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Counting numbers are numbers that can be used to count whole amounts of things: 1, 2, 3, etc. (no fractions)

You can divide 120 by 1 and get an answer of 120 with no remainder. Likewise you can divide 120 by 120 and get an answer of 1 with no remainder. Most of the time the answers will be in pairs.

Here are all of the counting numbers that leave no remainder when divided into 120:

1, 120, 2, 60, 3, 40, 4, 30, 5, 24, 6, 20, 8, 15, 10, 12.

Q: How many counting numbers leave no remainder when divided into 120?

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The infinitely many numbers that leave a remainder when divided by 9. The infinitely many numbers that leave a remainder when divided by 9. The infinitely many numbers that leave a remainder when divided by 9. The infinitely many numbers that leave a remainder when divided by 9.

There are 11 such numbers.

Twelve prime numbers smaller than 100 (7, 11, 19, 23, 31, 43, 47, 59, 67, 71, 79, and 83) leave the remainder 3 when divided by 4.

No, prime numbers can only be divided (and leave whole numbers) by themselves and 1 but 202 can obviously be divided by both 101 and by 2.

All even numbers greater than 2 are composite because they are all divisible by 2. Therefore, from 3 onwards, all prime numbers are odd. Consider three consecutive odd numbers. They must be of the form 2n+1, 2n+3 and 2n+5 where n is an integer. Since n is an integer, n will leave a remainder of 0, 1 or 2 when it is divided by 3. Suppose n leaves a remainder of 0 when divided by 3. Therefore n = 3k for some integer k. Then 2n+3 = 2*(3k) + 3 = 6k + 3 = 3*(2k+1). That is, middle of the three consecutive odd numbers is divisible by 3 and so it is not a prime. Now, suppose n leaves a remainder of 1 when divided by 3. Therefore n = 3k+1 for some integer k. Then 2n+1 = 2*(3k+1) + 1 = 6k+2+1 = 6k+3 = 3*(2k+1). That is, first of the three consecutive odd numbers is divisible by 3 and so it is not a prime. Finally, suppose n leaves a remainder of 2 when divided by 3. Therefore n = 3k+2 for some integer k. Then 2n+5 = 2*(3k+2) + 5 = 6k+4+5 = 6k+9 = 3*(2k+3). That is, last of the three consecutive odd numbers is divisible by 3 and so it is not a prime. Thus for any three consecutive odd numbers greater than 3, one of them is divisible by 3 and therefore the three cannot all be prime.

Related questions

The infinitely many numbers that leave a remainder when divided by 9. The infinitely many numbers that leave a remainder when divided by 9. The infinitely many numbers that leave a remainder when divided by 9. The infinitely many numbers that leave a remainder when divided by 9.

1, 2, 3, 5, 6, 10, 15 and 30.

All numbers that leave a remainder when divided by 35.

This is an extremely poor question. There are 28 pairs of numbers and, in less than 10 minutes, I could find reasons why 17 pair were different from the other numbers in the set. A bit more time and I am sure I could do the rest. So here they are: 53 and 72: leave a remainder of 15 when divided by 19. 53 and 77: leave a remainder of 5 when divided by 12. 53 and 82: leave a remainder of 24 when divided by 29. 53 and 87: leave a remainder of 2 when divided by 17. 53 and 95: leave a remainder of 11 when divided by 21. 53 and 97: leave a remainder of 9 when divided by 11. 67 and 95: leave a remainder of 11 when divided by 28. 67 and 87: leave a remainder of 7 when divided by 17. 67 and 97: leave a remainder of 7 when divided by 30. 72 and 82: leave a remainder of 0 when divided by 2 (are even). 72 and 87: leave a remainder of 0 when divided by 3 (multiples of 3). 72 and 95: leave a remainder of 3 when divided by 23. 72 and 97: leave a remainder of 22 when divided by 25. 77 and 95: leave a remainder of 5 when divided by 9. 77 and 97: leave a remainder of 17 when divided by 20. 82 and 95: leave a remainder of 4 when divided by 13. 87 and 95: leave a remainder of 7 when divided by 8.

0.1818

Eight.

They are the numbers that when divided by 2 leave a remainder of 1

The factors of 96 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, and 96.

0.1522

There are 11 such numbers.

0.1818

0.2857