CAT 2003 Slot 2 — QA Question 20
Consider the sets Tn = {n, n + 1, n + 2, n + 3, n + 4}, where n = 1, 2, 3, … , 96.
How many of these sets contain 6 or any integral multiple thereof (i.e., any one of the numbers 6, 12, 18, ...)?
Answer & solution
Correct answer: 80
80
- B
81
- C
82
- D
83
n = 1, 2, 3, … , 96 and Tn = {n, n + 1, n + 2, n + 3, n + 4}
n could be either 6k or 6k + 1 or 6k + 2 or 6k + 3 or 6k + 4 or 6k + 5.
When n = 6k, then set Tn will definitely contain a multiple of 6 as it contains n
When n = 6k + 5, then set Tn will contain a multiple of 6 as it contains n + 1 = 6k + 6
When n = 6k + 4, then set Tn will contain a multiple of 6 as it contains n + 2 = 6k + 6
When n = 6k + 3, then set Tn will contain a multiple of 6 as it contains n + 3 = 6k + 6
When n = 6k + 2, then set Tn will contain a multiple of 6 as it contains n + 4 = 6k + 6
However, for every n = 6k + 1, the set Tn will not contain any multiple of 6There will be 16 such sets for k = 0 to 15, for which Tn will not contain a multiple of 6
∴ (96 – 16) = 80 sets contain multiples of 6
Hence, option (a).