Math Problem Statement

If n is the digit at hundred place in 299156 then find out the total number of factors of 6n?

Solution

To solve this, let’s go through each step:

  1. Identify the Hundreds Place Digit: In the number 299156, the hundreds place digit is 1. So, n=1n = 1.

  2. Calculate 6n6n: 6n=6×1=66n = 6 \times 1 = 6

  3. Find the Factors of 6:

    • The prime factorization of 66 is: 6=2×36 = 2 \times 3
    • Since 6=21×316 = 2^1 \times 3^1, the formula to find the number of factors for any number with prime factorization p1a1×p2a2××pkakp_1^{a_1} \times p_2^{a_2} \times \ldots \times p_k^{a_k} is: (a1+1)(a2+1)(ak+1)(a_1 + 1)(a_2 + 1) \ldots (a_k + 1)
    • Here, a1=1a_1 = 1 and a2=1a_2 = 1, so the number of factors of 66 is: (1+1)(1+1)=2×2=4(1+1)(1+1) = 2 \times 2 = 4
  4. Conclusion: The total number of factors of 6n=66n = 6 is 4.

Let me know if you want more details or have any questions.


Related Questions:

  1. What are the factors of a number with a prime factorization of p12×p23p_1^2 \times p_2^3?
  2. How do you find the number of factors of a number using prime factorization?
  3. If a number is a product of three distinct primes, how many factors will it have?
  4. What is the total number of factors of 12?
  5. How do you find the prime factorization of a composite number?

Tip:

To find the number of factors of any number, use its prime factorization and apply the formula: multiply one more than each exponent in the prime factorization.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Place Value
Multiplication
Factors
Prime Factorization

Formulas

Finding factors of a number using prime factorization
Formula for total number of factors: (a1 + 1)(a2 + 1)...(ak + 1) for prime factorization p1^a1 * p2^a2 * ... * pk^ak

Theorems

Factor Count Theorem (using prime factorization)

Suitable Grade Level

Grades 6-8