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# Help....

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If  a/b is the probability that the reciprocal of a randomly selected positive odd integer less than 2010 gives a terminating decimal, with a and b being relatively prime positive integers, what is a+b?

Oct 4, 2020

#1
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Do you understand  your question and what is being asked of you to find out? I'm not quite certain but here is my attempt. Let me know your thinking.

1 - How many "positive odd integers" are there between 1 and 2010? There are: 2010 / 2 =1005. right?

2 -Of those 1005 "positive odd integers", how many do you think have their reciprocals end in a terminating decimal?

3 - My understanding, correct me if I'm wrong, is that only: 5^1 =5,  5^2 =25, 5^3 =125, and 5^4 =625, or a total of 4 integers have that property.

4 - What is the probability of randomly picking one of these 4 integers out of a total of 1005 "positive odd integers"?

5 -  Well, It is: 4 / 1005, which are "relatively prime", that is, their GCD = 1

6 - Therefore, a =4 and b=1005 and a + b =4 + 1005 =1009 .

7 - Did you understand all of the above and what do you think? Let me know if I went astray somewhere !.

Oct 4, 2020
#2
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Yep! I got the sawm but it was wrong..

Guest Oct 5, 2020