To solve the problem where \( N \) is a four-digit number with a sum of its digits equal to 26, we first need to determine how to check divisibility by 9. According to the rule of divisibility, a number is divisible by 9 if the sum of its digits is divisible by 9. 1. Sum of Digits: The sum of the digits of \( N \) is given as 26. 2. Finding Remainder: When 26 is divided by 9, we calculate: \[ 26 \mod 9 = 8 \] This tells us that when \( N \) is taken as it is, it leaves a remainder of 8 when divided by 9. 3. Required Condition: To make the new number (after subtraction) divisible by 9, we need to subtract a number such that the resulting number gives a remainder of 0 when divided by 9. From the current remainder of 8, we would need to subtract: \[ 8 \text{ (the current remainder)} - x \equiv 0 \mod 9 \] This implies that \( x \) must equal 8 (since \( 8 - 8 = 0 \)). Thus, the smallest natural number that must be subtracted from \( N \) to make it divisible by 9 is 8. The answer to the question is thus option A) 8【6:0†source】.