If 21168 = 2a × 3b × 7c, where a, b and c are natural numbers, then what is…

Concept: By the Fundamental Theorem of Arithmetic, every natural number greater than 1 has a unique prime-power representation. To find each exponent, repeatedly divide the number by the smallest applicable prime until it no longer divides evenly, and count how many times that prime divides out.

Application: Factor 21168 one prime at a time.

1. Divide by 2 repeatedly: 21168 ÷ 2 = 10584, ÷ 2 = 5292, ÷ 2 = 2646, ÷ 2 = 1323. Since 1323 is odd, 2 divides out exactly 4 times, so a = 4.

2. Divide the remaining 1323 by 3 repeatedly: 1323 ÷ 3 = 441, ÷ 3 = 147, ÷ 3 = 49. Since 49 is not divisible by 3, 3 divides out exactly 3 times, so b = 3.

3. Divide the remaining 49 by 7 repeatedly: 49 ÷ 7 = 7, ÷ 7 = 1. Since 1 cannot be divided further, 7 divides out exactly 2 times, so c = 2.

4. Substitute a = 4, b = 3 and c = 2 into the required expression: 4a − 5b + c = 4(4) − 5(3) + 2 = 16 − 15 + 2 = 3.

Cross-check: Multiplying the obtained powers back together must reproduce 21168: 24 × 33 × 72 = 16 × 27 × 49 = 21168, which matches the number given in the question, confirming a = 4, b = 3 and c = 2.