no that's wrong slightly excessive numbers are actuallly where all n's factors add up to one more than n- i don't know where 2n came from!
so if n is odd all the factors must add to one more...even!
answer to first one is that the factors and one must add up to be even. therefore you must have a factor which only counted once to allow the total to be even so therefore a sqaure
answer to first one is that the factors and one must add up to be even. therefore you must have a factor which only counted once to allow the total to be even so therefore a sqaure
i do hope you realize you have just broken my heart! if i change it to diclude odd squares eg, 9 with the argument they would not be big enough? no that wouldn't work with numbers like 81 where their square root has factors in turn eg 81 = 1 + 9 + 3 + 27!aaaahhhhhh!ill keep working on it!:)!
hang on shall i start from the beginning?!
right. if the slightly excessive number was odd it would have only odd factors so when you add the factors together you would get 1 + odd + odd and as the odd numbers would always be in pairs ie there would be an even number of them, the total of all...
so...do they/ don't they exist?
i don't think they do as they would have to be even with an even number of (odd*even) factor pairs.
And if you assume this number is in the form of 2p where p is odd and has only two (odd*even) facots pairs, it must have the factors (2*psqaured) and (p * 2p)...
please please help me quick!
hi i was practisin a gcse maths paper and need some help with last question;
x and y are two positive irrational numbers. x + y is rational and so it x times y.
a) by writing the 1/x + 1/y as a single fraction explain why 1/x + 1/y is always rational.
b)...
no, I am only at secondary school but it makes more sense for the answer to be 0.
the answer is only 1 for other intergers to the power of 0 becuse say take 3
3to the 3 = 27 3 to the 2=9 3 to the 1= 3 you are dividing by three each time so it makes sense for the nxt to be 0 .
you do not...