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Find the number of fractions less than 1 such that, in lowest terms, the denominator is 1000.

 Oct 1, 2019
 #1
avatar+33 
+2

oh wait, its fine, after a while i figured out the answer was 400

 Oct 1, 2019
 #2
avatar+2847 
+2

How did you solve it?

 

Is it counting primes? wink

CalculatorUser  Oct 1, 2019
 #3
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+2

This is the "closed form" of the sequence that will generate all 400 numerators in the question:

 

 a(n) = 1/4 (10 n + (1 - i) (-i)^n + (1 + i) i^n + (-1)^(n + 1) - 5)

 

Basically, the 400 numbers are ALL the odd numbers between 1 and 1000 of which there are 500 such numbers, exluding 100 numbers that end in 5.

 Oct 2, 2019
edited by Guest  Oct 2, 2019
 #4
avatar+2847 
+2

Nice job, Guest!

CalculatorUser  Oct 3, 2019
 #5
avatar+33 
+1

actually, I eliminated all the integers divisible by 1000.

 Oct 4, 2019

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