Finding the Greatest Number with the Same Remainder

The problem asks us to find the greatest number that divides 36, 64, and 92, leaving the same remainder in each case. Let the required greatest number be \(d\) and the common remainder be \(r\).

According to the division algorithm, if a number \(N\) is divided by \(d\) and leaves a remainder \(r\), then \(N = qd + r\), where \(q\) is the quotient and \(0 \le r < d\). This can be rewritten as \(N - r = qd\). This means \((N - r)\) is perfectly divisible by \(d\).

So, for the given numbers 36, 64, and 92, we have:

• \(36 = q_1 d + r \implies 36 - r = q_1 d\)
• \(64 = q_2 d + r \implies 64 - r = q_2 d\)
• \(92 = q_3 d + r \implies 92 - r = q_3 d\)

This means \(d\) is a common divisor of \((36 - r)\), \((64 - r)\), and \((92 - r)\).

A key property when dealing with the same remainder is that if a number \(d\) divides \(A\) and \(B\) with the same remainder, then \(d\) must divide their difference \((B - A)\) exactly. This is because \((A - r)\) and \((B - r)\) are both divisible by \(d\), and the difference \((B - A) = (B - r) - (A - r)\) must also be divisible by \(d\).

Applying this property to the given numbers 36, 64, and 92, the greatest number \(d\) that divides them with the same remainder must be a common divisor of the differences between these numbers:

• Difference between 64 and 36: \(64 - 36\)
• Difference between 92 and 64: \(92 - 64\)
• Difference between 92 and 36: \(92 - 36\)

Let's calculate these differences:

• \(64 - 36 = 28\)
• \(92 - 64 = 28\)
• \(92 - 36 = 56\)

The greatest number \(d\) that divides 36, 64, and 92 with the same remainder is the Greatest Common Divisor (GCD) of these differences: GCD(28, 28, 56).

Now, let's find the GCD of 28, 28, and 56.

We can find the GCD by finding the prime factorization of each number:

• Prime factorization of 28: \(28 = 2 \times 2 \times 7 = 2^2 \times 7^1\)
• Prime factorization of 56: \(56 = 2 \times 2 \times 2 \times 7 = 2^3 \times 7^1\)

The common prime factors are 2 and 7. The lowest power of 2 common to 28 and 56 is \(2^2\). The lowest power of 7 common to 28 and 56 is \(7^1\).

GCD(28, 56) = \(2^2 \times 7^1 = 4 \times 7 = 28\).

Since GCD(28, 28, 56) is the same as GCD(28, GCD(28, 56)), we have GCD(28, 28, 56) = GCD(28, 28) = 28.

So, the greatest number that divides 36, 64, and 92 leaving the same remainder is 28.

Let's verify this by dividing 36, 64, and 92 by 28:

• \(36 \div 28\): \(36 = 1 \times 28 + 8\). Remainder = 8.
• \(64 \div 28\): \(64 = 2 \times 28 + 8\). Remainder = 8.
• \(92 \div 28\): \(92 = 3 \times 28 + 8\). Remainder = 8.

In each case, the remainder is 8. Thus, our answer is correct.

Revision Table: Greatest Number with Same Remainder

Additional Information: GCD and Number Properties

The Greatest Common Divisor (GCD), also known as the Highest Common Factor (HCF), of two or more integers is the largest positive integer that divides each of the integers without a remainder.

There are several methods to find the GCD:

• Prime Factorization Method: Find the prime factorization of each number. The GCD is the product of the lowest powers of all common prime factors.
• Euclidean Algorithm: For two numbers \(a\) and \(b\), where \(a > b\), GCD(\(a, b\)) = GCD(\(b\), \(a \pmod b\)). Repeat until the remainder is 0. The last non-zero remainder is the GCD.

In this problem, the core idea is that if a number divides multiple numbers leaving the same remainder, it must exactly divide the differences between those numbers. This is a useful property for solving problems involving common remainders.