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December 21, 2014

December 21, 2014

Posted by **Sumare** on Tuesday, July 16, 2013 at 12:34am.

- Math -
**Bosnian**, Tuesday, July 16, 2013 at 12:46am2002 = 2 * 7 * 11 * 13

- Math -
**Sumare**, Tuesday, July 16, 2013 at 1:08amI meant what are the steps to finding the Answer.

- Math -
**MathMate**, Tuesday, July 16, 2013 at 8:13amFrom 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).

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