To determine which value of $x$ makes the 8-digit number $4x3074yy$ not divisible by 24, we need to break down the divisibility rules for 24. Since $24 = 8 \times 3$, we need to check two divisibility conditions:

1. Divisibility by 8: The last three digits of the number must be divisible by 8.
2. Divisibility by 3: The sum of all digits must be divisible by 3.

Step 1: Divisibility by 8

For the number to be divisible by 8, the last three digits, $74yy$, must be divisible by 8. Thus, the number formed by the last three digits, $74yy$, must satisfy the condition for divisibility by 8.

Let's consider different values of $y$ and check if $74yy$ is divisible by 8. We’ll try $y$ values from 0 to 9:

• For $y = 0$, $740 \div 8 = 92.5$ (not divisible by 8)
• For $y = 1$, $741 \div 8 = 92.625$ (not divisible by 8)
• For $y = 2$, $742 \div 8 = 92.75$ (not divisible by 8)
• For $y = 3$, $743 \div 8 = 92.875$ (not divisible by 8)
• For $y = 4$, $744 \div 8 = 93$ (divisible by 8)
• For $y = 5$, $745 \div 8 = 93.125$ (not divisible by 8)
• For $y = 6$, $746 \div 8 = 93.25$ (not divisible by 8)
• For $y = 7$, $747 \div 8 = 93.375$ (not divisible by 8)
• For $y = 8$, $748 \div 8 = 93.5$ (not divisible by 8)
• For $y = 9$, $749 \div 8 = 93.625$ (not divisible by 8)

Thus, $74yy$ is divisible by 8 only when $y = 4$.

Step 2: Divisibility by 3

Now, we need to check if the sum of the digits is divisible by 3. The sum of the digits is: $4 + x + 3 + 0 + 7 + 4 + y + y = 18 + x + 2y$ For divisibility by 3, $18 + x + 2y$ must be divisible by 3.

We know that $y = 4$ works for divisibility by 8, so we substitute $y = 4$ into the sum: $18 + x + 2(4) = 18 + x + 8 = 26 + x$ For divisibility by 3, $26 + x$ must be divisible by 3. We check different values of $x$ to find which do not work:

• For $x = 0$, $26 + 0 = 26$ (not divisible by 3)
• For $x = 1$, $26 + 1 = 27$ (divisible by 3)
• For $x = 2$, $26 + 2 = 28$ (not divisible by 3)
• For $x = 3$, $26 + 3 = 29$ (not divisible by 3)
• For $x = 4$, $26 + 4 = 30$ (divisible by 3)
• For $x = 5$, $26 + 5 = 31$ (not divisible by 3)
• For $x = 6$, $26 + 6 = 32$ (not divisible by 3)
• For $x = 7$, $26 + 7 = 33$ (divisible by 3)
• For $x = 8$, $26 + 8 = 34$ (not divisible by 3)
• For $x = 9$, $26 + 9 = 35$ (not divisible by 3)

Conclusion:

The values of $x$ that make the number divisible by 24 are $x = 1, 4, 7$, and the values that do not work are $x = 0, 2, 3, 5, 6, 8, 9$.

Thus, 0 is not a possible value for $x$.

Would you like any further details or clarifications? Here are some related questions:

1. How do you apply the divisibility rule for 8 in other cases?
2. How can you determine divisibility by 24 when more digits are involved?
3. What happens if the number is only divisible by 3 but not 8?
4. How do you quickly check divisibility by 3 for larger numbers?
5. Can you apply this method for numbers with more than two variables like $x$ and $y$?

Tip: When dealing with divisibility by 3, remember the rule that the sum of the digits of the number must be divisible by 3.