How many integers less than 1000 have no factors (other than 1) in com

Question

How many positive integers less than 1000 have no factors (other than 1) in common with 1000 ?

A. 400
B. 401
C. 410
D. 420
E. 421

Bunuel · Accepted Answer

First of all it should be "how many  positive integers  less than 1000 have no factors (other than 1) in common with 1000", as if we consider negative integers answers will be: infinitely many.

1000=2^3*5 ^3  so basically we are asked to calculate the # of positive integrs less than 1000, which are not multiples of 2 or/and 5.

Multiples of 2 in the range 0-1000, not inclusive -  \frac{998-2}{2}+1=499 ;
Multiples of 5 in the range 0-1000, not inclusive -  \frac{995-5}{5}+1=199 ;
Multiples of both 2 and 5, so multiples of 10 -  \frac{990-10}{10}+1=99 .

Total # of positive integers less than 1000 is 999, so # integers which are not factors of 2 or 5 equals to  999-(499+199-99)=400 .

Answer: A.

gmacforjyoab · Answer

consider integers between 1 and 100 - half of them are even - hence 50 integers are multiples of 2 ( which also includes even multiples of 5) + 10 odd multiples of 5 = 60
Hence 40 integers that are not multiples of 2 and/or 5 -
hence considering integers between 1 and 1000 - there are 40*10 = 400 integers which do not have common multiple with 1000 other than 1.

nonameee · Answer

Bunuel, why can't we simply divide 1000 by 2 to find the number of multiples of 2? My reasoning is that every second number is a multiple of 2 so there must be exactly 500 numbers.

Thanks.

Bunuel · Answer

There are 100/2=500 multiple of 2 in the range 1-1000 INCLUSIVE. As we need numbers LESS than 1000 which are also multiples of 2 then we should subtract 1 from that number. So there are total of 500-1=499 multiples of 2 in the range 0-1000, not inclusive.

rohantiwari · Answer

The question asks for the number of integers less than 1000 and other than 1.
Isnt one included in the 400 integers that you are claimimg to be the answer?
Answer should be 399 if we exclude 1.
Please correct me in case i missed something.

KarishmaB · Answer

The question does not ask you to exclude 1.

Every positive integer less than 1000 has one common factor with 1000. What is it? It is 1. 
1 is a common factor between any two positive integers.

If the question were: How many positive integers less than 1000 have no factors in common with 1000 ?
Then the answer would be 0. There are no positive integers which have no common factors with 1000. All the positive integers have a common factor and that is 1. But the question wants to know the number of positive integers which have no common factor other than 1 (1 will always be a common factor). Basically, it is looking for positive integers which are co-prime with 1000.

sumitchawla · Answer

What about the prime numbers Bunuel ?? For ex : 7. Neither its a multiple of 2, nor 5 and it does not has any common factors with 1000 (except 1)
So, shouldn't the answer include prime numbers between 1-999 as well. And if YES, how do we calculate the number of primer numbers from 1-999 ???
Plz clarfily.

Thanks.

Bunuel · Answer

We counted multiples of 2 or 5 in the range 0-1000, not inclusive and then subtracted that from total number of integers in the range 0-1000. The number we get contains all numbers which are not multiples of 2 or 5, thus all primes (apart from 2 and 5) in that range too.

Hope it's clear.

JeffTargetTestPrep · Answer

Since 1,000 breaks down to prime factors of twos and fives, we need to find all the numbers less than 1,000 that do not contain those factors. To do so, let’s find all the numbers less than 1000 that contain factors of two’s and five’s. Note that all even numbers (multiples of 2) and all multiples of 5 must be accounted for.

Number of even numbers less than 1000:

(998 - 2)/2 + 1 = 499

Number of multiples of five less than 1000:

(995 - 5)/5 + 1 = 199

We must find the double-counted numbers, also called overlap numbers, which are numbers that are multiples of both 2 and 5. To find the overlap, we need to determine the number of multiples of 5 and 2 (or of 10) less than 1000:

(990 - 10)/10 + 1 = 99

Thus, the number of multiples of 2 or multiples of 5 less than 1000 is:

499 + 199 - 99 = 599

Finally, the number of numbers less than 1000 that ARE NOT multiples of 2 or 5 is:

999- 599 = 400

Answer: A

Terabyte · Answer

The answer is pretty simple, in my opinion. You don't need almost any calculations. 1000 has only 2 distinct factors - 2 and 5. It is obvious that there are 500 odd and even numbers from 1 to 1000. Exclude 1000 (even number) you will have 500 odd numbers and 499 even numbers. And there are 200 multiples of 5 between 1 and 1000 (1000/5). 100 of them are odd numbers (5, 15, 25 etc.) and 100 of them are even numbers (10, 20, 30 etc.). Exclude 1000 - you will have 100 and 99 respectively. (p.s. you don't even need to exclude 1000 since we will not count 1000 anyway).

