2 and 3 can divide the numbers tooo....................................................

Number Theory Level 3

If three distinct numbers are chosen randomly from the first 100 100 natural numbers, the the probability that all three of them are divisible by both 2 2 and 3 3 is p / q p/q . Then the value of p + q p+q is ?????

{p and q are coprime numbers}


The answer is 1159.

Arpit Dhimanj by Arpit Dhimanj , 23, India

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1 solution

Ryan Tamburrino
Jun 30, 2014

If a number is divisible by 2 and 3, it is divisible by 6. The number chosen must be divisible by 6. There are 16 numbers between 1 and 100 that have 6 as a factor, because 6(16) = 96. Therefore, the probability of choosing one of these numbers is 16/100. Since we must pick 3 distinct numbers, the probability for the next two picks are 15/99 and 14/98. Multiplying these probabilities and simplifying gives 4/1155 = p/q. Thus, p + q = 1159.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Should have mentioned that p and q are coprime numbers....I got 28/8085 and didn't simplify.....got it on second attempt

Tanya Gupta - 6 years, 11 months ago

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As a matter of fact it is wise to assume simplification unless mentioned otherwise.

Nishant Sharma - 6 years, 11 months ago

Other Way we can think is- desired outcome= 16(C)3 Total outcome= 100(C)3

Probability= 16(C)3 / 100(C)3 =4/1155=p/q

Saurabh Chaudhary - 6 years, 11 months ago

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