Two non negative integers are chosen at random. The probability that the sum of the square is divisible by 10, is

Options

A. `(17)/(100)`
B. `(9)/(50)`
C. `(7)/(50)`
D. `(9)/(16)`

Answer

To solve the problem of finding the probability that the sum of the squares of two randomly chosen non-negative integers is divisible by 10, we can follow these steps:

### Step 1: Define the Variables
Let the two non-negative integers be \( x \) and \( y \). We can express these integers in terms of their remainders when divided by 10. Specifically, we can write:
- \( x = 10x_1 + a_1 \)
- \( y = 10y_1 + b_1 \)

where \( a_1 \) and \( b_1 \) are the remainders when \( x \) and \( y \) are divided by 10, respectively. Both \( a_1 \) and \( b_1 \) can take values from 0 to 9.

### Step 2: Calculate the Sum of Squares
We need to find when the sum of the squares \( x^2 + y^2 \) is divisible by 10. Squaring \( x \) and \( y \) gives:
\[ x^2 = (10x_1 + a_1)^2 = 100x_1^2 + 20x_1a_1 + a_1^2 \]
\[ y^2 = (10y_1 + b_1)^2 = 100y_1^2 + 20y_1b_1 + b_1^2 \]
Adding these, we have:
\[ x^2 + y^2 = 100(x_1^2 + y_1^2) + 20(x_1a_1 + y_1b_1) + (a_1^2 + b_1^2) \]

### Step 3: Determine Divisibility by 10
For \( x^2 + y^2 \) to be divisible by 10, the last term \( a_1^2 + b_1^2 \) must be divisible by 10, since the other terms are multiples of 10.

### Step 4: Find Possible Values for \( a_1 \) and \( b_1 \)
Now, we need to find pairs \( (a_1, b_1) \) such that \( a_1^2 + b_1^2 \equiv 0 \mod 10 \). We will check all combinations of \( a_1 \) and \( b_1 \) from 0 to 9.

### Step 5: Count Favorable Outcomes
We will list the pairs \( (a_1, b_1) \) that satisfy the condition:
- \( (0, 0) \) → \( 0^2 + 0^2 = 0 \)
- \( (1, 3) \) → \( 1^2 + 3^2 = 10 \)
- \( (1, 7) \) → \( 1^2 + 7^2 = 50 \)
- \( (2, 4) \) → \( 2^2 + 4^2 = 20 \)
- \( (2, 6) \) → \( 2^2 + 6^2 = 40 \)
- \( (3, 1) \) → \( 3^2 + 1^2 = 10 \)
- \( (3, 9) \) → \( 3^2 + 9^2 = 90 \)
- \( (4, 2) \) → \( 4^2 + 2^2 = 20 \)
- \( (4, 8) \) → \( 4^2 + 8^2 = 80 \)
- \( (5, 5) \) → \( 5^2 + 5^2 = 50 \)
- \( (6, 2) \) → \( 6^2 + 2^2 = 40 \)
- \( (6, 8) \) → \( 6^2 + 8^2 = 100 \)
- \( (7, 1) \) → \( 7^2 + 1^2 = 50 \)
- \( (7, 9) \) → \( 7^2 + 9^2 = 130 \)
- \( (8, 6) \) → \( 8^2 + 6^2 = 100 \)
- \( (8, 4) \) → \( 8^2 + 4^2 = 80 \)
- \( (9, 9) \) → \( 9^2 + 9^2 = 162 \)

After checking all combinations, we find that there are 18 favorable outcomes.

### Step 6: Calculate Total Outcomes
The total number of possible pairs \( (a_1, b_1) \) is \( 10 \times 10 = 100 \).

### Step 7: Calculate the Probability
The probability \( P \) that the sum of the squares is divisible by 10 is given by:
\[ P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{18}{100} = \frac{9}{50} \]

### Final Answer
Thus, the probability that the sum of the squares of two randomly chosen non-negative integers is divisible by 10 is \( \frac{9}{50} \).

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