Understanding the Divisibility Rule for 72

The problem asks for a possible value of the sum $X + Y$ given a 9-digit number, 937X728Y6, which is divisible by 72. A number is divisible by 72 if and only if it is divisible by both 8 and 9, since 8 and 9 are coprime factors of 72 ($72 = 8 \times 9$).

Applying the Divisibility Rule for 8

For a number to be divisible by 8, its last three digits must form a number divisible by 8. In our case, the last three digits are 8Y6.

We need to find the value(s) of the digit Y such that the number $8Y6$ is divisible by 8. Let's test the possibilities:

• If Y = 0, 806 is not divisible by 8.
• If Y = 1, 816 is divisible by 8 ($816 \div 8 = 102$). So, Y = 1 is a possibility.
• If Y = 2, 826 is not divisible by 8.
• If Y = 3, 836 is not divisible by 8.
• If Y = 4, 846 is not divisible by 8.
• If Y = 5, 856 is divisible by 8 ($856 \div 8 = 107$). So, Y = 5 is a possibility.
• If Y = 6, 866 is not divisible by 8.
• If Y = 7, 876 is not divisible by 8.
• If Y = 8, 886 is not divisible by 8.
• If Y = 9, 896 is divisible by 8 ($896 \div 8 = 112$). So, Y = 9 is a possibility.

Therefore, the possible values for Y are 1, 5, or 9.

Applying the Divisibility Rule for 9

For a number to be divisible by 9, the sum of its digits must be divisible by 9. The digits of the number 937X728Y6 are 9, 3, 7, X, 7, 2, 8, Y, and 6.

The sum of the digits is:

The sum $S$ must be divisible by 9. This means $42 + X + Y$ must be a multiple of 9.

Finding Possible Values for X + Y

We know that X and Y are digits, meaning they can range from 0 to 9. We also know that Y can be 1, 5, or 9. Let's check each possibility for Y to find the corresponding value of X and then the sum $X + Y$.

Case 1: Y = 1

• The sum of digits is $42 + X + 1 = 43 + X$.
• We need $43 + X$ to be a multiple of 9. The smallest multiple of 9 greater than or equal to 43 is 45 ($9 \times 5 = 45$).
• If $43 + X = 45$, then $X = 45 - 43 = 2$.
• In this case, $X = 2$ and $Y = 1$. The sum is $X + Y = 2 + 1 = 3$.
• The next multiple of 9 is 54 ($9 \times 6 = 54$). If $43 + X = 54$, then $X = 11$, which is not a single digit. So, this is not possible.

Case 2: Y = 5

• The sum of digits is $42 + X + 5 = 47 + X$.
• We need $47 + X$ to be a multiple of 9. The smallest multiple of 9 greater than or equal to 47 is 54 ($9 \times 6 = 54$).
• If $47 + X = 54$, then $X = 54 - 47 = 7$.
• In this case, $X = 7$ and $Y = 5$. The sum is $X + Y = 7 + 5 = 12$.
• The next multiple of 9 is 63 ($9 \times 7 = 63$). If $47 + X = 63$, then $X = 16$, which is not a single digit.

Case 3: Y = 9

• The sum of digits is $42 + X + 9 = 51 + X$.
• We need $51 + X$ to be a multiple of 9. The smallest multiple of 9 greater than or equal to 51 is 54 ($9 \times 6 = 54$).
• If $51 + X = 54$, then $X = 54 - 51 = 3$.
• In this case, $X = 3$ and $Y = 9$. The sum is $X + Y = 3 + 9 = 12$.
• The next multiple of 9 is 63 ($9 \times 7 = 63$). If $51 + X = 63$, then $X = 12$, which is not a single digit.

Identifying the Possible Value of X + Y

From our analysis, the possible values for the sum $X + Y$ are 3 and 12. The question asks for one of the possible values of $X + Y$. Looking at the given options:

• 12
• 8
• 9
• 5

The value 12 is among the possible sums we calculated and is listed as an option.