I = CAN ÷ NO. Given, N = 4, O = 2, I < 10 Find C + I.

Concept

In a cryptarithmetic (letter-arithmetic) puzzle, each letter stands for one digit consistent with the given arithmetic relation. Known digits are substituted first, then a units-digit (mod 10) check on the operation — together with any digit-count constraint on the result — narrows the unknown digits before a final direct-computation check.

Application

1. Substitute the given digits: N = 4 and O = 2, so the two-digit number NO = 42.

2. Since I = CAN ÷ NO with I < 10, rearranging gives CAN = 42 × I, and CAN must be a genuine three-digit number.

3. The units digit of CAN is N = 4, so the units digit of 42 × I must be 4. Checking the units digit of 2 × I for I = 0 to 9 shows this holds only for I = 2 and I = 7.

4. Testing I = 2: 42 × 2 = 84, only two digits, so it cannot equal the three-digit CAN. This value of I is rejected.

5. Testing I = 7: 42 × 7 = 294, a three-digit number whose units digit is 4, matching N. This fixes C = 2 and confirms I = 7.

6. Therefore, C + I = 2 + 7 = 9.

Cross-check

Reversing the operation confirms consistency: 294 ÷ 42 = 7, matching the value taken for I, and the units digit of 294 is indeed 4, matching N. Both governing constraints hold, so C + I = 9 is the only consistent value.