Test WW - B

useless · Long-term Member

Test WW - B

bradp · Senior Member

19) 9/ [(5^4)(2)] = 9/[(625)(2)] = 9/1250
from here, 1250 * 8 = 10,000 (125 * 8 = thousand)
so (9/1250) * (8/8) = 72/10,000 = .0072
to get the right decimal places, think of 72/100 as .72, 72/1000 is then .072, so this is .0072

30) 93015 / 351 using the method posted on this site for figuring out how many digits are answer (5 digits - 3 digits = 2 digits + 1 because 930 > 351, for more info: http://math-magic.com/approx/div_num.htm )

so answer has 3 digits
now, approximate into 93/ 35 = 2 +(23/35) is about 2 +(2/3)
answer has 3 digits = 266
range is 252-278

i like the math-magic method for figuring # of digits because you can then ignore all the zeros and stuff

31) sum of positive integral divisors of 30

there is a method here: http://math-magic.com/misc/pos_int_div2.htm

good method, but it takes practice - so for small numbers like 30 i add them up manually

1,30 10,3 5,6 15,2
31 + 13 + 11 + 17 = 72 (i dont try to remember the sum of each pair of factors, but rather add them up as i go)

but I need to force myself to learn the method
so i'll start now

30 = 2^1 * 3^1 * 5^1

if exponent of the factor is 1, add 1 to the number and multiply by the others (if the exponent is more than 1, it's different)

3*4*6 = 12 * 6 = 72

a bit easier

fht · Long-term Member

In Mr. White's word, KNOW YOUR ONESIES. FYI in case you don't know, Onesies are the value and percentage of fractions 1 over 2,3,4,5,6,7,8,9,10,11,12,14,15,16,18,20,25,27,30 and so on. 1/125 is one of them, so by memorization = .008, so 9/1250=.0008*9=.0072. For 2 thru 12, also remember all the values and percentages of fractions in its simplest form(2/12 can be reduced 1/6,but things 5/12 or 7/12 need to be memorized).

93015/351: round to 91000/350=9100/35=1300/5=260 and because 91000 is merely a bit more than 2% off 93000((93000-91000)/93000=2/93<5%), so just put 260.

Sum of PIDs: In pseudo-Java,

public int findPID(){

int i, product=1; //also iterable List primeFactors (pF) and powerOfPrimeFactors (pPF)

for(pF p, pPF k; p.hasNext(), k.hasNext(); p.next(), k.next()){

i=Math.pow(p,k+1); i=(i-1)/(p-1); product*=i;

}return product;

}

Best Wishes

-- Edited by fht at 20:19, 2006-05-24

fht · Long-term Member

And the reason why when the power of the prime factor(p) is 1 you just add 1 to the factor is because following the process (p^2-1)/(p-1) = (p-1)(p+1)/(p-1) = p+1

Best Wishes