Last updated on May 26th, 2025
The divisibility rule is a way to find out whether a number is divisible by another number without using the division method. In real life, we can use the divisibility rule for quick math, dividing things evenly, and sorting things. In this topic, we will learn about the divisibility rule of 945.
The divisibility rule for 945 is a method by which we can find out if a number is divisible by 945 or not without using the division method. To determine if a number is divisible by 945, it must be divisible by 3, 5, and 7, since 945 = 3 × 5 × 7 × 9. Check whether 2835 is divisible by 945 with the divisibility rule.
Step 1: Check divisibility by 3. A number is divisible by 3 if the sum of its digits is divisible by 3. For 2835, 2 + 8 + 3 + 5 = 18, which is divisible by 3.
Step 2: Check divisibility by 5. A number is divisible by 5 if it ends in 0 or 5. Since 2835 ends in 5, it is divisible by 5.
Step 3: Check divisibility by 7. Use the divisibility rule of 7. Multiply the last digit by 2, and subtract it from the rest of the number. For 2835, 5 × 2 = 10, and 283 - 10 = 273. Repeat the process: 3 × 2 = 6, and 27 - 6 = 21. Since 21 is a multiple of 7, 2835 is divisible by 7.
Since 2835 is divisible by 3, 5, and 7, it is divisible by 945.
Learning the divisibility rule will help kids master division. Let’s learn a few tips and tricks for the divisibility rule of 945.
Memorize the prime factors of 945 (3, 5, 7) to quickly check divisibility.
To determine divisibility by 945, a number must be divisible by each of its prime factors (3, 5, and 7).
Students can use the division method as a way to verify and cross-check their results. This will help them verify their results and also learn.
The divisibility rule of 945 helps us quickly check if a given number is divisible by 945, but common mistakes like calculation errors lead to incorrect results. Here we will understand some common mistakes that will help you avoid them.
Can 4725 be divisible by 945?
Yes, 4725 is divisible by 945.
To check if 4725 is divisible by 945, it must be divisible by 3, 5, and 7, since 945 = 3 × 5 × 7 × 9.
1) Divisibility by 3: Sum the digits: 4 + 7 + 2 + 5 = 18, and 18 is divisible by 3.
2) Divisibility by 5: The last digit is 5, which is a multiple of 5.
3) Divisibility by 7:
Double the last digit: 5 × 2 = 10.
Subtract from remaining: 472 - 10 = 462.
Repeat: 2 × 2 = 4, 46 - 4 = 42.
42 is a multiple of 7.
Since 4725 is divisible by 3, 5, and 7, it is divisible by 945.
Is 1890 divisible by 945?
No, 1890 is not divisible by 945.
For 1890 to be divisible by 945, it must pass the tests for divisibility by 3, 5, and 7.
1) Divisibility by 3: Sum the digits: 1 + 8 + 9 + 0 = 18, and 18 is divisible by 3.
2) Divisibility by 5: The last digit is 0, which is a multiple of 5.
3) Divisibility by 7:
Double the last digit: 0 × 2 = 0.
Subtract from remaining: 189 - 0 = 189.
Double the last digit of 189: 9 × 2 = 18.
Subtract: 18 - 18 = 0.
0 is a multiple of 7.
Though 1890 is divisible by 3, 5, and 7, it must also be divisible by 9 (since 945 = 3^3 × 5 × 7), but 1 + 8 + 9 + 0 = 18, and 18 is not divisible by 9. Therefore, 1890 is not divisible by 945.
Check if -2835 is divisible by 945.
Yes, -2835 is divisible by 945.
To determine if -2835 is divisible by 945, we check divisibility by 3, 5, and 7.
1) Divisibility by 3: Sum the digits: 2 + 8 + 3 + 5 = 18, and 18 is divisible by 3.
2) Divisibility by 5: The last digit is 5, which is a multiple of 5.
3) Divisibility by 7:
Double the last digit: 5 × 2 = 10.
Subtract from remaining: 283 - 10 = 273.
Double last digit of 273: 3 × 2 = 6.
Subtract: 27 - 6 = 21.
21 is a multiple of 7.
Since -2835 is divisible by 3, 5, and 7, it is divisible by 945.
Determine if 1134 is divisible by 945.
No, 1134 is not divisible by 945.
To check if 1134 is divisible by 945, it must be divisible by 3, 5, and 7.
1) Divisibility by 3: Sum the digits: 1 + 1 + 3 + 4 = 9, and 9 is divisible by 3.
2) Divisibility by 5: The last digit is 4, which is not a multiple of 5.
Since 1134 fails the divisibility test for 5, it cannot be divisible by 945.
Verify the divisibility of 2835 by 945.
Yes, 2835 is divisible by 945.
To verify divisibility by 945, check divisibility by 3, 5, and 7.
1) Divisibility by 3: Sum the digits: 2 + 8 + 3 + 5 = 18, and 18 is divisible by 3.
2) Divisibility by 5: The last digit is 5, which is a multiple of 5.
3) Divisibility by 7:
Double the last digit: 5 × 2 = 10.
Subtract from remaining: 283 - 10 = 273.
Double last digit of 273: 3 × 2 = 6.
Subtract: 27 - 6 = 21.
21 is a multiple of 7.
Since 2835 is divisible by 3, 5, and 7, it is divisible by 945.
Hiralee Lalitkumar Makwana has almost two years of teaching experience. She is a number ninja as she loves numbers. Her interest in numbers can be seen in the way she cracks math puzzles and hidden patterns.
: She loves to read number jokes and games.