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Secondary 1 | Maths
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Date Posted: 3 years ago
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Hi!
(a) To find highest common factor (HCF), you want to multiply the prime factors which are common to the numbers.
180 = 2^2 × 3^2 × 5
784 = 2^4 × 7^2
The largest factors shared by both numbers is 2^2.(since 180 cannot be divided by 7, and 784 cannot be divided by 3 or 5)
So the HCF is 2^2 = 4.
(b) Given the expressions of 180 and 784 as prime factors (shown above), we may express 180x784 as...
180 × 784 = 2^6 × 3^2 × 5 × 7^2
In order for 180×784×m to be cube-rootable, all their prime factors need to be in powers of 3 (or multiples of 3).
So 180 × 784 × m would be 2^6 × 3^3 × 5^3 × 7^3,
m = 3 × 5^2 × 7 = 525
(c)
180 = 2^2 × 3^2 × 5
16 = 2^4
Lowest common multiple (LCM) = 2^4 × 3^2 × 5
= 720
Hence,
180n = 720
n = 4
Hope this helps!
(a) To find highest common factor (HCF), you want to multiply the prime factors which are common to the numbers.
180 = 2^2 × 3^2 × 5
784 = 2^4 × 7^2
The largest factors shared by both numbers is 2^2.(since 180 cannot be divided by 7, and 784 cannot be divided by 3 or 5)
So the HCF is 2^2 = 4.
(b) Given the expressions of 180 and 784 as prime factors (shown above), we may express 180x784 as...
180 × 784 = 2^6 × 3^2 × 5 × 7^2
In order for 180×784×m to be cube-rootable, all their prime factors need to be in powers of 3 (or multiples of 3).
So 180 × 784 × m would be 2^6 × 3^3 × 5^3 × 7^3,
m = 3 × 5^2 × 7 = 525
(c)
180 = 2^2 × 3^2 × 5
16 = 2^4
Lowest common multiple (LCM) = 2^4 × 3^2 × 5
= 720
Hence,
180n = 720
n = 4
Hope this helps!
Thank uuuu
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