# SOLUTION: The total number of prime divisors of the natural number 2002 is: (A) 2 (B) 3 (C) 4 (D) 5 (E) None of the above.

Algebra ->  Algebra  -> Divisibility and Prime Numbers -> SOLUTION: The total number of prime divisors of the natural number 2002 is: (A) 2 (B) 3 (C) 4 (D) 5 (E) None of the above.      Log On

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 Click here to see ALL problems on Divisibility and Prime Numbers Question 238461: The total number of prime divisors of the natural number 2002 is: (A) 2 (B) 3 (C) 4 (D) 5 (E) None of the above.Answer by jsmallt9(3296)   (Show Source): You can put this solution on YOUR website!Since 2002 is even it is clearly divisible by 2: 2002 = 2 * 1001 Now we look for factors of 1001, if any. Since it is odd, 2 will not divide evenly. And since the digits add up to 2 (1+0+0+1 = 2), it is not divisible by 3. And since 1001 does not end in a 5 or 0, it is not divisible by 5. The next prime is 7. There is no rule for divisibility by 7 so we just have to divide by 7 to see if it works. And it does! Now we have: 2002 = 2 * 7 * 143 Now we look for prime factors of 143, if any. 7 does not work on this so we move on to the next prime which is 11. (If there is a rule for divisibility by 11 I am not aware of it.) Dividing 143 by 11 we find that it divides evenly giving us: 2002 = 2 * 7 * 11 * 13 The four factors are all prime so your answer is (C).