How do I find the prime factorization of number 2002?
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).
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).
To find the prime factorization of a number, such as 2002, you can follow these steps:
1. Begin by dividing the number by the smallest prime number, which is 2. If the number is divisible by 2, continue dividing it until it is no longer divisible.
2002 ÷ 2 = 1001
2. Since 1001 is not divisible by 2, move on to the next prime number, which is 3.
1001 ÷ 3 = 333.6667
3. Since 333 is not divisible by 3, move on to the next prime number, which is 5.
333 ÷ 5 = 66.6
4. Similarly, 66 is not divisible by 5, so move on to the next prime number, which is 7.
66 ÷ 7 = 9.4286
5. Now, continue the process with the number 9.
9 ÷ 3 = 3
6. Finally, you are left with the prime number 3.
We have found all the prime factors: 2, 7, 3, and 3.
Therefore, the prime factorization of 2002 is 2 × 7 × 3 × 3, or simply written as 2 × 7 × 3^2.