How many 3 digit positive integers exist that when divided

Question

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

jamifahad · Accepted Answer

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

bholakc · Answer

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.

jitbec · Answer

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

crackjack · Answer

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?

AccipiterQ · Answer

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

Bunuel · Answer

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.

AccipiterQ · Answer

Thank you, are there other problems like this one to practice on?

mvictor · Answer

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.

gracie · Answer

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.

Abhishek009 · Answer

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

Abhishek009 · Answer

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

ProfX · Answer

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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shouri · Answer

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

TheNightKing · Answer

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!

Trang Do · Answer

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

ScottTargetTestPrep · Answer

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 +  . Here, the largest multiple is 999, the smallest multiple is 103, and the common difference is 7. We get:

\Rightarrow  1 +

\Rightarrow  1 + 896/7

\Rightarrow  1 + 128 = 129

Answer: E

sarthak1701 · Answer

128 is a trap choice, the first 3-digit that'll leave the remainder 5 is 103 and not 110.

a1 = 103

d = 7

an = 999 (994 +5)

an = a1 + (n-1)d

999 = 103 + 7n - 7

Solving for n we get n = 129

Ans E

How many 3 digit positive integers exist that when divided

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How many 3 digit positive integers exist that when divided

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