Final answer:
To find the smaller integer for which the difference of the reciprocals of two consecutive integers equals 1/420, we solve the equation 1/n - 1/(n+1) = 1/420, which leads us to conclude that the pair of consecutive integers is (20,21), making 20 the smaller integer.
Step-by-step explanation:
The student is asking to find the smaller of two positive consecutive integers where the difference of their reciprocals is 1/420. Let's assume the smaller integer is 'n'; therefore, the next consecutive integer would be 'n+1'. The reciprocals of these integers would be 1/n and 1/(n+1), respectively. According to the problem, the difference of these reciprocals is given by:
1/n - 1/(n+1) = 1/420
To solve for 'n', we will find a common denominator, which is n*(n+1), and rewrite the equation as:
((n+1) - n) / (n*(n+1)) = 1/420
This simplifies to:
1 / (n*(n+1)) = 1/420
From this, we can deduce that n*(n+1) = 420. Factorizing 420 gives us pairs of factors such as (1,420), (2,210), (3,140), (4,105), (5,84), (6,70), (7,60), (10,42), (12,35), (14,30), (15,28), and (20,21). Among these pairs, only the pair (20,21) consists of consecutive integers. Therefore, the smaller integer is 20.