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# Help!

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Let m and n be positive integers such that m = 24n + 51. What is the largest possible value of the greatest common divisor of 2m and 3n?

Apr 21, 2018

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Let m and n be positive integers such that m = 24n + 51. What is the largest possible value of the greatest common divisor of 2m and 3n?

m=3(8n+17)

 n 1 2 3 4 5 6 7 8 9 m 3*25 3*33 3*41 3*49 3*57 3*65 3*73 3*81 3*89 3n 3 6 9 12 15 18 21 24 27 2m 2*3*25 2*3*33 2*3*41 2*3*49 2*3*57 2*3*65 2*3*73 2*3*81 2*3*89 HCF 3 6 3 6 3 6 3 6 3

Well

if n is odd the HCF is 3

if n is even the HCF is 6

Apr 21, 2018
#2
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(n*3=102, 2m=48*n+102=16*3*34+102=

16*102+102=102*17

HCF of n3 and 2m is 102

Guest Apr 21, 2018
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Guest Apr 21, 2018
#4
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Hey there,

Try substituting 24n+51 into m for 2m

Then, use the Euclidean Algorithm and repeatedly subtract 3n from 2(24n+51)

Or check the NT message board. I had the same problem :)

Apr 22, 2018
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Sorry guest, I have only just seen your post.  I could easily have missed it.

If you were a member you could have sent me a private message with the address of this thread and that way i could not have missed it.

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GOOD EXAMINATION GUEST!  You are right. :)

Let m and n be positive integers such that m = 24n + 51.

What is the largest possible value of the greatest common divisor of 2m and 3n?

m=3(8n+17)

Ltt's consider where  n is a multiple of 17

let n = 17k where k is a positive integer.

n=17k

m=3(8*17k+17)

m=3(9*17k)       error   m=3*17(8k+1)

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3n=3*17   *  k

2m=3*17   * 2(8k+1)

If k is even then 2 is another factor.

Assume k is even and let k=2c

3n=3*17 *2  *  c

2m=3*17*2  (16c+1)

I do not think that c and 16c+1 can have any common factors.

Which means that the highest possible common factor is 3*17*2 = 102

So if n is a multiple of 34 the highest common factor is 102.

If n is a multiple of 17 but NOT of 2 then the highest common factor is 51

if n is not a multiple of 17 and is odd the HCF is 3

if n is not a multiple of 17even the HCF is 6

I think this is right.....

Apr 23, 2018
edited by Melody  Apr 24, 2018
#6
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Thanks melody :)

I think the largest possible common divisor is 102 (please correcr me if im wrong):

2m=2*(24n+51)=48n+102. If 3n and 2m have a common divisor,  then 2m-16*(3n)=102 has that divisor too, meaning that if 2m and 3n have a common divisor D then 102 is divisible by D, therefore the max value of D is 102.

Guest Apr 24, 2018
#7
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***please correct me if i'm wrong

Guest Apr 24, 2018
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Yes i think that you are right.