UPSC CAPF Assistant Commandant Model Paper 4

GIVEN: $12^x$ divides the product of the first 50 natural numbers. CONCEPT: 12 = 4 × 3 Whenever the expression is divisible by 12, there must have a pair of ‘4 and 3’ in the expression. So we need to count the total possible pairs of ‘4 and 3’ in the expression to get the maximum value x such that the expression is divisible by $12^x$ FORMULA USED: Power of 2 in X! = X / 2 + X / 4 + X / 8 + X / 16 + X / 32 …… (All values in integer) And Power of 3 in X! = X / 3 + X / 9 + X / 27 + X / 81.…...... (All values in integer) CALCULATION: Product of the first 50 natural numbers = [1 × 2 × 3 × 4 × 5 × 6 ……… × 48 × 49 × 50] = 50! Now, Power of 4 in 50! = (Power of 2 in 50!) / 2 Power of 2 in 50! = 50 / 2 + 50 / 4 + 50 / 8 + 50 / 16 + 50 / 32 = 25 + 12 + 6 + 3 + 1 = 47 So, Power of 4 in 50! = 47 / 2 = 23 (Nearest integer) And Power of 3 in 50! = 50 / 3 + 50 / 9 + 50 / 27 = 16 + 5 + 1 = 22 Now, The number of possible pairs of ‘4 and 3’ in the expression = 22 (minimum of two) ∴ Maximum value of ‘x’ will be 22 such that the expression is divisible by $12^x$