Difficulty: Medium
Correct Answer: 270
Explanation:
Introduction / Context:
This question is a classic application of the division algorithm and basic algebra. You are given the difference between two numbers and the result of dividing the larger by the smaller (including quotient and remainder). From this information, you need to determine the smaller number. Such questions test your understanding of how quotient, remainder, dividend and divisor relate to each other, which is very important in number system problems.
Given Data / Assumptions:
Concept / Approach:
The division algorithm states that for any integers dividend L, divisor S and remainder R with 0 ≤ R < S, we can write L = S * Q + R, where Q is the quotient. Here, Q = 6 and R = 15, so L = 6S + 15. We also know that the difference L − S equals 1365. By substituting L from the division equation into the difference equation, we can solve a simple linear equation to find S, the smaller number.
Step-by-Step Solution:
Step 1: From the division algorithm, write L = 6S + 15.
Step 2: From the difference condition, L − S = 1365.
Step 3: Substitute L = 6S + 15 into L − S = 1365, giving (6S + 15) − S = 1365.
Step 4: Simplify the equation: 6S + 15 − S = 5S + 15. So we have 5S + 15 = 1365.
Step 5: Subtract 15 from both sides: 5S = 1365 − 15 = 1350.
Step 6: Divide both sides by 5: S = 1350 / 5 = 270. So the smaller number is 270.
Verification / Alternative check:
Once we get S = 270, we can find L using L = 6S + 15 = 6 * 270 + 15 = 1620 + 15 = 1635. Check the difference: L − S = 1635 − 270 = 1365, which matches the given difference. Also check the division: 1635 divided by 270 gives quotient 6 and remainder 15 because 270 * 6 = 1620 and 1635 − 1620 = 15. Both conditions are satisfied, confirming that 270 is correct.
Why Other Options Are Wrong:
Option 1270 and 720: If these were the smaller number, substituting into the equations does not produce a difference of 1365 and the given quotient and remainder.
Option 350 and 285: Similar substitutions for these values do not satisfy both the difference and the division conditions simultaneously.
Common Pitfalls:
A frequent mistake is to miswrite the division equation, for example as L = 6 + S + 15 instead of L = 6S + 15. Another error is to set up the difference equation incorrectly, such as S − L = 1365 instead of L − S. When dealing with quotient and remainder problems, always write the correct division algorithm formula and keep track of which number is larger. Double checking by substituting back is a good way to avoid algebra slips.
Final Answer:
The smaller number is 270.
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