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secondary 4 | E Maths
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MM
MM

secondary 4 chevron_right E Maths chevron_right Singapore

Help to solve part b and c

Date Posted: 5 years ago
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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
done 3 Upvotes
clear 0 Downvotes
Matthew Fan
Matthew Fan's answer
39 answers (A Helpful Person)
1st
Matthew Fan
Matthew Fan
5 years ago
There was a typo: minimally one more factor of 2, not 5
MM
MM
5 years ago
Want more clarification aboit parr b
Matthew Fan
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?
MM
MM
5 years ago
120 is correct
MM
MM
5 years ago
Why. 100 to 119 eliminate
MM
MM
5 years ago
Ok thank you
Matthew Fan
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.