Help!

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)

Well

if n is odd the HCF is 3

if n is even the HCF is 6

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 :)

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.....