How many positive whole-number divisors does 196 have?

Solutions

Solutions

This is a common type of counting problems in MATHCOUNTS.

Find all the prime numbers in 196. 196 should be the prodcut of all the numbers you will have found.

\[196 = 2 \times 98 = 2 \times 2 \times 49 = 2 \times 2 \times 7 \times 7 = 2^2 \times 7^2.\]

A divior of 196 can only have 2 or 7 as its prime factors. If a number has other prime factors, it cannot be a divisor of 196. For exmaple, 6 has a prime factor 3 and is not a divisor of 196. We now calculate how many ways we can use 2 and 7 to build divisors of 196.

Because the exponent of 2 is 2, there are 3 way’s to include 2: do not include 2, include 2 once, or include 2 twice. You cannot include 2 more than twice. For example, if you include 2 three times, you will get a number that is a multiple of 8 and it is not a divisor of 196.

The exponent of 7 happens to be 2, too. There are 3 way’s to include 7: do not include 7, include 7 once, or include 7 twice.

Therefore, 196 has \(3 \times 3 = 9\) positive divisors.

The following table shows all the divisors of 196 and the number of 2’s and 7’s in them.

2 7 divisors
0 0 \(2^0 \times 7^0 = 1\)
0 1 \(2^0 \times 7^1 = 7\)
0 2 \(2^0 \times 7^2 = 49\)
1 0 \(2^1 \times 7^0 = 2\)
1 1 \(2^1 \times 7^1 = 14\)
1 2 \(2^1 \times 7^2 = 98\)
2 0 \(2^2 \times 7^0 = 4\)
2 1 \(2^2 \times 7^1 = 28\)
2 2 \(2^2 \times 7^2 = 196\)

More advanced stuff

Any integer \(n\) that is grater than 1 (staring from 2) has a prime factor and \(n\) can be reprsented with a prodcut of prime factors, like

\[n = p_1 ^ {e_1} \times p_2 ^ {e_2} \times p_3 ^ {e_3} \ldots\]

where \(p_1, p_2, p_3, \ldots\) are distinct prime numbers and \(e_1, e_2, e_3, \ldots\) are their exponents. And there is only one (unique) way to do that. This is called unique factorization of an integer and there is a theorem called unique factorization theorem.

If a number \(d\) is a divisor of \(n\) and \(d\) is larger than 1, any prime factor of \(d\) must be one of \(p_1, p_2, p_3, \ldots\).

To find the number of positive divisors of \(n\), you just need to find out the number of ways you can pick prime numbers from the set of \(p_1, p_2, p_3, \ldots\), under the constraints that you can pick \(p_1\) up to \(e_1\) times, \(p_2\) up to \(e_2\) times, and so on.

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