O-Level Additional Math

[MathIzzzFun]

WTK:

Help please!


Find the greatest number that will divide 171, 255 and 304 so as to leave the same remainder in each case.

Kindly show the workings. Thank you.

Let the greatest number be n, and the remainder be R

171 = An + R
255 = Bn + R
304 = Cn + R
where A,B,C are integers

(B-A)n = 255 - 171 = 84 = 12 x 7
(C-B)n = 304 - 255 = 49 = 7 x 7
(C-A)n = 304 - 171 = 133 = 19 x 7

n = 7

The greatest number is 7

cheers.

[WTK]

Thanks MathIzzzFun.

[Alarmchain]

MathIzzzFun:

WTK:

Help please!


Find the greatest number that will divide 171, 255 and 304 so as to leave the same remainder in each case.

Kindly show the workings. Thank you.

Let the greatest number be n, and the remainder be R

171 = An + R
255 = Bn + R
304 = Cn + R
where A,B,C are integers

(B-A)n = 255 - 171 = 84 = 12 x 7
(C-B)n = 304 - 255 = 49 = 7 x 7
(C-A)n = 304 - 171 = 133 = 19 x 7

n = 7

The greatest number is 7

cheers.

Hi MathIzzzFun,

From your results above, we get
1) B - A = 12
2) C - B = 7
3) C - A = 19

Can I trouble you to show me the steps, from these 3 sets of equations, how do we end up with A = 24, B = 36 and C = 43?

Thank you.

[MathIzzzFun]

Alarmchain:

MathIzzzFun:

[quote=\"WTK\"]Help please!


Find the greatest number that will divide 171, 255 and 304 so as to leave the same remainder in each case.

Kindly show the workings. Thank you.

Let the greatest number be n, and the remainder be R

171 = An + R
255 = Bn + R
304 = Cn + R
where A,B,C are integers

(B-A)n = 255 - 171 = 84 = 12 x 7
(C-B)n = 304 - 255 = 49 = 7 x 7
(C-A)n = 304 - 171 = 133 = 19 x 7

n = 7

The greatest number is 7

cheers.

Hi MathIzzzFun,

From your results above, we get
1) B - A = 12
2) C - B = 7
3) C - A = 19

Can I trouble you to show me the steps, from these 3 sets of equations, how do we end up with A = 24, B = 36 and C = 43?

Thank you.[/quote]
171/7= 24r3, A= 24

do the same for 255 n 304

cheers

[Alarmchain]

Yes, that was what I did.


But I was wondering how the same values for A, B and C can be gotten from the 3 simultaneous equations.

Thanks.

[MathIzzzFun]

Alarmchain:

Yes, that was what I did.


But I was wondering how the same values for A, B and C can be gotten from the 3 simultaneous equations.

Thanks.

255/7 = 36R3, B = 36
304/7 = 43R3, C = 43

or with A = 24 and
1) B - A = 12
2) C - B = 7

you can also get B= 36, C = 43

cheers.

[jesschan]

Hi! Can anyone recommend a good math practice book for Sec One Maths (IP school)?

[Alarmchain]

MathIzzzFun:

Alarmchain:

Yes, that was what I did.


But I was wondering how the same values for A, B and C can be gotten from the 3 simultaneous equations.

Thanks.

255/7 = 36R3, B = 36
304/7 = 43R3, C = 43

or with A = 24 and
1) B - A = 12
2) C - B = 7

you can also get B= 36, C = 43

cheers.

My apologies, MathIzzzFun. I may have caused some confusion with my unclear question.

With 3 unknowns, we usually can find the unknowns using algebra, if we have 3 simultaneous equations.

In this instance, the 3 equations were:
1) B - A = 12
2) C - B = 7
3) C - A = 19

Yes, if we know that A= 24, we can substitute it into equations 1 and 2 to find the values of B and C. However, we got A by doing 171 / 7 first, which is perfectly ok for this problem.

But my question is, are we able to find A, B and C with just the above 3 simultaneous equations, without resorting to finding any one of them initially, by using 171 / 7 or 255 / 7 or 304 / 7.

Very sorry, my question is not just about this problem per se, but to correct a possible misunderstanding on my part, that Y unknowns can be found if we have Y number of simultaneous equations.

If it cannot be done, does it mean that when solving unknowns using simultaneous equations, it is not always true that all the Y unknowns can be found even if we have Y number of simultaneous equations given.

Again, very sorry for any confusion caused.

[MathIzzzFun]

Alarmchain:

MathIzzzFun:

[quote=\"Alarmchain\"]Yes, that was what I did.


But I was wondering how the same values for A, B and C can be gotten from the 3 simultaneous equations.

Thanks.

255/7 = 36R3, B = 36
304/7 = 43R3, C = 43

or with A = 24 and
1) B - A = 12
2) C - B = 7

you can also get B= 36, C = 43

cheers.

My apologies, MathIzzzFun. I may have caused some confusion with my unclear question.

With 3 unknowns, we usually can find the unknowns using algebra, if we have 3 simultaneous equations.

In this instance, the 3 equations were:
1) B - A = 12
2) C - B = 7
3) C - A = 19

Yes, if we know that A= 24, we can substitute it into equations 1 and 2 to find the values of B and C. However, we got A by doing 171 / 7 first, which is perfectly ok for this problem.

But my question is, are we able to find A, B and C with just the above 3 simultaneous equations, without resorting to finding any one of them initially, by using 171 / 7 or 255 / 7 or 304 / 7.

Very sorry, my question is not just about this problem per se, but to correct a possible misunderstanding on my part, that Y unknowns can be found if we have Y number of simultaneous equations.

If it cannot be done, does it mean that when solving unknowns using simultaneous equations, it is not always true that all the Y unknowns can be found even if we have Y number of simultaneous equations given.

Again, very sorry for any confusion caused.[/quote]In this case, although there are 3 equations :
1) B - A = 12
2) C - B = 7
3) C - A = 19

there are only 2 independent equations
eg with 1) and 2), we can get 3)
or with 2) and 3), we can get 1)
this is why the variables cannot be solved with the 3 equations.

In short, to solve for Y unknowns, you need Y independent equations.

cheers.

[Alarmchain]

Thank you so much! It is very clear now.


BTW, is there a way for a child to see and immediately tell that equations given are dependent or independent, like some kind of rule of thumb thingy? Unlike this example where the dependence was quite easy to visualise, not sure if there are such "tricks" or "tips" to help a child for more complicated examples.

Thanks again!