Post a New Question

Math

posted by .

How do I find the prime factorization of number 2002?

  • Math -

    2002 = 2 * 7 * 11 * 13

  • Math -

    I meant what are the steps to finding the Answer.

  • Math -

    From the knowledge of divisibility, we know that 2 is a factor (all even numbers).
    That reduces
    2002=2*1001

    Again, the difference of sums of alternate digits equals zero (1+0)-(0+1)=0, so it is divisible by 11.
    That makes
    2002=2*11*91
    We should (try to) know all prime factorization for numbers below 100, which in this case is 7*13=91.
    This completes the prime factorization of 2002 to
    2002=2*11*7*13

    Rule for 7,11 and 13 for large numbers.
    By the way,the given example is an excellent demonstration of another property:
    If the number is grouped in 3 digits from the right, difference of alternate groups has the same mod as the number itself with respect to divisors 7,11 & 13.

    For example to factorize 78169,
    we know from divisibility rules that it is not divisible by 2,3,5,11.
    Now form groups of 3 digits: 78 & 169.
    Take the difference 169-78=91.
    Since we know that 91 is divisible by 7 and 13 (but not 11), we conclude that 78169 is divisible by 7 and 13. The prime factorization is then:
    78169=7*13*859
    (note: 859 is prime).

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question