Let A be a set of all 4-digit natural numbers whose exactly one digit is 7. Then the probability that a randomly chosen element of A leaves remainder 2 when divided by 5 is :

Solution

Maths sir · 3 steps

Remainder 2 on division by 5 means the last digit is 2 or 7, so split the count on the last digit.

1. Size of A. If the 7 is the first digit, the other three digits avoid 7: 9³ = 729 ways. If the 7 sits in one of the other 3 places, the first digit has 8 choices (not 0, not 7) and the remaining two have 9 each: 3 × 8 × 9 × 9 = 1944. So n(A) = 729 + 1944 = 2673.
2. Condition on remainder. A number leaves remainder 2 when divided by 5 exactly when its last digit is 2 or 7. Count each case separately.

   Last digit	Where the single 7 goes	Count
   7	Last place, so the first three digits avoid 7	8 × 9 × 9 = 648
   2	First place	9 × 9 = 81
   2	Second or third place	2 × 8 × 9 = 144
   Total	All three cases together	648 + 81 + 144 = 873

   P = 873/2673 = 97/297

3. So. Dividing top and bottom by 9 gives 97/297, which is option (d).

The idea to keep. Turn a remainder condition into a last-digit condition, then count case by case. Always handle the first digit separately, because it cannot be 0.

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