Question

Find the greatest number that will divide \(43,\ 91,\ 183\) so as to leave the same remainder in each case.

• A. \(4\)
• B. \(7\)
• C. \(9\)
• D. \(14\)

Hint:

For “same remainder” problems: subtract the numbers pairwise to eliminate the remainder, then find the HCF of the resulting differences. This HCF is the required divisor.

Answer 1

The Correct Option is A

Step 1 (Understand the problem).
If a number \(N\) divides several numbers leaving the same} remainder \(R\), then when we subtract one number from another, the remainder is eliminated}.
This means \(N\) must exactly divide the differences of the given numbers.
So our task reduces to: \[ N = \text{HCF of all pairwise differences of the given numbers}. \] Step 2 (List the given numbers).
We are given: \(43,\ 91,\ 183\). Step 3 (Find the differences).
Subtract pairwise: \[ 91 - 43 = 48 \] \[ 183 - 91 = 92 \] \[ 183 - 43 = 140 \] So, the differences are \(48,\ 92,\ 140\). Step 4 (Find the HCF of the differences).
First, find \(\gcd(48, 9(b)\): Prime factors: \(48 = 2^4 \times 3\) \(92 = 2^2 \times 23\) Common factors: \(2^2 = 4\) \(\gcd(48, 9(b) = 4\).
Now, find \(\gcd(4, 140)\): \(140 = 2^2 \times 5 \times 7\) Common factors with \(4 = 2^2\) are \(2^2 = 4\). Hence, \(\gcd(4, 140) = 4\). Step 5 (Interpretation).
The HCF of the differences is \(4\). This means \(4\) is the largest number that divides \(43,\ 91,\ 183\) leaving the same remainder in each case. Step 6 (Verification).
If we divide each number by \(4\): \[ 43 \div 4 = 10 \ \text{remainder} \ 3 \] \[ 91 \div 4 = 22 \ \text{remainder} \ 3 \] \[ 183 \div 4 = 45 \ \text{remainder} \ 3 \] Indeed, the remainder \(3\) is the same in all cases, confirming our answer. Step 7 (Common mistake alert).
Some students mistakenly check each option directly without removing the remainder effect, or confuse with the concept of “dividing exactly” — here, the same remainder condition is key. \[ \boxed{4 \ \text{(Option (a)}} \]