Divisibility Rules for Solving Number Problems

The problem asks us to find the value of \( (x^2 + y) \) for the largest possible value of \(y\), given that the 7-digit number \(x8942y4\) is divisible by 56. We are told that \(x\) and \(y\) are natural numbers.

A number is divisible by 56 if and only if it is divisible by both 7 and 8. Therefore, the 7-digit number \(x8942y4\) must satisfy the divisibility rules for both 7 and 8.

Checking Divisibility by 8

The divisibility rule for 8 states that a number is divisible by 8 if the number formed by its last three digits is divisible by 8.

In the number \(x8942y4\), the last three digits are \(2y4\). So, the number \(2y4\) must be divisible by 8. Since \(y\) is a natural number, \(y\) can be any digit from 1 to 9.

Let's test the possible values for \(y\):

• If \(y = 1\), the number is 214. \(214 \div 8 = 26\) with a remainder. Not divisible by 8.
• If \(y = 2\), the number is 224. \(224 \div 8 = 28\). Divisible by 8.
• If \(y = 3\), the number is 234. \(234 \div 8 = 29\) with a remainder. Not divisible by 8.
• If \(y = 4\), the number is 244. \(244 \div 8 = 30\) with a remainder. Not divisible by 8.
• If \(y = 5\), the number is 254. \(254 \div 8 = 31\) with a remainder. Not divisible by 8.
• If \(y = 6\), the number is 264. \(264 \div 8 = 33\). Divisible by 8.
• If \(y = 7\), the number is 274. \(274 \div 8 = 34\) with a remainder. Not divisible by 8.
• If \(y = 8\), the number is 284. \(284 \div 8 = 35\) with a remainder. Not divisible by 8.
• If \(y = 9\), the number is 294. \(294 \div 8 = 36\) with a remainder. Not divisible by 8.

The possible values for \(y\) that make \(2y4\) divisible by 8 are 2 and 6.

The problem asks for the largest value of \(y\). Comparing 2 and 6, the largest value is \(y = 6\). So we will proceed with \(y = 6\).

Checking Divisibility by 7

Now, with \(y = 6\), the 7-digit number becomes \(x894264\). This number must be divisible by 7.

We can use a divisibility rule for 7. One method is to repeatedly subtract twice the last digit from the number formed by the remaining digits until we get a small number whose divisibility by 7 is easy to check.

• Start with the number \(x894264\).
• Take the number formed by the first six digits: \(x89426\). Subtract twice the last digit (4): \(x89426 - 2 \times 4 = x89426 - 8 = x89418\).
• Take the number formed by the first five digits: \(x8941\). Subtract twice the last digit (8): \(x8941 - 2 \times 8 = x8941 - 16 = x8925\).
• Take the number formed by the first four digits: \(x892\). Subtract twice the last digit (5): \(x892 - 2 \times 5 = x892 - 10 = x882\).
• Take the number formed by the first three digits: \(x88\). Subtract twice the last digit (2): \(x88 - 2 \times 2 = x88 - 4 = x84\).

For the original number \(x894264\) to be divisible by 7, the resulting number \(x84\) must be divisible by 7.

Here, \(x\) is a natural number and represents a single digit, so \(x\) can be any digit from 1 to 9.

Let's test the possible values for \(x\) in \(x84\):

• If \(x = 1\), the number is 184. \(184 \div 7 = 26\) with a remainder. Not divisible by 7.
• If \(x = 2\), the number is 284. \(284 \div 7 = 40\) with a remainder. Not divisible by 7.
• If \(x = 3\), the number is 384. \(384 \div 7 = 54\) with a remainder. Not divisible by 7.
• If \(x = 4\), the number is 484. \(484 \div 7 = 69\) with a remainder. Not divisible by 7.
• If \(x = 5\), the number is 584. \(584 \div 7 = 83\) with a remainder. Not divisible by 7.
• If \(x = 6\), the number is 684. \(684 \div 7 = 97\) with a remainder. Not divisible by 7.
• If \(x = 7\), the number is 784. \(784 \div 7 = 112\). Divisible by 7.
• If \(x = 8\), the number is 884. \(884 \div 7 = 126\) with a remainder. Not divisible by 7.
• If \(x = 9\), the number is 984. \(984 \div 7 = 140\) with a remainder. Not divisible by 7.

The only natural number value for \(x\) that makes \(x84\) divisible by 7 is \(x = 7\).

Calculating (x² + y)

We found the largest value of \(y\) that satisfies the divisibility by 8 condition is \(y = 6\). For this value of \(y\), we found the value of \(x\) that satisfies the divisibility by 7 condition is \(x = 7\).

Now we need to calculate the value of \( (x^2 + y) \).

Substitute the values \(x = 7\) and \(y = 6\):

\( x^2 + y = 7^2 + 6 \)

\( 7^2 = 49 \)

\( x^2 + y = 49 + 6 \)

\( x^2 + y = 55 \)

Thus, the value of \( (x^2 + y) \) for the largest value of \(y\) is 55.

Revision Table: Key Concepts

Additional Information: Understanding Divisibility

Divisibility rules are shortcuts to determine if a number is divisible by another number without performing the actual division. They are based on the properties of numbers and place values.

For composite numbers like 56, we can break down the divisibility rule into rules for its coprime factors. Since \(56 = 7 \times 8\) and \(gcd(7, 8) = 1\), a number is divisible by 56 if and only if it is divisible by both 7 and 8.

The divisibility rule for 8 focuses on the last three digits because \(1000\) is divisible by 8 (\(1000 \div 8 = 125\)). Any number \(N\) can be written as \(1000 \times Q + R\), where \(R\) is the number formed by the last three digits. If \(R\) is divisible by 8, then \(N\) is divisible by 8.

The divisibility rule for 7 used in the solution is one of several methods. Another common method involves grouping digits in threes from the right, alternatingly adding and subtracting the groups (with appropriate sign multipliers for larger numbers), but the method used above is simpler for checking smaller numbers derived from the larger one.

Understanding these divisibility rules is crucial for solving problems involving number properties in various mathematical and quantitative aptitude tests.