How many three-digit positive integers are there such that when divide

Question

How many three-digit positive integers are there such that when divided by 7 or 8 give a remainder of 4?

A. 15
B. 16
C. 17
D. 18
E. 19

CrackverbalGMAT · Answer

LCM of 7 and 8 = 56

Smallest 3 digit number (> 100) divisible by 56 = 112

To get a remainder of 4, the number we need is 112 + 4 = 116

Largest 3 digit number divisible by 56 is 952. To get a remainder of 4, the number is 952 + 4 = 956

The number of multiples of 56 =  \frac{Largest \space Number - Smallest \space Number}{Divisor} + 1 = \frac{956 - 116}{56} + 1 = \frac{840}{56} + 1 = 15 + 1 = 16

Option B

Arun Kumar

rocky620 · Answer

LCM of 7 & 8 is 56

We are looking for a 3 digit number, which, when divided by 56 give leave a remainder of 4. SO it will be of a form 56x + 4, in which x is an integer.

The smallest such number is 112+4=116 = (56*2)+4
The largest such number is 952+4=956 = (56*17)+4

So the total such numbers will be 17-2+1 = 16

Answer Option B

incomunity · Answer

This is officially my first post! Hello to everybody!
I have quite a different answer which can be wrong, please correct me.
LCM 7 & 8 is 56. 
3 digit --> Min 100
 Max 999

So : 999/56 = 17,8 
 100/56 = 1,7

17,8-1,7 =~ (almost) 16.

Ans B.

CrackverbalGMAT · Answer

Hi Incommunity. Nothing wrong in the answer you've got, because the smallest and the largest values (100 and 999) are both not divisible by 56.

Suppose the same question had asked how many multiples of 56 are there   from   112 till 952 and you had done it as 952/56 - 112/56, then your answer would have come as 15, which would be incorrect, in which case you would want to use the counting principle as shown.

Hope this helps

Arun Kumar

NoobGmatTaker · Answer

Bunuel 
It says divisible by 7 or 8 => (/ by 7 having rem as 4) Union (/ by 8 having rem as 4)
Which means divisible by 7 having rem as 4 + divisible by 8 having rem as 4 - divisible by 56 having rem as 4

Is my understanding incorrect

adewale223 · Answer

The solutions I think are for 7 and 8. 7 or 8 means 7 union 8.

Posted from my mobile device

Kinshook · Answer

How many three-digit positive integers are there such that when divided by 7 or 8 give a remainder of 4?

Let the three-digit positive integers be x.

x = 7k + 4 = 8k` + 4
7k = 8k`
k` = 7k/8 ; Since k` is a positive integer, k is a multiple of 8

100 < x = 7k + 4 < 1000
96 < 7k < 996
16 <= k <= 136
14 <= k`<= 119

x = {116, ...956}; It is an arithmetic series with common different = LCM(7,8) = 56

Number of three-digit positive integers = (956-116)/56+1 = 16

IMO B

How many three-digit positive integers are there such that when divide

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How many three-digit positive integers are there such that when divide

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