2002 = 2 * 7 * 11 * 13
I meant what are the steps to finding the Answer.
From the knowledge of divisibility, we know that 2 is a factor (all even numbers).
Again, the difference of sums of alternate digits equals zero (1+0)-(0+1)=0, so it is divisible by 11.
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
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:
(note: 859 is prime).
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