How many pairs (A, B) are possible in the number 479865AB if the number is divisible by 9 and it is given that the last digit of the number is odd?
5
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:
- \(39 + A + B = 45\)
- \(39 + A + B = 54\)
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
| Condition | Requirement | Analysis |
|---|---|---|
| Divisibility by 9 | Sum of digits must be a multiple of 9 | \(4+7+9+8+6+5+A+B = 39+A+B\) must be a multiple of 9 (45 or 54) |
| Last digit is odd | B must be 1, 3, 5, 7, or 9 | Possible values for B are restricted to {1, 3, 5, 7, 9} |
| Possible sums (A+B) | Derived from divisibility rule | \(A+B=6\) (from \(39+A+B=45\)) or \(A+B=15\) (from \(39+A+B=54\)) |
| Valid pairs (A,B) for A+B=6 and B odd | A is a digit (0-9) | (5,1), (3,3), (1,5) - Total 3 pairs |
| Valid pairs (A,B) for A+B=15 and B odd | A is a digit (0-9) | (8,7), (6,9) - Total 2 pairs |
| Total pairs (A,B) | Sum of valid pairs from both cases | 3 + 2 = 5 pairs |
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.
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