The number of points having position vector `a hat i + b hat j + c hatk `, where `1 leq a,b,cleq10` and `a,b,c in N`, such that `2^a + 3^b+5^c` is a multiple of 4 is (A) 1000 (B) 500 (C) 250 (D) 125

Options

A. 70
B. 140
C. 210
D. 280

Explanation

To solve the problem, we need to find the number of points with position vectors \( \hat{i} + \hat{j} + \hat{k} \) where \( 1 \leq a, b, c \leq 10 \) and \( a, b, c \in \mathbb{N} \) such that \( 2^a + 3^b + 5^c \) is a multiple of 4.

### Step 1: Analyze the expression \( 2^a + 3^b + 5^c \)

We need to determine when \( 2^a + 3^b + 5^c \equiv 0 \mod 4 \).

- **For \( 2^a \):**
- If \( a = 1 \), then \( 2^a \equiv 2 \mod 4 \).
- If \( a \geq 2 \), then \( 2^a \equiv 0 \mod 4 \).

- **For \( 3^b \):**
- \( 3^1 \equiv 3 \mod 4 \)
- \( 3^2 \equiv 1 \mod 4 \)
- \( 3^3 \equiv 3 \mod 4 \)
- \( 3^4 \equiv 1 \mod 4 \)
- Thus, \( 3^b \equiv 3 \mod 4 \) if \( b \) is odd, and \( 3^b \equiv 1 \mod 4 \) if \( b \) is even.

- **For \( 5^c \):**
- \( 5 \equiv 1 \mod 4 \) for any \( c \).
- Thus, \( 5^c \equiv 1 \mod 4 \).

### Step 2: Set up cases based on values of \( a \)

#### Case 1: \( a = 1 \)

In this case, \( 2^a \equiv 2 \mod 4 \). We need:
\[ 2 + 3^b + 5^c \equiv 0 \mod 4 \]
This simplifies to:
\[ 3^b + 5^c \equiv 2 \mod 4 \]
Since \( 5^c \equiv 1 \mod 4 \), we have:
\[ 3^b + 1 \equiv 2 \mod 4 \]
This implies:
\[ 3^b \equiv 1 \mod 4 \]
Thus, \( b \) must be even. The even values for \( b \) in the range \( 1 \leq b \leq 10 \) are \( 2, 4, 6, 8, 10 \) (5 options). \( c \) can take any value from 1 to 10 (10 options).

So, the total combinations for this case:
\[ 1 \text{ (for } a = 1) \times 5 \text{ (even } b) \times 10 \text{ (any } c) = 50 \]

#### Case 2: \( a \geq 2 \)

In this case, \( 2^a \equiv 0 \mod 4 \). We need:
\[ 0 + 3^b + 5^c \equiv 0 \mod 4 \]
This simplifies to:
\[ 3^b + 1 \equiv 0 \mod 4 \]
Thus:
\[ 3^b \equiv 3 \mod 4 \]
So \( b \) must be odd. The odd values for \( b \) in the range \( 1 \leq b \leq 10 \) are \( 1, 3, 5, 7, 9 \) (5 options). \( a \) can take values from 2 to 10 (9 options), and \( c \) can take any value from 1 to 10 (10 options).

So, the total combinations for this case:
\[ 9 \text{ (for } a \geq 2) \times 5 \text{ (odd } b) \times 10 \text{ (any } c) = 450 \]

### Step 3: Total combinations

Adding both cases together:
\[ 50 + 450 = 500 \]

Thus, the total number of points is \( \boxed{500} \).