A taxi driver states that his cab number is divisible by numbers 2, 3, 4, 5…

Solution: find the number N that leaves remainder 1 when divided by 2, 3, 4, 5, and 6, and is exactly divisible by 11.

Step 1: Compute the least common multiple (LCM) of 2, 3, 4, 5, and 6.

• LCM(2,3,4,5,6) = 60 (since 60 = 2^2 × 3 × 5).

Step 2: Represent the required numbers.

• Any number that leaves remainder 1 when divided by each of 2,3,4,5,6 must be of the form N = 60k + 1.

Step 3: Impose divisibility by 11 and solve for k.

• Require 60k + 1 ≡ 0 (mod 11). This is equivalent to 60k ≡ −1 ≡ 10 (mod 11).

• Compute 60 mod 11 = 5, so 5k ≡ 10 (mod 11).

• The multiplicative inverse of 5 modulo 11 is 9 (since 5 × 9 ≡ 1 mod 11). Multiply both sides by 9: k ≡ 9 × 10 ≡ 90 ≡ 2 (mod 11).

• Take the smallest positive k = 2, giving N = 60 × 2 + 1 = 121.

Step 4: Verify the result.

• 121 − 1 = 120 is divisible by 2, 3, 4, 5 and 6, so 121 leaves remainder 1 upon division by each of them.

• 121 is divisible by 11 (121 ÷ 11 = 11), so it satisfies the divisibility-by-11 condition.

Why the other given numbers fail:

• 61: Although 61 = 60 × 1 + 1 so it leaves remainder 1 for the divisors 2,3,4,5,6, it is not divisible by 11 (61 ÷ 11 leaves remainder).

• 91: 91 does not leave remainder 1 for every divisor listed (for example, 91 ÷ 4 leaves remainder 3), so it fails the remainder condition.

• 145: 145 is divisible by 5 (remainder 0), so it does not leave remainder 1 when divided by 5.

Answer: 121