we take into consideration 103, then we have 129.

let x+1=number of 3 digit integers leaving remainder of 5
lowest integer=103
highest integer=999
103+7x=999
x=128
x+1=129
E.

Define the SET

S = { 103 , 110 , 117.......992 , 999 }

No of digits will be \(\frac{( 999 - 103 )}{7} +1\) = 129

Hence correct answer will be (A) 129

Answer must be (E) 129, please post your approach to identify the part where you went wrong..

I already found it. I just divided 1000 by 7 thinking all multiples of 7 plus 5 will give me the answer (also subtracted one at the end I think). However, we only need the 3 digit numbers. I always make mistakes like this


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Is this a feasible solution?: a 3 digit number can either leave a remainder of 0 / 1 / 2 / 3 / 4 / 5 or 6 when divided by 7. So the answer will be 1/6th of 900 (3 digit numbers) = 128 + 1 (for 999) = 129

I guess yes it is.
But you have to be really comfortable with what you are doing. Chances are less that such an approach will strike you during the exam but if you're confident then it should work.

Also if you don't add that 1 you're still boomed.

Thank you!

I suggest another solution.
Because the 3-digit positive integers divide 7 get remainder of 5, so they will follow this term: 7x+5 (a)
3-digit positive integer will run from 100 to 999 (b)
From (a) and (b), we will have: 100 ≤ 7x+5 ≤ 999
Solve the inequality: 13.5 ≤ x ≤ 142 => x will run from 14 to 142 => there are (142 - 14 + 1) = 129 numbers
Ans: E

Let's first determine the smallest 3 digit positive integer that leaves a remainder of 5 when divided by 7. If we divide 100 by 7, we get a quotient of 14 and a remainder of 2. Thus, if we divide 100 + 3 = 103 by 7, we will get a remainder of 5. This is the smallest three digit integer to leave a remainder of 5 when divided by 7.

Next, we determine the largest 3 digit positive integer that leaves a remainder of 5 when divided by 7. To do that, we divide 1000 by 7. We will get a quotient of 142 and a remainder of 6. Since 1000 leaves a remainder of 6, 1000 - 1 = 999 will leave a remainder of 5 when divided by 7.

Finally, we need to determine the number of multiples of 7 between 103 and 999, inclusive. To do that, we can use the formula number of multiples = 1 + [(largest multiple - smallest multiple)/common difference]. Here, the largest multiple is 999, the smallest multiple is 103, and the common difference is 7. We get:

\(\Rightarrow\) 1 + [(largest multiple - smallest multiple)/common difference]

\(\Rightarrow\) 1 + [(999 - 103)/7]

\(\Rightarrow\) 1 + 896/7

\(\Rightarrow\) 1 + 128 = 129

Answer: E

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