Any number larger than 30 that ends in 2 or 7.
32, 37, 42, 47 and so on.
127 is the least prime number greater than 25 that will have a remainder of 2 when divided by 25.
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, 3 does not go into 2 evenly. In mathematical terms, this can be expressed as 2 divided by 3 results in a quotient of 0 with a remainder of 2. This is because 3 is greater than 2, so it cannot be divided into 2 without leaving a remainder.
When you divide by 5, the remainder can be 0, 1, 2, 3, or 4, but no more than 4.
A number between 1 and 500 is composite if it can be divided, without remainder, by a number other than 1 and itself.
Restate the question to, "What multiple of 7 is one more than a multiple of 15?" The answer is 91.
37 is the number.
Then divide the remainder again by the divisor until you get a remainder smaller than your divisor or an remainder equal to zero. The remainder in a division question should never be larger than the "divisor", but the remainder often is larger than the "answer" (quotient). For example, if 435 is divided by 63, the quotient is 22 and the remainder is 57.
3 is the remainder when 23 is divided by 10. 2 x 10 = 20, and 23 is 3 more than 20.
The LCM of 2, 3, 4 and 5 is 60. Since you need a remainder of 1 just add 1. So the answer is 61. Or any number that is 1 more than a multiple of 60.
3 divided by 2 has a remainder of 1. Which is 1 less than 2.
The remainder of any division MUST be smaller than the divisor. So no number divided by 10 or 7 can leave a remainder of 12. So review your question and post it when it makes sense.
Nothing. The remainder has to be less than the divisor.
58, 118, 178, ... That is all numbers that are 60n - 2 where n = 1, 2, 3, ... It has a remainder of 3 when divided by 5, means the last digit must be 3 or 8. It also has a remainder of 2 when divided by 4, means that the number must be even, so the last digit must be 8. It also has a remainder of 1 when divided by 3, means it must be 1 more than a multiple of 3 that ends in 7. So it must be 28, 58, 88, 118, ..., (each number 30 more than the last) but 28, 88, ... are divisible by 4, so only 58, 118, ... (each number 60 more than the last) need be considered. It also has a remainder of 4 when divided by 6, means it must be 4 more than a multiple of 6 that ends in 4. So it must be 58, 118, ...
24 is answer.
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.
0.0385