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Secondary 1 | Maths
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Pls help with part b. I don't understand the model answer as well.
Date Posted: 4 years ago
Views: 337
a) 504 =2^3 x 3^2 x 7
b) 6= 2 x 3
24 = 2^3 x 3
x = 3^2 x 7=63
The factors of the 3 numbers you have must be the same as 504. Since 6 is a factor of 24, you can treat the qn as the LCM of 24 and x is 504. 24 alr has the factor 2^3 and 3 so x must have the factors 3^2 and 7.
b) 6= 2 x 3
24 = 2^3 x 3
x = 3^2 x 7=63
The factors of the 3 numbers you have must be the same as 504. Since 6 is a factor of 24, you can treat the qn as the LCM of 24 and x is 504. 24 alr has the factor 2^3 and 3 so x must have the factors 3^2 and 7.
Thanks! I understand now.
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hope you understand my working! the circled numbers are how you get LCM. Since LCM in this case has prime factors of 3^2 and 7 which is absent in the list of prime factors of 6 and 24, you can confirm that x contains both 3^2 and 7. To obtain the smallest x value, 2 must be included since it is a common prime factor!
Date Posted:
4 years ago
Thanks, but the model answer is 63.
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