To solve the problem, we need to determine the values of \(x\) and \(y\) such that the seven-digit number \(54x29y6\) is divisible by 72. A number is divisible by 72 if it is divisible by both 8 and 9.
### Step 1: Check for divisibility by 8
For a number to be divisible by 8, the last three digits must form a number that is divisible by 8. The last three digits of our number are \(29y6\).
We can express this as \(29y6\) and check the possible values of \(y\) (from 0 to 9) to see which makes \(29y6\) divisible by 8.
- If \(y = 0\): \(2906 \div 8 = 363.25\) (not divisible)
- If \(y = 1\): \(2916 \div 8 = 364.5\) (not divisible)
- If \(y = 2\): \(2926 \div 8 = 365.75\) (not divisible)
- If \(y = 3\): \(2936 \div 8 = 367\) (divisible)
- If \(y = 4\): \(2946 \div 8 = 368.25\) (not divisible)
- If \(y = 5\): \(2956 \div 8 = 369.5\) (not divisible)
- If \(y = 6\): \(2966 \div 8 = 370.75\) (not divisible)
- If \(y = 7\): \(2976 \div 8 = 372\) (divisible)
- If \(y = 8\): \(2986 \div 8 = 373.25\) (not divisible)
- If \(y = 9\): \(2996 \div 8 = 374.5\) (not divisible)
From this, we find that \(y\) can be 3 or 7.
### Step 2: Check for divisibility by 9
Next, we need to check the sum of the digits for divisibility by 9. The sum of the digits is:
\[
5 + 4 + x + 2 + 9 + y + 6 = 26 + x + y
\]
This sum must be divisible by 9.
#### Case 1: \(y = 3\)
\[
26 + x + 3 = 29 + x
\]
We need \(29 + x\) to be divisible by 9. The possible values of \(x\) (from 0 to 9) give us:
- If \(x = 0\): \(29 + 0 = 29\) (not divisible)
- If \(x = 1\): \(29 + 1 = 30\) (not divisible)
- If \(x = 2\): \(29 + 2 = 31\) (not divisible)
- If \(x = 3\): \(29 + 3 = 32\) (not divisible)
- If \(x = 4\): \(29 + 4 = 33\) (divisible)
- If \(x = 5\): \(29 + 5 = 34\) (not divisible)
- If \(x = 6\): \(29 + 6 = 35\) (not divisible)
- If \(x = 7\): \(29 + 7 = 36\) (divisible)
- If \(x = 8\): \(29 + 8 = 37\) (not divisible)
- If \(x = 9\): \(29 + 9 = 38\) (not divisible)
So, for \(y = 3\), \(x\) can be 4 or 7.
#### Case 2: \(y = 7\)
\[
26 + x + 7 = 33 + x
\]
We need \(33 + x\) to be divisible by 9:
- If \(x = 0\): \(33 + 0 = 33\) (divisible)
- If \(x = 1\): \(33 + 1 = 34\) (not divisible)
- If \(x = 2\): \(33 + 2 = 35\) (not divisible)
- If \(x = 3\): \(33 + 3 = 36\) (divisible)
- If \(x = 4\): \(33 + 4 = 37\) (not divisible)
- If \(x = 5\): \(33 + 5 = 38\) (not divisible)
- If \(x = 6\): \(33 + 6 = 39\) (divisible)
- If \(x = 7\): \(33 + 7 = 40\) (not divisible)
- If \(x = 8\): \(33 + 8 = 41\) (not divisible)
- If \(x = 9\): \(33 + 9 = 42\) (not divisible)
So, for \(y = 7\), \(x\) can be 0, 3, or 6.
### Step 3: Apply the condition \(x > y\)
Now we have:
- For \(y = 3\), \(x\) can be 4 or 7 (both valid since \(x > y\)).
- For \(y = 7\), \(x\) can be 0, 3, or 6 (none valid since \(x\) cannot be greater than \(y\)).
Thus, the only valid pairs are:
1. \( (x, y) = (4, 3) \)
2. \( (x, y) = (7, 3) \)
### Step 4: Calculate \(2x + 3y\)
Now we calculate \(2x + 3y\) for both pairs.
1. For \( (x, y) = (4, 3) \):
\[
2x + 3y = 2(4) + 3(3) = 8 + 9 = 17
\]
2. For \( (x, y) = (7, 3) \):
\[
2x + 3y = 2(7) + 3(3) = 14 + 9 = 23
\]
Since both pairs are valid, we conclude that the value of \(2x + 3y\) can be either 17 or 23, but since \(x\) must be greater than \(y\), we take \( (x, y) = (7, 3) \).
### Final Answer
Thus, the value of \(2x + 3y\) is \(23\).
