I am having issues working out questions similiar to the below. I understand the general concept, but is there something I am missing to be easily about to see that the prime factors of 899 are 23 and 31?
- Express the following as a product of prime factors: (a) 26; (b) 100; (c) 27; (d) 71; (e) 64; (f) 87; (g) 437; (h) 899.
ANSWERS: 5. (a) 2 × 13; (b) 2 × 2 × 5 × 5; (c) 3 × 3 × 3; (d) 71 × 1; (e) 2 × 2 × 2 × 2 × 2 × 2; (f) 3 × 29; (g) 19 × 23; (h) 29 × 31.
The trick to $899$ is to realize it's "close" to $900$, a perfect square.
More specifically,
$$900 = 30^2 \text{ so } 899 = 30^2 - 1$$
but then, using the difference of squares formula $(a+b)(a-b) = a^2 - b^2$, and that $1^2 = 1$, we have
$$899 = (30-1)(30+1) = 29 \cdot 31$$