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man111 (54)

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(1)  a three digit no. is choosen at random . find the probability that a three digit  choosen no. has exactly 3 factor.

    
ultimator (401)

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If it has 3 factors, then it means
1)it is a perfect square
2) It has no factors except 1, itself and its square root
3) The root must be prime (not split into other factors)

From this, we see that 11square (121), 13 square (169), etc r possible.

The sq. roots of the numbers are 11, 13, 17 , 19, 23, 29 and 31.
There are 7 such numbers out of 100 to 999

So, probability = 7 / 900
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rudra.panda (2760)

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yup the answer is 7/900
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sboosy (3065)

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 \mbox{At this point I would simply like to add .as to why a number has} \\ \\ \mbox{to be a perfect square if it has 3 factors more mathematically} \\ \\ \mbox{Any number can be expressed as} \ (2^x)(3^y)(5^z)... \ \mbox{Similarly other prime numbers too} \\ \\ \mbox{Then the rule says, the number of factors are} \ (x+1)(y+1)(z+1) .... \\ \\ \mbox{Now,according to sum} \ 3 = (x+1)(y+1) ... \\ \\ \mbox{Only possibility is any one of them 3 and the other 1} \\ \\ \Rightarrow x=2,y=0 \\ \\ \mbox{which simply means that the required number is some prime number

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