1, 2, 3, 5, 6, 10, 15 and 30.
They are the numbers that when divided by 2 leave a remainder of 1
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11 different numbers leave a remainder of 1 when divided into 61.
Any number greater than 199 will divide into both zero times and leave a remainder (of the original number); there is no upper limit to the numbers greater than 199. -------------------------------- The largest number to divide into both and leave the SAME remainder is 56 (leaving a remainder of 31 in both cases).
The first five numbers which when divided by 5 leave a remainder of 4 are: 4 = 4/5 = 0 remainder 4 9 = 9/5 = 1 remainder 4 14 = 14/5 = 2 remainder 4 19 = 19/5 = 3 remainder 4 24 = 24/5 = 4 remainder 4 The pattern continues in this way.
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 16. 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.
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.
Eight.
They are the numbers that when divided by 2 leave a remainder of 1
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The factors of 96 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, and 96.
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There are 11 such numbers.
Well, isn't that a happy little math problem! When a number is divided by another number, we get a quotient and a remainder. In this case, we are looking for natural numbers that leave a remainder of 5 when divided into 47. So, we start by listing some numbers that fit this description: 5, 10, 15, 20, 25, 30, 35, 40, and 45. That's 9 different numbers in total that will leave a remainder of 5 when divided into 47.
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