Table of Contents
What is the rule for divisibility by 7?
Divisibility rules for numbers 1–30
Divisor | Divisibility condition | Examples |
---|---|---|
7 | Subtracting 2 times the last digit from the rest gives a multiple of 7. (Works because 21 is divisible by 7.) | 483: 48 − (3 × 2) = 42 = 7 × 6. |
Subtracting 9 times the last digit from the rest gives a multiple of 7. | 483: 48 − (3 × 9) = 21 = 7 × 3. |
What is the divisibility rule for 99?
Hence a number is divisible by 99 if and only if it is divisible by both 9 and 11. Use the divisibility rules of 9 and 11 to check whether a number is divisible by 9 and 11 and hence if it is divisible by 99.
How do you divide by 7?
Dividing by 7
- Take the last digit in a number.
- Double and subtract the last digit in your number from the rest of the digits.
- Repeat the process for larger numbers.
- Example: Take 357. Double the 7 to get 14. Subtract 14 from 35 to get 21, which is divisible by 7, and we can now say that 357 is divisible by 7.
How to find if a number is divisible by 99?
For a number to be divisible by 99 it should also be divisible by all its factors, i.e by 9 and 11. → A number is divisible by 9 if the sum of the digits is divisible by 9. ⇒ 3572404 isn’t divisible by 99. → Take the alternating sum of the digits in the number, read from left to right. If that is divisible by 11 , so is the original number.
How do you find the sum of numbers from $1-100$?
I found: sum of numbers from $1-100$ sum of numbers from $1-100$ divisible by $3$ sum of numbers from $1-100$ divisible by $7$ Then subtracted first sum by last $2$ sums as mentioned above but there are certain numbers that appear in both tables.
Is 3572404 divisible by 99?
⇒ 3572404 isn’t divisible by 99. → Take the alternating sum of the digits in the number, read from left to right. If that is divisible by 11 , so is the original number. ∴ 114345 is divisible by 99. Was this answer helpful?
What is the LCM of $3$ & $7$?
If you are familiar with unions and intersections of sets , then it is not a difficult problem. Your answer should be:Total sum-(sum of multiple of $3 +$ sum of multiple of $7 -$ sum of multiple of $21)$ Since $21$ is LCM of $3$ & $7$