See Matthew Fan's answer to MM's Secondary 4 E Maths Singapore question

Matthew Fan's answer to MM's Secondary 4 E Maths Singapore question.

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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

Date Posted: 5 years ago

Comments 7 add a comment

Matthew Fan

5 years ago

There was a typo: minimally one more factor of 2, not 5

5 years ago

Want more clarification aboit parr b

Matthew Fan

5 years ago

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?

5 years ago

120 is correct

5 years ago

Why. 100 to 119 eliminate

5 years ago

Ok thank you

Matthew Fan

5 years ago

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.