Question 626199
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(2) who am i, i have 3 digits. i can only be divided by myself and one. the sum of my digits is 11 and i am under 150. 
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<pre>
They want you find a 3-digit prime number between 100 and 150, with the sum of digits 11.


First of all, it is clear that the first digit of this number must be 1.

Hence, the sum of two remaining digits must be 10.


Next, the last (the ones) digits can not be even number, since, otherwise, the number in this interval is not a prime.


Also, it is clear that the last digit can not be 5, since, otherwise, the number is divisible by 5.


Taking this in account, we see that the last (= the ones) digit should be 1, or 3, or 7, or 9.



If the ones digit is 1, then the tens digit is 10-1 = 9, and the number itself is 191.

       This does not work, since 191 is greater than 150.



If the ones digit is 3, then the tens digit is 10-3 = 7, and the number itself is 173.

       This does not work, since 173 is greater than 150.



If the ones digit is 7, then the tens digit is 10-7 = 3, and the number itself is 137.

       This works, since 137 is a prime number (it is not divisible by 2, 3, 5, 7, 11,
       and it is just enough to see that 137 is a prime).



If the ones digit is 9, then the tens digit is 10-9 = 1, and the number itself is 119.

       This does not work, since 119 = 7*17 is not a prime number.


Thus, we analyzed all possible cases and by the method of exclusion PROVED
that there is one and only one possible answer, which is the number of  {{{highlight(highlight(137))}}}.   
</pre>

Solved.



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The solution method is a logical analysis in three steps.


First, &nbsp;we filter out many possible numbers based on simple divisibility properties.


Then we analyze the remaining &nbsp;4 &nbsp;cases and exclude three of them that do not work.


Finally, &nbsp;we check that the remaining number satisfies all the conditions.


This way we find a &nbsp;UNIQUE &nbsp;solution to the problem.