Coprime.

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by shovan85 » Fri Oct 29, 2010 10:25 am
IMO 160.

Number of factors of [600 = 2^3 * 3 * 5^2] = 24

Just listing the beginning 1,2,3,4,5,6,8,10,12..... 24 factors

Outta these 2, 3, and 5 are Prime. So rest other factors can be multiple of these 3 primes.

We need to find the co-primes (No which does not have a common factor with 600)

All multiples of 2 will have a common factor with 600. So 2*300 = 600 We can discard 300
Remaining 600-300 = 300

All multiples of 3 will have a common factor with 600. So 3*200 / 2 = 100 discarded (divided by 2 because we have already taken care of multiple of 2, This only includes 3,9,15, etc. but not 6,12,...)
Remaining 300-100 = 200

All multiples of 5. I could not find way for this :(
discard 5,25,35,55,65,85,95
105,115,145,155,175,185 = 13 in 200 then 39 in 600.

Thus 200 -39 = 161.

I think 600 has been included in 161. So discard that 160.

Not sure ;)
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by Rahul@gurome » Fri Oct 29, 2010 10:50 am
goyalsau wrote:How many numbers are less than and prime to 600?

120
140
160
180
200
The solution provided by shovan85 is correct and easy to understand.
May be this is irrelevant to this problem, but there is a standard method to calculate the number of relatively prime numbers less than a given number. This method is helpful for larger numbers.

The method is as follows:
  • 1. Find the exponential prime factorization of the number.
    2. Taking each term separately, change the term to 2 numbers:
    • a. Subtract 1 from the base for the first number.
      b. Subtract 1 from the exponent and evaluate the expression for the second number. (If the exponent is 1, the second number is always 1.)
    3. Multiply all the numbers together found in step 2.
The method gives rise to a formula which is also known as Euler's Totient Function. Anyone interested in the proof may refer to: https://www.rohitn.com/numbers/numbers_EulerTotient.aspx.

For this problem,
600 = (2^3)*(3^1)*(5^2)
  • For 2 : First number = 2 - 1 = 1; Second number = 2^(3 - 1) = 4
    For 3 : First number = 3 - 1 = 2; Second number = 1 (As exponent is 1)
    For 5 : First number = 5 - 1 = 4; Second number = 5^(2 - 1) = 5
Therefore, number of integers less than and coprime to 600 = [(1)*(4)]*[(2)*(1)]*[(4)*(5)] = 160.

The correct answer is C.
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by shovan85 » Fri Oct 29, 2010 11:10 am
Rahul@gurome wrote:
goyalsau wrote:How many numbers are less than and prime to 600?


120
140
160
180
200
The solution provided by shovan85 is correct and easy to understand.
May be this is irrelevant to this problem, but there is a standard method to calculate the number of relatively prime numbers less than a given number. This method is helpful for larger numbers. The method is as follows:
  • 1. Find the exponential prime factorization of the number.
    2. Taking each term separately, change the term to 2 numbers:
    • a. Subtract 1 from the base for the first number.
      b. Subtract 1 from the exponent and evaluate the expression for the second number. (If the exponent is 1, the second number is always 1.)
    3. Multiply all the numbers together found in step 2.
For this problem,
600 = (2^3)*(3^1)*(5^2)
  • For 2 : First number = 2 - 1 = 1; Second number = 2^(3 - 1) = 4
    For 3 : First number = 3 - 1 = 2; Second number = 1 (As exponent is 1)
    For 5 : First number = 5 - 1 = 4; Second number = 5^(2 - 1) = 5
Therefore, number of integers less than and coprime to 600 = [(1)*(4)]*[(2)*(1)]*[(4)*(5)] = 160.

The correct answer is C.
This is a quick approach. Thanks.

However, can you please help me with this...

The way i found out the multiple of 5 but not multiple 2 and 3. How to do this? I m sure there is a method that would have let me know 39 multiples of 5 are less than 600 but not multiple of 2 and 3.

Please help!!!
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by Rahul@gurome » Fri Oct 29, 2010 11:58 am
shovan85 wrote: All multiples of 5. I could not find way for this :(
discard 5,25,35,55,65,85,95
105,115,145,155,175,185 = 13 in 200 then 39 in 600.

I think 600 has been included in 161. So discard that 160.

Not sure ;)
Though your approach is correct, two things are wrong in your calculation/understanding. One of them includes how to find out the multiple of 5 but not multiple 2 and 3. This can be done using inclusion/exclusion principle of set theory.

