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secondary 4 | E Maths
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Help to solve part b and c
Date Posted: 4 years ago
Views: 486
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1440=10*144=2*5*12*12=2*5*2*2*3*2*2*3=2^5*3^2*5.
b) observe that 120 is a factor. It remains to confirm 101 to 119 do not work, which has to be done by brute force
c) Note that any cube can only have prime factors of multiples of three. 1440*5=2^5*3^2*5^2. Minimially we would need 1 more factor of 5, 1 more factor of 3 and 1 more factor of 5, to get
1440*5*(2*3*5)= (2^2*3*5)^3
b) observe that 120 is a factor. It remains to confirm 101 to 119 do not work, which has to be done by brute force
c) Note that any cube can only have prime factors of multiples of three. 1440*5=2^5*3^2*5^2. Minimially we would need 1 more factor of 5, 1 more factor of 3 and 1 more factor of 5, to get
1440*5*(2*3*5)= (2^2*3*5)^3
There was a typo: minimally one more factor of 2, not 5
Want more clarification aboit parr b
I made an error in part b, in fact we would have to eliminate 100 to 119 instead. As for clarification I am not sure what you want. Do you want me to show that 120 is a factor of 1440 or what?
120 is correct
Why. 100 to 119 eliminate
Ok thank you
For example, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119 are not possible because they are odd. 102 has a factor of 17, 104 has a factor of 13, 106 has a factor of 53, 108 has 3 factors of 3, 110 has a factor of 11, 112 has a factor of 7, 114 has a factor of 19, 116 has a factor of 29, 118 has a factor of 59.
Was this given as school homework? It seems too tedious for the student to explain why 100 to 119 doesnt work. And if it is school homework could you do me a favour and share with me what was the intended solution? Thank you very much.
Was this given as school homework? It seems too tedious for the student to explain why 100 to 119 doesnt work. And if it is school homework could you do me a favour and share with me what was the intended solution? Thank you very much.
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