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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
 #1
avatar+99214 
+2

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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+1

What about n=34?

 

(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
 #3
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+1

melody please respond

Guest Apr 21, 2018
 #4
avatar+164 
+1

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
 #5
avatar+99214 
+2

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. 

-----

 

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)

 

----

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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+1

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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+1

***please correct me if i'm wrong

 

lol i already made a mistake 

Guest Apr 24, 2018
 #8
avatar+99214 
0

Yes i think that you are right.

I made a careless error in my last answer.

I have now edited that answer and hopefully it is now correct.

My answer does correspond with what you have said :)

Melody  Apr 24, 2018
edited by Melody  Apr 24, 2018

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