O-Level Additional Math

[2DMommy]

Please help with this q :


Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (eg 123123). The number formed will always be divisible by 7,11 and 13.
A) explain why
B) what are the other factors of any such number formed ?

Thks !

[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.

[Jtutor]

2DMommy:

Please help with this q :


Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (eg 123123). The number formed will always be divisible by 7,11 and 13.
A) explain why
B) what are the other factors of any such number formed ?

Thks !

Hi 2DMommy,

A) 7x11x13 = 1001
1001x100 = 100100 (answer)
B) Other factors are 2^2 x 5^2

Hope the above is correct.

Cheers!
Jtutor

[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)?