answer this question and make a race with me RIGHT NOW! [ROUND 3]

[MrCoolness 547]

Consider a positive integer, a, such that a is between 1 and 100.

 

Defining a function G(a) to be the smallest positive integer that is divisible by each of the positive integers up to 1,2,3,...,a

Example: G(3) = 6, since the lowest common multiple of (1,2,3) is lcm(1,2,3) = 6

Example2: G(5) = 60, since the lcm(1,2,3,4,5) = 60.

 

How many positive integers, a, exist such that G(a) = G(a+4)?

 

 

Explain your answer.

 

STAFF AREN'T ALLOWED TO HELP

[Sean 3760]

Guess I'll write another code.

Sent from my SGH-I747 using Tapatalk

[Mimic 3307]

STAFF AREN'T ALLOWED TO HELP

 

Well that's unfortunate.

[Destin 1958]

Well fuck.

I saw your original post. Can't hide from me

[MrCoolness 547]

If you guys need the solution to 1 of the cases that G(a) = G(a+4) I can provide it.

 

 

Or Mr. Jacob can if he wants to

[SpartanSakaro 1977]

read the question wrong gg

Edited May 2, 2014 by SpartanSakaro

[Ika 320]

still working

Edited May 2, 2014 by Ika

[Destin 1958]

I read the question as g(a) = g(a+4) as in g(6) = g(10) ((I know those 2 don't equal as g(10) = 10 and g(6) = 60)) not g(a) = g(a) + 4  but I could be wrong

Edited May 2, 2014 by Destin

[MrCoolness 547]

I read the question as g(a) = g(a+4) as in g(6) = g(10) ((I know those 2 don't equal as g(10) = 10 and g(6) = 60)) not g(a) = g(a) + 4  but I could be wrong

 

Yes, g(6) = g(10) is the idea, but it has to be such that g(a) = g(a+4) and in that case it does not.

[MrCoolness 547]

LCM of a where a = (1,2,...a) will = the product of all the prime numbers in this set

therefore if a+1,2,3,4 does not include a prime, then the LCM of a+4 will = a.

so none of a+1,2,3,4 must be prime; we need a prime number with 4 or more numbers "above" it that aren't prime

numbers that fit this criteria <100: 23,24,31,32,47,48,53,54,61,62,73,74,83,84,89,90,91,92

total: 18

 

edit: nope, not correct.

 

Very close to finding the answer don't stop!

[Destin 1958]

Yes, g(6) = g(10) is the idea, but it has to be such that g(a) = g(a+4) and in that case it does not.

yea i was just telling people its g(a) = g(a+4) not g(a) + 4

[Ika 320]

revised answer:

 

LCM of a where a = (1,2,...a) will = the product of the prime numbers in that set, including factors of prime numbers (e.g. LCM(a=5) = 1*2*3*(2^2)*5.

^ simplified - the LCM of two numbers involves breaking them down into prime numbers, then multiplying the prime numbers with the largest exponents.

example: Take 16 and 150: 16 = 2^4; 150 =2 * 3 * 5^2. The LCM is: 5^2 * 2^4 * 3 = 1200.

therefore if a+1,2,3,4 does not include a prime or an exponent of a prime (which would make the LCM larger than the LCM of a, due to the way we find LCM above), then the LCM of a+4 will = a (this is the answer we are looking for).

so none of a+1,2,3,4 must be prime; we need a prime number with 4 or more numbers "above" it that aren't prime OR an exponent of a prime (e.g. 32 = 2^5). Look at the list of prime numbers in 1-104 and find a block of 4 numbers that can "fit" in between two primes e.g. 23,24,25,26.

numbers that fit this criteria <100 (disregarding exponents for now):

23,24,31,32,47,48,53,54,61,62,73,74,83,84,89,90,91,92,97

total: 19

relevant prime factors to exclude:
2: 2,4,8,16,32,64
3: 9,27,81
4: 16,64
5: 25
6: 36
7: 49

new set:
32,53,54,73,74,83,84,89,90,91,92 = 11 numbers

Edited May 2, 2014 by Ika

[MrCoolness 547]

revised answer:

 

LCM of a where a = (1,2,...a) will = the product of the two largest prime numbers in this set

therefore if a+1,2,3,4 does not include a prime, then the LCM of a+4 will = a.

so none of a+1,2,3,4 must be prime; we need a prime number with 4 or more numbers "above" it that aren't prime OR an exponent of a prime (e.g. 32 = 2^5)

numbers that fit this criteria <100 (disregarding prime factors):

31,32,47,48,53,54,61,62,73,74,83,84,89,90,91,92

total: 18

relevant prime factors to exclude:

2: 2,4,8,16,32,64

3: 9,27,81

4: 16,64

5: 25

6: 36

7: 49

new set:

53,54,61,62,73,74,83,84,90,91,92 = 11 numbers

 

YAY!!! THANK YOU IKA!!

CONGRATS!!!

Hurray to everyone for participating and look forward to tomorrow's question!

 

 

wait, numbers are off a bit.

[Ika 320]

REVISED NUMBERS BECAUSE I'M DUMB

 

32,53,54,73,74,83,84,89,90,91,92 = 11 numbers

Edited May 2, 2014 by Ika

[Wintergreen 1603]

22

[MrCoolness 547]

second place goes to wintergreen!

[Wintergreen 1603]

second place goes to wintergreen!

Mad

[Siesan 1802]

but if the integer of 100 is 50, doesn't that mean that 50 divided in half is 25?

 

so the answer is 25

[SpartanSakaro 1977]

Mad

bad

[Fetterolf 33]

I feel like coolness is doing this to help him with homework for something

[Ika 320]

I feel like coolness is doing this to help him with homework for something

 

well he has the answers, sooo