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The problem asks to find the number of possible values of \( N = 5a623b \), where \( a \) and \( b \) are single-digit whole numbers, given that \( N \) is divisible by 44.

Since \( 44 = 4 \times 11 \), the number \( N \) must be divisible by both 4 and 11.

### Step 1: Divisibility rule for 4
For \( N \) to be divisible by 4, the last two digits of the number must form a number divisible by 4. The last two digits of \( N \) are \( 23b \), so the number formed by these digits, \( 23b \), must be divisible by 4.

Let’s check possible values of \( b \) to see when \( 23b \) is divisible by 4:
- \( 230 \div 4 = 57.5 \) (not divisible)
- \( 231 \div 4 = 57.75 \) (not divisible)
- \( 232 \div 4 = 58 \) (divisible)
- \( 233 \div 4 = 58.25 \) (not divisible)
- \( 234 \div 4 = 58.5 \) (not divisible)
- \( 235 \div 4 = 58.75 \) (not divisible)
- \( 236 \div 4 = 59 \) (divisible)
- \( 237 \div 4 = 59.25 \) (not divisible)
- \( 238 \div 4 = 59.5 \) (not divisible)
- \( 239 \div 4 = 59.75 \) (not divisible)

So, \( b = 2 \) or \( b = 6 \) satisfies the condition for divisibility by 4.

### Step 2: Divisibility rule for 11
For a number to be divisible by 11, the alternating sum of its digits must be divisible by 11. The digits of \( N \) are \( 5, a, 6, 2, 3, b \).

The alternating sum is:
\[
S = 5 - a + 6 - 2 + 3 - b = 12 - a - b
\]
For divisibility by 11, \( 12 - a - b \) must be divisible by 11. Thus, we need:
\[
12 - a - b \equiv 0 \pmod{11} \quad \text{or} \quad 12 - a - b \equiv 11 \pmod{11}
\]
This gives us the condition:
\[
12 - a - b = 0 \quad \text{or} \quad 12 - a - b = 11
\]
We now solve this for \( b = 2 \) and \( b = 6 \).

### Case 1: \( b = 2 \)
\[
12 - a - 2 = 0 \implies 10 - a = 0 \implies a = 10 \quad (\text{not possible since } a \text{ must be a single-digit})
\]
\[
12 - a - 2 = 11 \implies 10 - a = 11 \implies a = -1 \quad (\text{not possible since } a \text{ must be non-negative})
\]

### Case 2: \( b = 6 \)
\[
12 - a - 6 = 0 \implies 6 - a = 0 \implies a = 6
\]
\[
12 - a - 6 = 11 \implies 6 - a = 11 \implies a = -5 \quad (\text{not possible since } a \text{ must be non-negative})
\]

Thus, the only valid solution is \( a = 6 \) and \( b = 6 \).

### Conclusion:
There is only **one** possible value for \( N \). The answer is **1**.

Would you like more details or have any further questions? Here are some related questions for further practice:
1. What is the divisibility rule for 8?
2. How can you verify if a number is divisible by both 5 and 6?
3. How many numbers between 1000 and 2000 are divisible by 44?
4. What happens to the divisibility rules for larger primes like 13 or 17?
5. Can the same number have multiple different digit arrangements and still be divisible by 44?

**Tip:** Always check the prime factorization of divisibility conditions carefully to break complex problems into smaller ones.

Given that, N = 5a623b, is divisible by 44. Find the number of possible values of N, where a and b are single-digit whole numbers.

Learn how to find the possible values of N = 5a623b, divisible by 44. This problem explores divisibility rules for 4 and 11, using modular arithmetic to narrow down possible digit values for a and b. Suitable for Grades 9-10.

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