JEE Main 2021 · 31 August · evening · Numerical answer · +4 / −0
The number of 4-digit numbers which are neither multiple of 7 nor multiple of 3 is ____________.
Answer and solution
Answer 5143
Solution
Maths sir · 3 steps
Count the numbers you do not want, the multiples of 3 or 7, and subtract them from all 9000.
- Total and method. 4-digit numbers run from 1000 to 9999, so there are 9000 of them. The multiples of k in this range number ⌊ 9999/k ⌋ - ⌊ 999/k ⌋.
Multiples of Up to 9999 Up to 999 4-digit count 3 3333 333 3000 7 1428 142 1286 21 (both 3 and 7) 476 47 429 - Remove the overlap. A multiple of 21 is counted once among the multiples of 3 and again among the multiples of 7. Subtract it once so it is counted only once.
n(3 ∪ 7) = 3000 + 1286 - 429 = 3857
- So. Numbers that are neither = 9000 - 3857 = 5143.
The common slip: 4714. Forgetting to subtract the 429 multiples of 21 gives 9000 - 4286 = 4714, because those numbers get removed twice.
The idea to keep. For "neither A nor B", find the count of "A or B" using n(A) + n(B) - n(A ∩ B) and subtract it from the total. For multiples of 3 and 7, the overlap is the multiples of their LCM, 21.
Answer from the NTA final answer key. Solution by PrepWiser.