Understanding the Problem: Finding the Greatest Number with Same Remainder

The question asks for the greatest number that divides three given numbers (43, 91, and 183) such that the remainder is the same in all three division cases.

When a number, say $N$, divides $a$, $b$, and $c$ leaving the same remainder, say $r$, it means that $a-r$, $b-r$, and $c-r$ are exactly divisible by $N$. This implies that $N$ must be a common factor of $(a-r)$, $(b-r)$, and $(c-r)$.

A property related to this is that if $N$ divides $a-r$, $b-r$, and $c-r$, then $N$ must also divide the differences between these numbers. The differences between $a-r$, $b-r$, and $c-r$ are the same as the differences between $a$, $b$, and $c$.

• $(b-r) - (a-r) = b - a$
• $(c-r) - (b-r) = c - b$
• $(c-r) - (a-r) = c - a$

Therefore, the greatest number $N$ that divides $a$, $b$, and $c$ leaving the same remainder is the Highest Common Factor (HCF) of the differences between the numbers: $(b-a)$, $(c-b)$, and $(c-a)$.

Calculating the Differences

The given numbers are 43, 91, and 183.

• Difference between 91 and 43: $91 - 43 = 48$
• Difference between 183 and 91: $183 - 91 = 92$
• Difference between 183 and 43: $183 - 43 = 140$

We need to find the HCF of 48, 92, and 140.

Finding the HCF of 48, 92, and 140

To find the HCF, we can use the prime factorization method.

Prime factorization of 48:

\(48 = 2 \times 24 = 2 \times 2 \times 12 = 2 \times 2 \times 2 \times 6 = 2 \times 2 \times 2 \times 2 \times 3 = 2^4 \times 3^1\)

Prime factorization of 92:

\(92 = 2 \times 46 = 2 \times 2 \times 23 = 2^2 \times 23^1\)

Prime factorization of 140:

\(140 = 2 \times 70 = 2 \times 2 \times 35 = 2 \times 2 \times 5 \times 7 = 2^2 \times 5^1 \times 7^1\)

Now, we identify the common prime factors and their lowest powers present in all three factorizations:

• The prime factor 2 is present in all three numbers. The lowest power of 2 is $2^2$.
• The prime factor 3 is only in 48.
• The prime factor 23 is only in 92.
• The prime factors 5 and 7 are only in 140.

The only common prime factor is 2, and its lowest power is $2^2$.

The HCF is the product of the common prime factors raised to their lowest powers.

HCF(48, 92, 140) = \(2^2 = 4\)

Verification

Let's check if dividing 43, 91, and 183 by 4 leaves the same remainder:

• \(43 \div 4\): \(43 = 4 \times 10 + 3\). Remainder is 3.
• \(91 \div 4\): \(91 = 4 \times 22 + 3\). Remainder is 3.
• \(183 \div 4\): \(183 = 4 \times 45 + 3\). Remainder is 3.

The remainder is indeed the same (3) in all three cases.

Conclusion

The greatest number that will divide 43, 91, and 183 so as to leave the same remainder in each case is 4.