Problem Analysis: Divisibility by 44

We are given a six-digit number $15x1y2$. The goal is to find the minimum possible value for the sum of its digits $x$ and $y$, denoted as $(x + y)$, such that the number is perfectly divisible by 44.

Divisibility Rules for 44

A number is divisible by 44 if and only if it is divisible by both 4 and 11. This is because $44 = 4 \times 11$, and 4 and 11 have no common factors other than 1.

Applying Divisibility by 4

The divisibility rule for 4 states that a number is divisible by 4 if the number formed by its last two digits is divisible by 4.

In our number $15x1y2$, the last two digits form the number $y2$. This means the two-digit number $10y + 2$ must be divisible by 4.

We check possible values for the digit $y$ (from 0 to 9):

• If $y=0$, $02$ is not divisible by 4.
• If $y=1$, $12$ is divisible by 4. (Possible)
• If $y=2$, $22$ is not divisible by 4.
• If $y=3$, $32$ is divisible by 4. (Possible)
• If $y=4$, $42$ is not divisible by 4.
• If $y=5$, $52$ is divisible by 4. (Possible)
• If $y=6$, $62$ is not divisible by 4.
• If $y=7$, $72$ is divisible by 4. (Possible)
• If $y=8$, $82$ is not divisible by 4.
• If $y=9$, $92$ is divisible by 4. (Possible)

Thus, the possible values for $y$ are $1, 3, 5, 7, 9$.

Applying Divisibility by 11

The divisibility rule for 11 states that a number is divisible by 11 if the alternating sum of its digits (starting from the rightmost digit) is either 0 or a multiple of 11.

For the number $15x1y2$, the digits are 1, 5, x, 1, y, 2.

Sum of digits at odd places (1st, 3rd, 5th from the right): $2 + 1 + 5 = 8$.

Sum of digits at even places (2nd, 4th, 6th from the right): $y + x + 1$.

The alternating sum is $(2 + 1 + 5) - (y + x + 1) = 8 - (x + y + 1) = 7 - (x + y)$.

This difference, $7 - (x + y)$, must be a multiple of 11. We can write this as:

where $k$ is any integer ($k \in \{..., -1, 0, 1, ...\}$).

Rearranging the equation to find the sum $(x+y)$:

Finding Possible Sums $(x+y)$

Since $x$ and $y$ are digits, they must be integers between 0 and 9, inclusive.

• The minimum possible value for $x + y$ is $0 + 0 = 0$.
• The maximum possible value for $x + y$ is $9 + 9 = 18$.

We now find the possible values of $(x + y)$ using the equation $x + y = 7 - 11k$, ensuring the sum is within the range [0, 18]:

• For $k = 0$: $x + y = 7 - 11(0) = 7$. This is within the range [0, 18].
• For $k = 1$: $x + y = 7 - 11(1) = -4$. This is outside the range (too small).
• For $k = -1$: $x + y = 7 - 11(-1) = 7 + 11 = 18$. This is within the range [0, 18].
• For $k = -2$: $x + y = 7 - 11(-2) = 7 + 22 = 29$. This is outside the range (too large).

The possible values for the sum $(x + y)$ are 7 and 18.

Combining Conditions for Minimum $(x+y)$

We need the minimum value of $(x + y)$, which could be 7 or 18. The smaller value is 7.

We need to confirm that there exist digits $x$ and $y$ that satisfy both the condition for divisibility by 4 (i.e., $y \in \{1, 3, 5, 7, 9\}$) and the condition for divisibility by 11 (i.e., $x + y = 7$ or $x + y = 18$), and also that $x$ is a valid digit ($0 \le x \le 9$).

Case 1: $(x + y) = 7$

• If $y = 1$, then $x = 7 - 1 = 6$. Valid digits ($x=6, y=1$). Number: $156112$.
• If $y = 3$, then $x = 7 - 3 = 4$. Valid digits ($x=4, y=3$). Number: $154132$.
• If $y = 5$, then $x = 7 - 5 = 2$. Valid digits ($x=2, y=5$). Number: $152152$.
• If $y = 7$, then $x = 7 - 7 = 0$. Valid digits ($x=0, y=7$). Number: $150172$.
• If $y = 9$, then $x = 7 - 9 = -2$. Not possible as $x$ cannot be negative.

We found valid pairs $(x, y)$ where the sum $x + y = 7$. So, 7 is an achievable sum.

Case 2: $(x + y) = 18$

• We check possible values for $y$ from $\{1, 3, 5, 7, 9\}$. Only $y=9$ can potentially lead to $x+y=18$.
• If $y = 9$, then $x = 18 - 9 = 9$. Valid digits ($x=9, y=9$). Number: $159192$.

We found a valid pair $(x, y)$ where the sum $x + y = 18$. So, 18 is also an achievable sum.

Final Minimum Value $(x+y)$

The possible values for the sum $(x + y)$ are 7 and 18. The minimum of these values is 7.