Find the HCF of the numbers A and B

[chwala]

Homework Statement

kindly see attached...

Relevant Equations

hcf

Kindly see attached:

Since n belongs to the class of Natural numbers, then we may have,
if $n=1, hcf (A,B)= hcf(15,8)=1$
$n=2, hcf(17,9)=1$
.
.
.
$n=8, hcf (29,15)=1$...

Therefore in my reasoning the correct solution is b. Is there a different approach to answering this question?
,

 

[haruspex]

Homework Statement:: kindly see attached...
Relevant Equations:: hcf

Kindly see attached:

View attachment 289497

Since n belongs to the class of Natural numbers, then we may have,
if $n=1, hcf (A,B)= hcf(15,8)=1$
$n=2, hcf(17,9)=1$
.
.
.
$n=8, hcf (29,15)=1$...

Therefore in my reasoning the correct solution is b. Is there a different approach to answering this question?
,

Any sum or difference of A and B would also be divisible by their HCF. This gives the answer in two lines.

 

[chwala]

Any sum or difference of A and B would also be divisible by their HCF. This gives the answer in two lines.

is that a general rule or one that works for this problem?
if $n=1$, sum is $23$, difference is $7$... hcf$=1$

 

[haruspex]

is that a general rule or one that works for this problem?
if $n=1$, sum is $23$, difference is $7$... hcf$=1$

But that's just n=1. Try it for unknown n.

 

[chwala]

But that's just n=1. Try it for unknown n.

That was my first approach yesterday and i told myself that could not be the right approach, as they have indicated that we have to have $n$, (its a condition )...i guess you want me to solve the simultaneous and eliminate $n$ from the picture.
and i am really not getting what you want me to do here, you are saying if i have two numbers say $20$ and $25$ being our $A$ and $B$, then,
$25+20=45$ clearly $45$ is divisible by the hcf value which is $=5$ in this case. Also,
$25-20 = 5$ also divisible by the hcf value which is $=5$...what exactly do you mean?
in our case, then $A+B=3n+20$
$A-B=n+6$...then?

 

[haruspex]

in our case, then $A+B=3n+20$

Try A-B.

 

[chwala]

I am trying to understand your view, try and do what haruspex? Substitute for $n?$

 

[haruspex]

I am trying to understand your view, try and do what haruspex? Substitute for $n?$

You tried A+B, which gave 3n+20. Try A-B instead.

The important fact, easy to prove, is that if c divides b and a then c divides a+b, a-b, a+2b, a-2b, 2a+b...

 

[chwala]

You tried A+B, which gave 3n+20. Try A-B instead.

The important fact, easy to prove, is that if c divides b and a then c divides a+b, a-b, a+2b, a-2b, 2a+b...

In post $5$, I had shown that $A-B=n+6$...

 

[haruspex]

In post $5$, I had shown that $A-B=n+6$...

Sorry, missed that.
So what is the HCF of n+6 and n+7? If it still isn't clicking, take another difference to eliminate n.

 

[chwala]

The hcf is just equal to $1$.

 

[Frabjous]

You can also check when A,B are odd, even.
or
B is also ℕ - {1,2,3,4,5,6,7}. The HCF is 1.

 

[mathwonk]

haruspex is using a version of the most important tool that exists for finding the HCF, called the euclidean algorithm. it is well worth learning. It occurs as Proposition 2, in Book VII of Euclid. In particular it works not just on this problem, but on essentially all such problems.

Howvever your method is also correct for this particular problem. Namely, if it is possible to find the HCF without knowing n, then the answer for any particular value of n must be the answer. Hence using just n=1 is sufficient. This also gives the answer in one or two lines, assuming the problem is correctly posed.

 

[Frabjous]

assuming the problem is correctly posed.

It’s a strange problem. I cannot tell what they are trying to teach. In my day they would have called this a new math problem. Maybe they wanted the OP to apply Euclid’s algorithm, but it does not look like the OP was taught that.

 

[chwala]

haruspex is using a version of the most important tool that exists for finding the HCF, called the euclidean algorithm. it is well worth learning. It occurs as Proposition 2, in Book VII of Euclid. In particular it works not just on this problem, but on essentially all such problems.

Howvever your method is also correct for this particular problem. Namely, if it is possible to find the HCF without knowing n, then the answer for any particular value of n must be the answer. Hence using just n=1 is sufficient. This also gives the answer in one or two lines, assuming the problem is correctly posed.

Thanks man! am pretty conversant with the Euclidean algorithm though... will probably try and re look at this from that perspective. cheers. Bingo

 

[haruspex]

assuming the problem is correctly posed.

Yes, it ought to offer the option "not enough information".