Solving for Possible (X, Y) Pairs in a 5-Digit Number Divisible by 3

The problem asks us to find the number of possible pairs of digits (X, Y) such that the five-digit number XY235 is divisible by 3, given the constraint that X and Y are digits satisfying X + Y ≤ 5.

Understanding Divisibility by 3

A fundamental rule in number theory is that a number is divisible by 3 if and only if the sum of its digits is divisible by 3.

For the number XY235, the digits are X, Y, 2, 3, and 5. The sum of these digits is:

Sum of digits \(= X + Y + 2 + 3 + 5 = X + Y + 10\)

For XY235 to be divisible by 3, the sum of its digits must be divisible by 3. So, \(X + Y + 10\) must be divisible by 3.

Applying the Divisibility Condition

If \(X + Y + 10\) is divisible by 3, it means:

\(X + Y + 10 \equiv 0 \pmod{3}\)

Since \(10 \equiv 1 \pmod{3}\), we can rewrite the condition as:

\(X + Y + 1 \equiv 0 \pmod{3}\)

This simplifies to:

\(X + Y \equiv -1 \pmod{3}\)

Since \(-1 \equiv 2 \pmod{3}\), the condition is:

\(X + Y \equiv 2 \pmod{3}\)

This means that the sum \(X + Y\) must be a number that leaves a remainder of 2 when divided by 3. Possible values for \(X + Y\) could be 2, 5, 8, 11, 14, etc.

Considering the Constraint on X and Y

We are also given the constraint that \(X + Y \le 5\). Combining this with the condition \(X + Y \equiv 2 \pmod{3}\), the possible values for the sum \(X + Y\) are restricted to:

• \(X + Y = 2\) (since \(2 \le 5\) and \(2 \equiv 2 \pmod{3}\))
• \(X + Y = 5\) (since \(5 \le 5\) and \(5 \equiv 2 \pmod{3}\))

Any other value for \(X + Y\) that satisfies \(X + Y \equiv 2 \pmod{3}\) (like 8, 11, etc.) would violate the constraint \(X + Y \le 5\).

Identifying Possible Pairs (X, Y)

We need to find pairs of digits (X, Y) that satisfy either \(X + Y = 2\) or \(X + Y = 5\). Remember that X and Y must be digits from 0 to 9. Additionally, since XY235 is a five-digit number, X must be the first digit, which means X cannot be 0.

So, X must be in \(\{1, 2, 3, 4, 5, 6, 7, 8, 9\}\), and Y must be in \(\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}\).

Case 1: \(X + Y = 2\)

We list pairs (X, Y) where X ≠ 0 and Y is a digit, summing to 2:

• If X = 1, then Y = \(2 - 1 = 1\). (1, 1) - Valid (X=1 ≠ 0, Y=1 is a digit).
• If X = 2, then Y = \(2 - 2 = 0\). (2, 0) - Valid (X=2 ≠ 0, Y=0 is a digit).
• X cannot be greater than 2, as Y would be negative.

Possible pairs for \(X + Y = 2\) are (1, 1) and (2, 0). There are 2 such pairs.

Case 2: \(X + Y = 5\)

We list pairs (X, Y) where X ≠ 0 and Y is a digit, summing to 5:

• If X = 1, then Y = \(5 - 1 = 4\). (1, 4) - Valid (X=1 ≠ 0, Y=4 is a digit).
• If X = 2, then Y = \(5 - 2 = 3\). (2, 3) - Valid (X=2 ≠ 0, Y=3 is a digit).
• If X = 3, then Y = \(5 - 3 = 2\). (3, 2) - Valid (X=3 ≠ 0, Y=2 is a digit).
• If X = 4, then Y = \(5 - 4 = 1\). (4, 1) - Valid (X=4 ≠ 0, Y=1 is a digit).
• If X = 5, then Y = \(5 - 5 = 0\). (5, 0) - Valid (X=5 ≠ 0, Y=0 is a digit).
• X cannot be greater than 5, as Y would be negative.

Possible pairs for \(X + Y = 5\) are (1, 4), (2, 3), (3, 2), (4, 1), and (5, 0). There are 5 such pairs.

Total Number of Possible Pairs

The total number of possible pairs (X, Y) is the sum of the number of pairs from Case 1 and Case 2.

Total pairs = (Pairs from \(X + Y = 2\)) + (Pairs from \(X + Y = 5\))

Total pairs = \(2 + 5 = 7\)

There are 7 possible pairs of values for (X, Y) that satisfy all the given conditions.

The total count of distinct valid (X, Y) pairs is 7.

Revision Table: Key Concepts

Additional Information: Divisibility Rules and Digit Puzzles

This problem combines number theory concepts (divisibility rules) with combinatorial aspects (counting pairs of digits under constraints). Understanding divisibility rules is crucial for solving many quantitative problems.

• Divisibility by 2: A number is divisible by 2 if its last digit is even (0, 2, 4, 6, 8).
• Divisibility by 4: A number is divisible by 4 if the number formed by its last two digits is divisible by 4.
• 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 9: A number is divisible by 9 if the sum of its digits is divisible by 9. This is similar to the rule for 3.
• Divisibility by 10: A number is divisible by 10 if its last digit is 0.

Problems involving finding unknown digits often require applying these divisibility rules along with any given constraints on the digit values or their sums/differences.

In this specific problem, carefully listing the possible values for \(X+Y\) based on both the divisibility rule and the inequality constraint was the key step before finding the valid (X,Y) pairs while remembering that X cannot be zero in a five-digit number XY235.