Multiple of 2 = 600/2 = 300
Multiple of 3 = 600/3 = 200
Multiple of 5 = 600/5 = 120

Multiple of 2 and 3 = 600/6 = 100
Multiple of 2 and 5 = 600/10 = 60
Multiple of 3 and 5 = 600/15 = 40

Multiple of 2, 3 and 5 = 600/30 = 20

Multiple of only 2 = Multiple of 2 - Multiple of 2 and 3 - Multiple of 2 and 5 + Multiple of 2, 3 and 5 = 300 - 100 - 60 + 20 = 160

In the same manner,
Multiple of only 3 = 200 - 100 - 40 + 20 = 80
Multiple of only 5 = 120 - 60 - 40 + 20 = 40

Others can be calculated in the same manner.

Thus, number of coprimes
= 600 - Multiple of { Only 2, Only 3, Only 5, Only 6, Only 10, Only 15, Only 30}
= 600 - Multiple of { 2, 3 but not 2, 5 but not 2 and 3}

@Shovan85: There are 40 such numbers. Your mistakes are:
  • (1) Including 105 (= 3*5*7, contains 3) and excluding 125(= 5*5*5, doesn't contain 2 or 3).
    (2) Exclusion of numbers like 215(= 5*43), 235(= 5*47) etc. This mistake is due to generalizing the fact that the numbers are uniformly distributed. Whereas this is impossible as this depends upon the distribution of prime numbers, which doesn't follow any known pattern.
    (3) The red line is another mistake. You have excluded 600 when you have discarded the first 300 numbers which is divisible by 2.
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by shovan85 » Fri Oct 29, 2010 8:32 pm
Rahul@gurome wrote: @Shovan85: There are 40 such numbers. Your mistakes are:
  • (1) Including 105 (= 3*5*7, contains 3) and excluding 125(= 5*5*5, doesn't contain 2 or 3).
    (2) Exclusion of numbers like 215(= 5*43), 235(= 5*47) etc. This mistake is due to generalizing the fact that the numbers are uniformly distributed. Whereas this is impossible as this depends upon the distribution of prime numbers, which doesn't follow any known pattern.
    (3) The red line is another mistake. You have excluded 600 when you have discarded the first 300 numbers which is divisible by 2.
Yes!! Thanks a lot. Got it now :)
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by gmatmachoman » Sat Oct 30, 2010 5:28 am
goyalsau wrote:How many numbers are less than and prime to 600?


120
140
160
180
200
This question can be reframed as " How many numbers that are less than 600 and not divisible by 2 or 3 or 5


Formula ( Got frm a friend)

N * ( 1- 1/a) * ( 1- 1/b) * ( 1-1/c)..... where a,b, c are the prime factors of N

SO N = 600 and a=2, b=3, c= 5

Applying the formula : 600 * 1/2 * 2/3 * 4/5
: 160

We can test this formula for small number like N = 10 and not divisible by 2 or 5

10 * ( 1-1/2) ( 1-1/5)

10 * 1/2 * 4/5

= 4

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by shovan85 » Sat Oct 30, 2010 5:51 am
gmatmachoman wrote: This question can be reframed as " How many numbers that are less than 600 and not divisible by 2 or 3 or 5

Formula ( Got frm a friend)

N * ( 1- 1/a) * ( 1- 1/b) * ( 1-1/c)..... where a,b, c are the prime factors of N

SO N = 600 and a=2, b=3, c= 5

Applying the formula : 600 * 1/2 * 2/3 * 4/5
: 160

We can test this formula for small number like N = 10 and not divisible by 2 or 5

10 * ( 1-1/2) ( 1-1/5)

10 * 1/2 * 4/5

= 4
Nice formula!!! Just an addition I tried out now ;)

We can test this formula for small number like N = 10 and not divisible by 2 or 3 or 5

10 * ( 1-1/2) (1-1/3) ( 1-1/5)

10 * 1/2 * 2/3 * 4/5

= 2.66

Then it will be 2 not 3. Means take the largest +ve integer below the fraction.
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by Rahul@gurome » Sat Oct 30, 2010 8:20 am
gmatmachoman wrote:Formula ( Got frm a friend)

N * ( 1- 1/a) * ( 1- 1/b) * ( 1-1/c)..... where a,b, c are the prime factors of N
This formula is same as the one I have mentioned before. Just a rearrangement of terms gives it a simple look! Easy to remember also!

Thanks.
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