What is the least number that must be subtracted from 627349 so that the remaining number is divisible by 15?

Introduction / Context:
This problem tests your knowledge of divisibility rules, particularly for 3 and 5, because a number divisible by 15 must be divisible by both 3 and 5. It also checks your ability to adjust a number by the smallest possible subtraction to meet those divisibility requirements.

Given Data / Assumptions:

• Original number N = 627349.
• We must subtract the smallest positive integer x so that N - x is divisible by 15.
• Divisible by 15 means divisible by both 3 and 5 simultaneously.

Concept / Approach:
Use divisibility rules: a number is divisible by 5 when its last digit is 0 or 5. A number is divisible by 3 when the sum of its digits is divisible by 3. We search for the smallest non negative integer x such that N - x satisfies both of these rules.

Step-by-Step Solution:
Step 1: Check divisibility by 5. The last digit of 627349 is 9. For N - x to end with 0 or 5, test small x.Step 2: If x = 4, the last digit becomes 9 - 4 = 5, which is valid for divisibility by 5.Step 3: Now check divisibility by 3. Sum of digits of N is 6 + 2 + 7 + 3 + 4 + 9 = 31.Step 4: Sum of digits of N - x will be 31 - x (because subtracting x reduces the total digit sum by x for small x). For N - x to be divisible by 3, 31 - x must be divisible by 3.Step 5: Compute 31 mod 3 = 1, so x must leave remainder 1 when divided by 3. Among small options, x = 1, 4, 7, ... satisfy this.Step 6: Combine conditions. From the divisibility by 5 check, x could be 4 or 9 (for last digit 5 or 0). From the divisibility by 3 condition, x must be congruent to 1 modulo 3. The smallest x that works for both is 4.

Verification / Alternative check:
Compute N - 4 = 627345. Last digit is 5, so it is divisible by 5.Digit sum of 627345 is 6 + 2 + 7 + 3 + 4 + 5 = 27, and 27 is divisible by 3, so 627345 is divisible by 3 and therefore by 15.

Why Other Options Are Wrong:
If x = 1, N - 1 ends in 8, so it is not divisible by 5.If x = 2, N - 2 ends in 7, again not divisible by 5.If x = 3, N - 3 ends in 6 and does not satisfy the condition for divisibility by 5.

Common Pitfalls:
Forgetting that divisibility by 15 requires both the 3 and 5 tests to hold at the same time.Only adjusting the last digit without checking the digit sum can lead to an incorrect choice.Working systematically through both conditions for small candidate values prevents these errors.

Final Answer:
The least number to be subtracted is 4.