Divisibility Check for Number 1071819

This question asks us to identify which number among the given options (47, 35, 64, 53) is a factor of 1071819. A number is considered divisible by another number if the division results in a whole number with no remainder.

Checking Divisibility Options

We can determine the divisibility by performing the division for each option or by using divisibility rules where applicable.

• Option 1: 47
• Option 2: 35
• Option 3: 64
• Option 4: 53

Applying Divisibility Rules and Calculations

Divisibility by 35

A number is divisible by 35 if it is divisible by both 5 and 7.

• For divisibility by 5, the number must end in 0 or 5. The number 1071819 ends in 9.
• Therefore, 1071819 is not divisible by 5, and consequently, it is not divisible by 35.

Divisibility by 64

A number is divisible by 64 if it is even. 64 is a power of 2 ($64 = 2^6$).

• The number 1071819 is an odd number because it ends in 9.
• Odd numbers are not divisible by any power of 2 greater than $2^0=1$.
• Therefore, 1071819 is not divisible by 64.

Divisibility by 47 and 53

For prime numbers like 47 and 53, specific divisibility rules are not commonly used. The most straightforward method is direct division.

Checking 1071819 ÷ 47

Performing the division calculation:

$1071819 \div 47 \approx 22804.66$

Since the result is not a whole number (it has a decimal part), 1071819 is not divisible by 47.

Checking 1071819 ÷ 53

Performing the division calculation:

We can use long division or a calculator to check this:

$1071819 \div 53$

Let's trace the steps:

1. Divide 107 by 53: $107 \div 53 = 2$ with a remainder of $1$.
2. Bring down the next digit (1), making it 11. Divide 11 by 53: $11 \div 53 = 0$ with a remainder of $11$.
3. Bring down the next digit (8), making it 118. Divide 118 by 53: $118 \div 53 = 2$ with a remainder of $12$ (since $53 \times 2 = 106$).
4. Bring down the next digit (1), making it 121. Divide 121 by 53: $121 \div 53 = 2$ with a remainder of $15$ (since $53 \times 2 = 106$).
5. Bring down the last digit (9), making it 159. Divide 159 by 53: $159 \div 53 = 3$ with a remainder of $0$ (since $53 \times 3 = 159$).

The division results in a whole number:

$1071819 \div 53 = 20223$

Since the division yields the integer 20223 with no remainder, the number 1071819 is confirmed to be divisible by 53.

Conclusion

After checking the divisibility of 1071819 by each of the provided options using divisibility rules and direct calculation, we found that only 53 divides 1071819 exactly.