If the probability that a randomly chosen 6-digit number formed by using digits 1 and 8 only is a multiple of 21 is p, then 96 p is equal to _______________.

Solution

Maths sir · 4 steps

Write every 8 as 1 + 7: then each such number is already a multiple of 7, and only the test for 3 is left.

1. Total count. Each of the 6 places holds either 1 or 8, so there are 2⁶ = 64 such numbers, all equally likely.
2. Divisible by 7. Since 8 = 1 + 7, every such number equals 111111 plus 7 times the sum of the place values where an 8 sits. Because 111111 = 7 × 15873, all 64 numbers are multiples of 7.
3. Divisible by 3. If k digits are 8, the digit sum is (6-k) + 8k = 6 + 7k. This is a multiple of 3 only when k is, so k = 0, 3 or 6. That gives C(6, 0) + C(6, 3) + C(6, 6) = 1 + 20 + 1 = 22 numbers.

   p = 22/64 = 11/32

4. So. 96p = 96 × 11/32 = 3 × 11 = 33.

The common slip: 30. Counting only the numbers with three 8s and missing k = 0 (111111) and k = 6 (888888) gives p = 20/64 and 96p = 30.

The idea to keep. When a number uses only two digits, write it as a fixed base plus multiples of the difference between the digits. One divisibility test often becomes automatic, and the other turns into a simple count.

Answer from the NTA final answer key. Solution by PrepWiser.