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?

MIRB16 Apr 21, 2018

#1**+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

Melody Apr 21, 2018

#4**+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 :)

Creeperhissboom Apr 22, 2018

#5**+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.

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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.....

Melody Apr 23, 2018

#6**+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