Every factor of 112 is of the form 2^a × 7^b, where a 0, 1, 2, 3, 4 and b 0, 1. The powers of 2 are 2^0 = 1, 2^1 = 2, 2² = 4, 2³ = 8, and 2^4 = 16. The powers of 7 are 7^0 = 1 and 7^1 = 7.

how to find factors of any numbers 112 easily step by step using prime factorisation

Answer: The factors of 112 are 1, 2, 4, 7, 8, 14, 16, 28, 56, and 112.

Step-by-step solution

Step 1: Perform prime factorisation of 112

To begin, divide $112$ successively by the smallest prime numbers. Dividing $112$ by $2$ gives $56$ , dividing by $2$ gives $28$ , dividing by $2$ gives $14$ , and dividing by $2$ gives $7$ , which is prime. Thus, the prime factorisation in exponent form is $2^4 \times 7^1$ .

Step 2: Determine the total number of factors

If a number has prime factorisation $p^a \times q^b$ , the total count of its factors is given by $(a + 1)(b + 1)$ . Here, the exponents of $2$ and $7$ are $4$ and $1$ respectively, giving $(4 + 1)(1 + 1) = 10$ factors.

Step 3: List all combinations of prime powers

Every factor of $112$ is of the form $2^a \times 7^b$ , where $a \in \{0, 1, 2, 3, 4\}$ and $b \in \{0, 1\}$ . The powers of $2$ are $2^0 = 1$ , $2^1 = 2$ , $2^2 = 4$ , $2^3 = 8$ , and $2^4 = 16$ . The powers of $7$ are $7^0 = 1$ and $7^1 = 7$ .

Step 4: Multiply the powers to find all factors

Now, multiply each power of $2$ by $1$ : $1, 2, 4, 8, 16$ . Then multiply each power of $2$ by $7$ : $7, 14, 28, 56, 112$ . Arranging these in ascending order gives all $10$ factors of $112$ : $1, 2, 4, 7, 8, 14, 16, 28, 56, 112$ .

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