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.

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(3758) (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).
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