Finding Pairs (A, B) for Divisibility by 9

The question asks us to find the number of possible pairs of digits (A, B) such that the six-digit number 479865AB is divisible by 9, and the last digit, B, is an odd number.

Understanding Divisibility by 9

A key concept here is the rule for divisibility by 9. A number is divisible by 9 if and only if the sum of its digits is divisible by 9.

Analyzing the Number 479865AB

The number is 479865AB. This means the digits are 4, 7, 9, 8, 6, 5, A, and B.

Let's find the sum of the known digits:

$$ \text{Sum of known digits} = 4 + 7 + 9 + 8 + 6 + 5 $$

$$ \text{Sum of known digits} = 39 $$

The sum of all digits in the number 479865AB is $39 + A + B$.

Applying the Divisibility Rule

For the number 479865AB to be divisible by 9, the sum of its digits, $39 + A + B$, must be a multiple of 9. Multiples of 9 are 9, 18, 27, 36, 45, 54, 63, and so on.

A and B are single digits, which means they can take any integer value from 0 to 9.

$$ 0 \le A \le 9 $$

$$ 0 \le B \le 9 $$

Therefore, the sum $A + B$ can range from $0 + 0 = 0$ to $9 + 9 = 18$.

So, the sum of all digits, $39 + A + B$, can range from $39 + 0 = 39$ to $39 + 18 = 57$.

The multiples of 9 between 39 and 57 (inclusive) are 45 and 54.

This gives us two possible cases for the sum of digits:

1. $39 + A + B = 45$
2. $39 + A + B = 54$

Considering the Last Digit is Odd

We are also given that the last digit, B, must be an odd number. The odd digits are 1, 3, 5, 7, and 9.

Case 1: Sum of Digits is 45

If $39 + A + B = 45$, then $A + B = 45 - 39 = 6$.

Now, we find pairs (A, B) such that $A + B = 6$ and B is an odd digit (1, 3, 5, 7, 9). Remember A must be a digit between 0 and 9.

• If $B = 1$, then $A + 1 = 6 \implies A = 5$. Pair (A, B) is (5, 1). This is valid as A = 5 is a digit.
• If $B = 3$, then $A + 3 = 6 \implies A = 3$. Pair (A, B) is (3, 3). This is valid as A = 3 is a digit.
• If $B = 5$, then $A + 5 = 6 \implies A = 1$. Pair (A, B) is (1, 5). This is valid as A = 1 is a digit.
• If $B = 7$, then $A + 7 = 6 \implies A = -1$. This is not valid as A must be a digit $\ge 0$.
• If $B = 9$, then $A + 9 = 6 \implies A = -3$. This is not valid as A must be a digit $\ge 0$.

From Case 1, the possible pairs (A, B) are (5, 1), (3, 3), and (1, 5). There are 3 such pairs.

Case 2: Sum of Digits is 54

If $39 + A + B = 54$, then $A + B = 54 - 39 = 15$.

Now, we find pairs (A, B) such that $A + B = 15$ and B is an odd digit (1, 3, 5, 7, 9). Remember A must be a digit between 0 and 9.

• If $B = 1$, then $A + 1 = 15 \implies A = 14$. This is not valid as A must be a digit $\le 9$.
• If $B = 3$, then $A + 3 = 15 \implies A = 12$. This is not valid.
• If $B = 5$, then $A + 5 = 15 \implies A = 10$. This is not valid.
• If $B = 7$, then $A + 7 = 15 \implies A = 8$. Pair (A, B) is (8, 7). This is valid as A = 8 is a digit.
• If $B = 9$, then $A + 9 = 15 \implies A = 6$. Pair (A, B) is (6, 9). This is valid as A = 6 is a digit.

From Case 2, the possible pairs (A, B) are (8, 7) and (6, 9). There are 2 such pairs.

Total Number of Possible Pairs (A, B)

Combining the valid pairs from both cases:

Total pairs = (Pairs from Case 1) + (Pairs from Case 2)

Total pairs = 3 + 2 = 5.

There are 5 possible pairs (A, B) such that the number 479865AB is divisible by 9 and B is an odd digit.

Revision Table: Divisibility by 9 and Odd Last Digit

Additional Information: Understanding Divisibility Rules

Divisibility rules are shortcuts to check if a number is exactly divisible by another number without performing the division. The rule for 9 is based on the sum of digits.

• Divisibility by 2: A number is divisible by 2 if its last digit is even (0, 2, 4, 6, 8).
• Divisibility by 3: A number is divisible by 3 if the sum of its digits is divisible by 3.
• Divisibility by 5: A number is divisible by 5 if its last digit is 0 or 5.
• Divisibility by 6: A number is divisible by 6 if it is divisible by both 2 and 3.
• Divisibility by 10: A number is divisible by 10 if its last digit is 0.

Understanding these rules helps solve problems involving digits and number properties efficiently, like the one involving finding pairs (A, B) for divisibility by 9 with an odd last digit.