Understanding the Problem: Finding the Greatest Number and Remainder

The question asks us to find a specific value, (x - y), where x is the greatest number that divides 1062, 1134, and 1182, leaving the same remainder, y, in each case.

When a number 'N' is divided by another number 'x' and leaves a remainder 'y', it means that \( N = qx + y \), where q is the quotient. Rearranging this, we get \( N - y = qx \). This shows that \( N - y \) is perfectly divisible by x.

In this problem, 1062, 1134, and 1182 all leave the same remainder 'y' when divided by 'x'. This means:

• \( 1062 - y \) is divisible by x
• \( 1134 - y \) is divisible by x
• \( 1182 - y \) is divisible by x

If two numbers are divisible by x, their difference is also divisible by x. The differences between these numbers will eliminate the remainder 'y':

• Difference 1: \( (1134 - y) - (1062 - y) = 1134 - 1062 = 72 \)
• Difference 2: \( (1182 - y) - (1134 - y) = 1182 - 1134 = 48 \)
• Difference 3: \( (1182 - y) - (1062 - y) = 1182 - 1062 = 120 \)

Since \( 1062 - y \), \( 1134 - y \), and \( 1182 - y \) are all divisible by x, their differences (72, 48, and 120) must also be divisible by x. The problem states that x is the greatest number that divides the original numbers with the same remainder. This implies that x is the Greatest Common Divisor (GCD), also known as the Highest Common Factor (HCF), of these differences.

Finding the Greatest Number (x): GCD Calculation

We need to find the GCD of 72, 48, and 120.

Let's find the prime factorization of each number:

• \( 72 = 2 \times 36 = 2 \times 6 \times 6 = 2 \times (2 \times 3) \times (2 \times 3) = 2^3 \times 3^2 \)
• \( 48 = 2 \times 24 = 2 \times 2 \times 12 = 2 \times 2 \times 2 \times 6 = 2 \times 2 \times 2 \times 2 \times 3 = 2^4 \times 3 \)
• \( 120 = 10 \times 12 = (2 \times 5) \times (2^2 \times 3) = 2^3 \times 3 \times 5 \)

To find the GCD, we take the lowest power of each common prime factor:

• Common prime factor 2: The lowest power is \( 2^3 \) (from 72 and 120).
• Common prime factor 3: The lowest power is \( 3^1 \) (from 48 and 120).
• Common prime factor 5: Not common to all three numbers.

So, the GCD is \( 2^3 \times 3^1 = 8 \times 3 = 24 \).

Thus, the greatest number x is 24.

\( x = 24 \)

Finding the Remainder (y)

Now that we know x = 24, we can find the remainder y by dividing any of the original numbers (1062, 1134, or 1182) by 24.

Let's divide 1062 by 24:

\( 1062 \div 24 \)

Using division:

\( 1062 = 24 \times q + y \)

We can estimate the quotient:

• \( 24 \times 40 = 960 \)
• \( 1062 - 960 = 102 \)
• Now divide 102 by 24: \( 24 \times 4 = 96 \)
• \( 102 - 96 = 6 \)

So, \( 1062 = 24 \times 44 + 6 \). The quotient is 44 and the remainder is 6.

Let's verify with another number, 1134:

\( 1134 \div 24 \)

• \( 24 \times 40 = 960 \)
• \( 1134 - 960 = 174 \)
• Now divide 174 by 24: \( 24 \times 7 = 168 \)
• \( 174 - 168 = 6 \)

So, \( 1134 = 24 \times 47 + 6 \). The remainder is 6.

The remainder y is 6.

\( y = 6 \)

Calculating the Final Value (x - y)

We have found \( x = 24 \) and \( y = 6 \).

The value we need to find is \( (x - y) \).

\( x - y = 24 - 6 \)

\( x - y = 18 \)

Summary of Steps

1. Understood that when numbers leave the same remainder 'y' upon division by 'x', the differences between these numbers are perfectly divisible by 'x'.
2. Calculated the differences between the given numbers: 72, 48, and 120.
3. Determined that the greatest number 'x' is the GCD of these differences.
4. Calculated the GCD of 72, 48, and 120 to find \( x = 24 \).
5. Divided one of the original numbers (1062) by x (24) to find the remainder \( y = 6 \). Verified with another number.
6. Calculated \( (x - y) \) which is \( 24 - 6 = 18 \).

Answer

The value of \( (x - y) \) is 18.

So, \( x = 24 \) and \( y = 6 \).

\( x - y = 24 - 6 = 18 \).

Revision Table: Greatest Number and Remainder

Additional Information: GCD and Division with Remainder

Greatest Common Divisor (GCD) / Highest Common Factor (HCF):

The GCD of two or more non-zero integers is the largest positive integer that divides the numbers without a remainder. It is a fundamental concept in number theory. Methods to find GCD include prime factorization and the Euclidean algorithm.

• Prime Factorization Method: Find the prime factorization of each number. The GCD is the product of the lowest powers of all common prime factors. As shown in this problem, \( GCD(72, 48, 120) = GCD(2^3 \times 3^2, 2^4 \times 3, 2^3 \times 3 \times 5) = 2^{\min(3,4,3)} \times 3^{\min(2,1,1)} = 2^3 \times 3^1 = 8 \times 3 = 24 \).
• Euclidean Algorithm: This method is based on the principle that the GCD of two numbers does not change if the larger number is replaced by its difference with the smaller number. More efficiently, it uses the division algorithm: \( GCD(a, b) = GCD(b, a \pmod{b}) \). For three numbers, \( GCD(a, b, c) = GCD(a, GCD(b, c)) \).

Division Algorithm and Remainder Theorem:

For any integer \( a \) and any positive integer \( b \), there exist unique integers \( q \) (quotient) and \( r \) (remainder) such that \( a = bq + r \), where \( 0 \le r < b \). In this problem, a is the original number (1062, 1134, or 1182), b is the divisor x, and r is the remainder y.

The key insight used here is that if \( a_1 \equiv y \pmod{x} \) and \( a_2 \equiv y \pmod{x} \), then \( a_1 - y \) is a multiple of x and \( a_2 - y \) is a multiple of x. Consequently, \((a_1 - y) - (a_2 - y) = a_1 - a_2 \) is also a multiple of x. This means x must be a divisor of the difference between \( a_1 \) and \( a_2 \). Since x is the *greatest* such number, it must be the GCD of all possible differences between the original numbers.