Find the greatest number that can divide 43, 91, and 183 in such a way that the remainder left in each case is the same. What is that greatest number?

Introduction / Context:
This question explores an important property of equal remainders. If several numbers leave the same remainder when divided by a positive integer d, then the differences between those numbers are divisible by d. We can therefore convert the problem into finding the HCF of these differences.

Given Data / Assumptions:

• Numbers: 43, 91, 183
• Each leaves the same remainder when divided by d
• We need the greatest possible value of d

Concept / Approach:
If each number leaves the same remainder r when divided by d, then:
43 = d * q1 + r91 = d * q2 + r183 = d * q3 + r
Subtracting these equations pairwise removes r, so d divides the differences:
d divides (91 - 43), (183 - 91), and (183 - 43)
Thus, d must be a common divisor of these differences, and the greatest such d is their HCF.

Step-by-Step Solution:
Step 1: Compute the differences.91 - 43 = 48183 - 91 = 92183 - 43 = 140Step 2: Find gcd(48, 92, 140).First, gcd(48, 92): 92 mod 48 = 44, 48 mod 44 = 4, 44 mod 4 = 0, so gcd(48, 92) = 4.Next, gcd(4, 140) = 4, since 140 mod 4 = 0.Step 3: Therefore, the greatest possible d is 4.

Verification / Alternative check:
Check with d = 4: 43 mod 4 = 3, 91 mod 4 = 3, and 183 mod 4 = 3. All three numbers leave the same remainder, so d = 4 is valid. Any larger value would need to divide all the differences 48, 92, and 140, but the HCF of those differences is 4, so no larger value can work.

Why Other Options Are Wrong:
7: 48 is not divisible by 7.9: 92 is not divisible by 9.11: 48 is not divisible by 11.13: 92 is not divisible by 13.

Common Pitfalls:
Trying to take the HCF of the original numbers instead of their differences.Not realizing that equal remainders imply divisibility of pairwise differences.Checking only one difference instead of verifying divisibility for all differences.

Final Answer:
4