Hence, we will just need to exclude odd multiples of 5 from a list of odd numbers from 1 to 999. We need only yellow space

https://image.ibb.co/bv2rwa/Untitled.png

harshbirajdar · Answer

Hey Guys,

If I solve this using the method of combinations, can someone please tell me where am I going wrong?

We have 3 spots to fill. The nos. cannot be divisible by 2, 5 or 10. The first and the second spots can have any values. The last spot can have 4 values (1, 3, 7, 9). As this will contain the number 001, we will subtract 1 from the count.

According to my method the answer is coming out to be (10*10*4) - 1 = 399. What am I missing?  VeritasKarishma  if you could give in your 2 cents too.

Thanks!

KarishmaB · Answer

The only thing you are missing is that you do not have to subtract out 1. 1 is acceptable to us.
The only factor that 1 has in common with 1000 is 1. The question stem allows 1 to be common.

"How many integers less than 1000 have no factors  (other than 1)  in common with 1000 ?"

Note that every one of the other 399 numbers that are acceptable to you also have 1 common with 1000.

Bambi2021 · Answer

This should be all the odd numbers from 1-999 minus the odd multiples of 5:

Odd numbers from 1-999 = 500
Odd multiples of 5 from 5-995 = ten per hundred = 10*10 = 100

500-100 = 400

Kinshook · Answer

Asked: How many integers less than 1000 have no factors (other than 1) in common with 1000 ?

1000 = 2^3*5^3
All even numbers are not eligible
All multiple of 5 are not eligible

Even numbers = (999-2)/2 + 1 = 499
Mulitiple of 5 = (995-5)/5 + 1 = 199
Even numbers and multiple of 5 = (990-10)/10 + 1 = 99

Number of integers less than 1000 have no factors (other than 1) in common with 1000 = 999 - (499+199-99) = 400

IMO A

mmdfl · Answer

I wrote down the numbers and saw the pattern:
 1  2  3  4 5 6  7  8  9  10
 11  12  13  14 15 16  17  18  19  20
 21  22  23  24 25 26  27  28  29  30
...  999

Each row has 4 of each 10 numbers not multiple of 2 or 5. So (4/10)*1000 = 400. => A)

ocelot22 · Answer

1000 is made up of prime factors 2 and 5. so any multiples of prime factors that dont include 2 and 5 work. Now, excluding 1 between 2 and 999
we have 999-2 + 1 = 998 integers. One could divide this difference by 3,7,9,11,..... but is too time consuming. Instead, we can count those integers that are divisible by 2 or 5 and then subtract from 998. So, the number of integers divisible by 2 or 5 = # divisible by 2 + # divisible by 5 - number divisible by 2 and 5, or 10

998|2 gives 499. 998| 5 gives 199. 998|10 gives 99. 499+199-99 = 599. 998 - 599+1 = 400. OE is A

RushabhD · Answer

The way I did it was:

Consider the integers to be "three-digit numbers" (counting 0 as a digit).

The integers cannot have 2 or 5 as factors. So the integer cannot end with an even number or with 5.

The first digit (or the hundredth place) can be any number from 0-9 i.e 10 possible numbers
The second digit (or the tenth place) can be any number from 0-9 i.e. 10 possible numbers
The third digit (or the unit place) can be 1, 3, 7, 9 i.e. 4 possible number

The number of integers that can be formed with these conditions are 10 x 10 x 4 = 400 (Answer)

Since we consider 0 as a digit even if placed in the hundredth or tenth place as well, one and two digit numbers also get accounted for.

Nipunh · Answer

1000=10^3=(2x5)^3
we need to basically eliminate 2,5 factors from the list of 1st 1000 no. excluding 1000.

lets eliminate even i.e no of odd terms in 1-999 is 500
ok now , let eliminate factors of 5 (we know since we have odd no.s only, we have 1,3,5,7,9 ..
we need to remove 5 from every place possible.. now possible unit digits are 1,3,7,9 i.e 4/5 of odd no.s
4/5 * 500 = 400 ans

whompp7 · Answer

Solved this using the fundamental counting principle.
We have to find out multiples of 2 and 5 under 1000;

If number is ABC, then number of digits multiple of 2 and 5 equals:

10 x 10 x 6

A: 1 to 10
B: 1 to 10
C: 2, 4, 5, 6 8, 0

Resultant 600 is to be subtracted from 999 to give 399. Add 1 because we have to count 1 as well. (thanks @KarishmaB!)
399+1= 400

Ans. A

How many integers less than 1000 have no factors (other than 1) in com

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