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secondary 4 | E Maths
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Help Jonathan has been failing amaths lmfao! Anyone can contribute an answer, even non-tutors.
q14 pls:)
So in order to have a perfect square, all your prime factors must be ___².
Since your question asks to find the smallest value of k, we multiply by the smallest possible values to make the prime factors become __²
You have 3. So just multiply by 3 to obtain 3².
You have 7. So just multiply by 7 to get 7².
You have 2³. So just multiply by 2 to get 2⁴.
Realise that 2⁴
= 2 x 2 x 2 x 2
= 2² x 2²
= (2²)²
= 4²
So x 3 x 7 x 2 → 42.
k = 42
168 x 42 = 2³ x 3 x 7 x 2 x 3 x 7
= 2⁴ x 3² x 7²
= (2² x 3 x 7) x (4 x 3 x 7)
= (4 x 3 x 7)²
= 84²
= 7056
Likewise, √7056 = 84 = 4 x 3 x 7
168 = 2³ x 3 x 7
In order for 168n to be a multiple of 240,
168n must contain all the prime factors of 240.
So 168n has to contain 2⁴, 3 and 5.
2⁴ : 168 only has 2³. So multiply by 2
to get 2⁴.
3 : 168 already has one 3.
5 : 168 does not have 5. So multiply by 5.
So x 2 x 5 →x 10
So n = 10
168n
168 x 10 = 1680
= 2³ x 3 x 7 x 2 x 5
= 2⁴ x 3 x 5 x 7
= 240 x 7
168 ÷ 240 = 0.7
168 is 0.7 times of 240. 0.7 is a decimal.
How to make 168 a multiple of 240?
Just multiply by 10.
168 x 10 ÷ 240
= 168 ÷ 240 x 10
= 0.7 x 10
= 7
So n is 10.
You can also see it as shifting the decimal place one time to get 0.7 to become a whole number.