The number of three digit PINs, in which the third digit is the sum of the first two digits, is
Part ACSIR NET September 2022pin-allows-leading-zero
Pin allows leading zero
- A.55✓
- B.9
- C.45
- D.11
The idea, in plain words
A PIN is a string of digits, not a number, so it is allowed to begin with zero. That single word changes the count, because it lets in every code starting with 0 that a three-digit number would exclude.
Solution
A PIN may begin with 0, so the first two digits a and b range over 0–9 subject only to a + b ≤ 9. The number of such pairs is C(11, 2) = 55, and each determines the third digit.
Why each option is right or wrong
Checked against 2 symbolic computations- A.Correct
True. A PIN is a string, not a number, so it may begin with 0. The first two digits a and b range over 0–9 subject only to a + b ≤ 9, and the third digit is then determined. The count of such pairs is C(11, 2) = 55 (verified). The trigger: 'PIN' and 'three-digit number' are different objects, and only the second forbids a leading zero.
- B.Wrong
False. 9 is the number of valid choices for b when a = 1, that is one row of the count rather than the whole triangle.
- C.WrongBoundary and endpoint
False, and this is the engineered near-miss: 45 is exactly the answer if the first digit is forbidden from being 0, giving 9 + 8 + ⋯ + 1 = 45 (verified). The 10 extra PINs are those starting 0, from 000 up to 099. Everything about the calculation is right except the reading of 'PIN'.
- D.Wrong
False. 11 is the number of admissible values of the digit sum, 0 through 9 plus one, rather than the number of PINs. Each sum value admits several (a, b) pairs.
1 of these 3 wrong options correspond to a named reasoning error — the same errors recur across subjects.