`x#y={{:(x+2y,if,|x+y|" is even"),(2x" @2 "y,if,|x+y|" is odd"):}`
`x"@"y={{:(3x-y,if,|x+y|" is even"),(2x#2y" ,if",|x+y|" is odd"):}`
Find the value of `2#(7" @ "(4#5))" @" 3)`:

Options

A. a. 204
B. b. 306
C. c. 408
D. d. 102

Answer

To solve the problem step by step, we need to evaluate the expression `2#(7 @ (4#5)) @ 3`. We will break it down into parts according to the definitions provided for the operations `#` and `@`.

### Step 1: Evaluate `4#5`
According to the definition of the operation `#`:
- If `|x + y|` is even, then `x#y = x + 2y`
- If `|x + y|` is odd, then `x#y = 2x @ 2y`

For `4#5`:
- Calculate `4 + 5 = 9`, which is odd.
- Therefore, we use the second case: `4#5 = 2*4 @ 2*5 = 8 @ 10`.

### Step 2: Evaluate `8 @ 10`
According to the definition of the operation `@`:
- If `|x + y|` is even, then `x@y = 3x - y`
- If `|x + y|` is odd, then `x@y = 2x # 2y`

For `8 @ 10`:
- Calculate `8 + 10 = 18`, which is even.
- Therefore, we use the first case: `8 @ 10 = 3*8 - 10 = 24 - 10 = 14`.

### Step 3: Substitute back into the expression
Now we substitute back into the original expression:
`2#(7 @ 14) @ 3`.

### Step 4: Evaluate `7 @ 14`
For `7 @ 14`:
- Calculate `7 + 14 = 21`, which is odd.
- Therefore, we use the second case: `7 @ 14 = 2*7 # 2*14 = 14 # 28`.

### Step 5: Evaluate `14 # 28`
For `14 # 28`:
- Calculate `14 + 28 = 42`, which is even.
- Therefore, we use the first case: `14 # 28 = 14 + 2*28 = 14 + 56 = 70`.

### Step 6: Substitute back into the expression
Now we substitute back into the expression:
`2#70 @ 3`.

### Step 7: Evaluate `2#70`
For `2#70`:
- Calculate `2 + 70 = 72`, which is even.
- Therefore, we use the first case: `2#70 = 2 + 2*70 = 2 + 140 = 142`.

### Step 8: Substitute back into the expression
Now we substitute back into the expression:
`142 @ 3`.

### Step 9: Evaluate `142 @ 3`
For `142 @ 3`:
- Calculate `142 + 3 = 145`, which is odd.
- Therefore, we use the second case: `142 @ 3 = 2*142 # 2*3 = 284 # 6`.

### Step 10: Evaluate `284 # 6`
For `284 # 6`:
- Calculate `284 + 6 = 290`, which is even.
- Therefore, we use the first case: `284 # 6 = 284 + 2*6 = 284 + 12 = 296`.

### Final Answer
Thus, the final value of `2#(7 @ (4#5)) @ 3` is **296**.

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