MedVision ad

need help with this q any help appreciated thanks (1 Viewer)

Run hard@thehsc

Well-Known Member
Joined
Oct 7, 2021
Messages
784
Gender
Male
HSC
2022
@amdspotter not sure, but I feel like this question cannot be solved using simultaneous equations especially part (a). It is given that x and y are positive integers. If this is the case, then using the fact that 10x + y = 13M (where M is an arbitrary integer), we can confirm that the only combination for the base case possible is that x = 1 and y = 3 for the base value of M = 1. And then we can go on forwards - someone correct me if I am wrong!!
 

Epicman69

Active Member
Joined
Jan 24, 2020
Messages
213
Gender
Male
HSC
2022
You can just split 8112 into the form of 10x + y which gives you 10 x 811 + 2, where you can now apply the x + 4y equation to give you 819, then you can just continue to repeat this process which gets you 117? and 39? Then you can just say 39 is obviously divisible by 3 (also because you can't recursively do it anymore it just gives you 39 over and over again) and by our iff statement this must mean that all the numbers above i.e 117 must be divisible by 13.
 

amdspotter

Member
Joined
Dec 9, 2020
Messages
126
Gender
Male
HSC
2022
You can just split 8112 into the form of 10x + y which gives you 10 x 811 + 2, where you can now apply the x + 4y equation to give you 819, then you can just continue to repeat this process which gets you 117? and 39? Then you can just say 39 is obviously divisible by 3 (also because you can't recursively do it anymore it just gives you 39 over and over again) and by our iff statement this must mean that all the numbers above i.e 117 must be divisible by 13.
ah this makes so much sense thanks m8
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top