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I am trying to count the number of prime numbers which end by digit "$3$" such as $3, 13, 23$, etc. and are below $10^6$.

The number of primes existing below $10^6$ is known empirically to be $78~ 498$. Because, at most 1 every 10 numbers ends by the digit 3, it means that, at most there could be $7850$. So this is the lowest bound I have found so far.

Is this lower bound correct? I thought so, but when trying to answer in a questionnaire, it is said my result is wrong: enter image description here

What is the problem?

Jose
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  • it seems you lost a 0 you start with 78.498 and finsh with 7.850. btw if the distribuition of last digit is unifor I will divide this number by 4 getting 19.000 primes – ALG Nov 12 '18 at 14:55
  • Why would you divide by $10$? Even if you (inappropriately) assume that the primes are as uniformly distributed as possible, the only digits a prime $>5$ can end with are ${1,3,7,9}$. – lulu Nov 12 '18 at 14:58

2 Answers2

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For primes greater than $5$ the last digit is always $1,3,7,$ or $9$. Any other last digit means the number is divisible by $2$ or $5$. They are approximately equally distributed so you would have about $19,500$

Ross Millikan
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  • Thank's! However, that is not an answer to my question. My question is what is wrong in my simple reasoning. – Jose Nov 12 '18 at 15:01
  • You give no justification to your division by $10$. You say "at most" 7850, which would imply that there must be 7849 or 7850 ending with each digit, but we know there are none or 1 ending in each of six digits. – Ross Millikan Nov 12 '18 at 15:05
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Oh! I found the mistake. My mistake was thinking that at most 1 of every 10 number in 78 498 prime numbers below $10^6$ could be 3. Because those do not include all numbers, but only the ones which are primes.

Jose
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