Understanding the Problem: Divisibility of a 5-Digit Number

We are given a 5-digit number, 688xy, where x and y are digits ranging from 0 to 9. This number is stated to be divisible by 3, 7, and 11. Our goal is to find the value of the expression $(5x + 3y)$.

For a number to be divisible by 3, 7, and 11, it must be divisible by their Least Common Multiple (LCM). Since 3, 7, and 11 are prime numbers, their LCM is simply their product.

$\text{LCM}(3, 7, 11) = 3 \times 7 \times 11 = 231$

So, the number 688xy must be divisible by 231.

Applying Divisibility Rules

Let's use the divisibility rule for 11 first, as it helps establish a relationship between the digits x and y.

Divisibility Rule for 11

For a number to be divisible by 11, the alternating sum of its digits, starting from the rightmost digit, must be divisible by 11.

For the number 688xy, the alternating sum is:

$y - x + 8 - 8 + 6$

This sum simplifies to $y - x + 6$. For 688xy to be divisible by 11, $(y - x + 6)$ must be a multiple of 11.

Since x and y are digits (0-9), the value of $(y - x)$ can range from $0 - 9 = -9$ to $9 - 0 = 9$.

Therefore, $(y - x + 6)$ can range from $-9 + 6 = -3$ to $9 + 6 = 15$.

The only multiple of 11 within the range [-3, 15] is 11 itself.

So, we must have:

$y - x + 6 = 11$

$y - x = 11 - 6$

$y - x = 5$

This gives us the relationship $y = x + 5$. Based on this relationship and the fact that x and y are digits (0-9), we can list the possible pairs of (x, y):

• If $x = 0$, $y = 0 + 5 = 5$. Pair: (0, 5). Number: 68805.
• If $x = 1$, $y = 1 + 5 = 6$. Pair: (1, 6). Number: 68816.
• If $x = 2$, $y = 2 + 5 = 7$. Pair: (2, 7). Number: 68827.
• If $x = 3$, $y = 3 + 5 = 8$. Pair: (3, 8). Number: 68838.
• If $x = 4$, $y = 4 + 5 = 9$. Pair: (4, 9). Number: 68849.
• If $x \ge 5$, $y$ would be $\ge 10$, which is not a single digit. So, these are all possible pairs.

Verifying with Divisibility by 3 and 7 (or 21)

Now we need to check which of these numbers are also divisible by 3 and 7. Equivalently, we check which number is divisible by 21 (LCM of 3 and 7).

Let's test each possible number:

• Number 68805:
  • Divisibility by 3: Sum of digits = $6+8+8+0+5 = 27$. 27 is divisible by 3. (Pass)
  • Divisibility by 7: Let's use the rule. Remove the last digit (5), multiply by 2 (10), and subtract from the remaining number (6880). $6880 - 10 = 6870$. Repeat: $687 - 2 \times 0 = 687$. Repeat: $68 - 2 \times 7 = 68 - 14 = 54$. 54 is not divisible by 7. (Fail)
• Number 68816:
  • Divisibility by 3: Sum of digits = $6+8+8+1+6 = 29$. 29 is not divisible by 3. (Fail)
• Number 68827:
  • Divisibility by 3: Sum of digits = $6+8+8+2+7 = 31$. 31 is not divisible by 3. (Fail)
• Number 68838:
  • Divisibility by 3: Sum of digits = $6+8+8+3+8 = 33$. 33 is divisible by 3. (Pass)
  • Divisibility by 7: $6883 - 2 \times 8 = 6883 - 16 = 6867$. Repeat: $686 - 2 \times 7 = 686 - 14 = 672$. Repeat: $67 - 2 \times 2 = 67 - 4 = 63$. 63 is divisible by 7 ($63 = 7 \times 9$). (Pass)
  • Divisibility by 11: We already know $y-x+6 = 8-3+6 = 5+6=11$, which is divisible by 11. (Pass)

  Since 68838 is divisible by 3, 7, and 11, this means $x=3$ and $y=8$ is the correct pair of digits.

• Number 68849:
  • Divisibility by 3: Sum of digits = $6+8+8+4+9 = 35$. 35 is not divisible by 3. (Fail)

Calculating the Expression (5x + 3y)

We found that $x=3$ and $y=8$. Now we can calculate the value of $(5x + 3y)$.

Value $= (5 \times 3) + (3 \times 8)$

Value $= 15 + 24$

Value $= 39$

Final Answer

The value of $(5x + 3y)$ is 39.

Revision Table: Key Concepts Reviewed

Additional Information: Solving Divisibility Problems

Problems involving divisibility rules for composite numbers (like 6, 10, 12, etc.) can often be solved by breaking them down into prime factors. For example, a number divisible by 6 must be divisible by both 2 and 3. A number divisible by 12 must be divisible by both 3 and 4.

When a number is divisible by multiple numbers that are coprime (have no common factors other than 1), it must be divisible by their product. For example, 3, 7, and 11 are pairwise coprime, so a number divisible by all three must be divisible by their product, 231.

Using the divisibility rules for individual prime factors (like 3, 7, 11 in this case) is often easier than checking divisibility by the LCM directly, especially when trying to find unknown digits.

In this problem, applying the divisibility rule for 11 first was helpful as it narrowed down the possibilities for the pair of digits (x, y) significantly, making it easier to test the remaining conditions (divisibility by 3 and 7).