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3. Compute the number of integers in [1000] relatively prime to...

Question

3. Compute the number of integers in [1000] relatively prime to

3. Compute the number of integers in [1000] relatively prime to



Answers

Estimate the expected number of integers with 1000 digits that need to be selected at random to find a prime, if the probability a number with 1000 digits is prime is approximately 1?2302.

In this question you have to find us some of all natural numbers between 250 and 1000 which are exactly divisible by three. Yeah. First number yeah. And Mhm. Last number it's not yeah. Between good decision. 2 15 and 1000. Mhm. Just which are Mhm. Okay. Exactly. Yeah. Thank you for having me. Individual by three. It was good. Uh 2 52 What and 999 respectively. Yeah. Mhm. Just mhm. People no sequences 2 52. You have to 2 55. 2 58. Up to 900 99. Given sequences in ap Because defense of consecutive term is constant and in the end is equal to three. Therefore even is equal to first turn credit. Is he going to 252 and D. Is equal to? Mhm. Mhm. Common difference. Okay. Okay that's good. Is equal to 2 55 minus 2 52 is equal to three plate. Mhm A. N. Is equal to just and tell you for therefore in is equal to 999. Good. We know that energy equal to even plus and minus one in two. D. Yeah. Subsequute even is equal to 252 reception And is equal to three. Move 2 52. Right entitling question. Mhm. But Mhm. Mhm. Mhm. Okay. In critical condition. Yeah. And minus one is equal to seven four seven upon three. Therefore N is equal to 2 15. But mhm. Yeah. But it's we know that some of in terms of an AP is given by And a .2 into first term plus. Laster here And is equal to 250 and first term is 2 52 and last term is 999 mm. Therefore as soon as equal to 15 six three seven, things like that. Mhm. Okay. Yeah, I'm looking with industry just Yeah, well over required some. Is it Clayton this 156 37 five. What would people think This is our final answer for this question.

So the number of plastic number is pitiable off. 3 17. 95. Noticed that the energy that are multiples off 30 city off. 3. 17. 51. We see the created 705 GCT off 17 30. Published president by However, the three d. C. D off 17. 35 is exceed 1000 at least 1 7085. So thes three D c. D guest in 1000, because it will self destruct them. So we have ah, Horan's off three in 17 1035. It's about 383 8 28 respectively. And then you have to simplify it. So 33% plus we get management management management that is equals stranded 90. So the number of positive integers less than 1000. And that leaves visible but 3. 17 or 35 This 1000 minus 3 98 Sequels toe 610 in three years

A number is divisible by three. Some of its digits is divisible by three. For example, four plus five is nine and nine is divisible by three. So this this is the visible by three. Same with 72 7 was to his nine and now it is visible by three. I never said add up. Two numbers that are divisible by three are 378 570 5 85 three and 300. The others are not divisible by three.

To be prime. A number must have two factors, namely two distinct factors. One and itself. Zero has many factors, and one only is one factor, namely itself. So it fails our test because of prime factor. A prime number has two factors. Next, let's eliminate anything that's a multiple of five. So anything that ends in five is a multiple of five and can be eliminated. That includes 15 55 105. Next, let's eliminate perfect squares. 49 is seven times seven, so that one's gone. After that, we should look for things that we know where multiples of three well eight plus seven is 15 and 15 is a multiple three. So that one's gone. Now 91 is seven times 13 so that one is gone as well. Next we see what's left 1923 and 31 on Lee of one and itself as factors. So their prime same goes for 59 97 and 100 and three


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