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How many numbers are there less than 121 which have exactly 4 factors and is also divisible by 2? |
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Answer» Step-by-step explanation: Every number x can be REPRESENTED in this manner. x=ab11∗ab22∗ab33x=a1b1∗a2b2∗a3b3 … where a1,a2,a3a1,a2,a3 are all prime numbers Number of factors of xx is (1+b1)∗(1+b2)∗(1+b3)(1+b1)∗(1+b2)∗(1+b3) … Now, we know that 4=2∗24=2∗2 or 1∗41∗4 So, all numbers which have 4 factors would EITHER be in the form of a3a3 where AA is a prime; Factors would be 1,a,a^2 & a^31,a,a^2 & a^3 a∗ba∗b where aa and bb are both primes; Factors would be1,a,b & ab1,a,b & ab Prime CUBES between 11 and 100100 are 88 and 2727. So, 2 cases. Cases for the second form: For a=2a=2, b=3,5,7,11,13,17,19,23,29,31,37,41,43,47→14b=3,5,7,11,13,17,19,23,29,31,37,41,43,47→14 cases For |
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