# In how many ways can the number 7056 be resolved into two factors?

Last updated date: 28th Mar 2023

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Answer

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Hint: In this question we need to find the number of ways in which the given number can be resolved into two factors. So, firstly we would be doing prime factorization of the given number and then use the formula for the same to resolve it into two factors. This would help us find the answer.

Complete step-by-step answer:

We have been given the number 7056. Now, if we do the prime factorization of the number we get,

$7056 = {2^4} \times {3^2} \times {7^2}$

So, the given number is of form ${a^p}{b^q}{c^r}.....$where a, b, c…. are prime numbers and p, q, r…. are all even numbers.

So, we can resolve the number into two factors in $\dfrac{1}{2}\left[ {\left( {p + 1} \right)\left( {q + 1} \right)\left( {r + 1} \right)....... + 1} \right]$ ways.

So, the number of ways in which the given number can be resolved into two factors is $\dfrac{1}{2}\left[ {\left( {4 + 1} \right)\left( {2 + 1} \right)\left( {2 + 1} \right) + 1} \right]$

$ = \dfrac{1}{2}\left[ {5 \times 3 \times 3 + 1} \right]$

$ = \dfrac{{46}}{2}$

$ = 23$

Hence 7056 can be resolved into two factors in 23 ways.

Note: Whenever we face such types of problems the key point to remember is that we need to have a good grasp over permutation and combinations. In these questions we should always find the prime factorization of the number and then use the formulas like given above. This helps in getting us the required condition and gets us on the right track to reach the answer.

Complete step-by-step answer:

We have been given the number 7056. Now, if we do the prime factorization of the number we get,

$7056 = {2^4} \times {3^2} \times {7^2}$

So, the given number is of form ${a^p}{b^q}{c^r}.....$where a, b, c…. are prime numbers and p, q, r…. are all even numbers.

So, we can resolve the number into two factors in $\dfrac{1}{2}\left[ {\left( {p + 1} \right)\left( {q + 1} \right)\left( {r + 1} \right)....... + 1} \right]$ ways.

So, the number of ways in which the given number can be resolved into two factors is $\dfrac{1}{2}\left[ {\left( {4 + 1} \right)\left( {2 + 1} \right)\left( {2 + 1} \right) + 1} \right]$

$ = \dfrac{1}{2}\left[ {5 \times 3 \times 3 + 1} \right]$

$ = \dfrac{{46}}{2}$

$ = 23$

Hence 7056 can be resolved into two factors in 23 ways.

Note: Whenever we face such types of problems the key point to remember is that we need to have a good grasp over permutation and combinations. In these questions we should always find the prime factorization of the number and then use the formulas like given above. This helps in getting us the required condition and gets us on the right track to reach the answer.

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