O-Level Additional Math
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Pls help me on the following Qin. Thanks
If 600 = 2^3 x 3 x 5^2. Find the smallest value of p such that 600p is a perfect cube. -
shin88:
600 = 2^3 x 3 x 5^2Pls help me on the following Qin. Thanks
If 600 = 2^3 x 3 x 5^2. Find the smallest value of p such that 600p is a perfect cube.
To be a perfect cube , the index of the prime factors of 600p must be multiple of 3 ,
600p = 2^3*3^3*5^3
p = 3^2*5 = 45 -
Thanks for your solution. I understand how u get 2^3x3^3x5^3 but how to get 3^2x5 ?
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shin88:
Thanks for your solution. I understand how u get 2^3x3^3x5^3 but how to get 3^2x5 ?
600 = 2^3 x 3 x 5^2
600p = 2^3 * 3^3 * 5^3
2^3 * 3 * 5^2 * p =2^3 * 3^3 * 5^3
p = 3^2*5 = 45 -
Thanks a lot, I finally understand :rahrah:
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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 ! -
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. -
2DMommy:
Hi 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 !
A) 7x11x13 = 1001
1001x100 = 100100 (answer)
B) Other factors are 2^2 x 5^2
Hope the above is correct.
Cheers!
Jtutor -
WTK:
Let the greatest number be n, and the remainder be RHelp 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.
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. -
Thanks MathIzzzFun.
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