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kevincan
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kevincan
How could this be done without Cicerone's formula, (which is very nice, by the way)?


I am sorry, I can't find Ciceone's formula. Can someone help?
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kevincan
How could this be done without Cicerone's formula, (which is very nice, by the way)?


This is the way it was sujjested..

8184 = 2^3*3^1*11^1*31^1

= (2^4)-1/(2-1) * (3^2)-1/(3-1)* (11^2)-1/(11-1) * (31^2)-1/31-1

= 15*4*some number*...

unit's digit is 0
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I am happy that people r using the formula given by me to answer these questions. I am sure that formulae like these will help us to save time in the exam,

keep rocking folks,

Regards,
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I get 1 or 7?

Factors of 8184 = 2^3 * 3^1 * 11^1 * 31^1

So either 2+3+11+31 = 47 So units digit is 7

OR

2+2+2+3+11+31 = 51 So units digit 1
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cicerone
I am happy that people r using the formula given by me to answer these questions. I am sure that formulae like these will help us to save time in the exam,

keep rocking folks,

Regards,


sure cicerone...it would definetely save a huge chunk of time
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AHHHHH, I've seen the light.

I read as sum of smallest factors, not ALL factors. D'oh!
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kevincan
If q is the sum of all factors of 8184, then what is the units digit of q?

(A) 0 (B) 1 (C) 4 (D) 7 (E) none of these


The factors are 31*11*3*2^3

If we take only fatcors that are not repetitive then it is 7.

If we can take all the 3 2's in 2^3 the it is 1
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trivikram, I think I fell into this trap. It asks for the sum of ALL of the factors so 8184, 4092.......1

I think!
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Thanks to Kevin for giving such wonderful questions



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