171 cannot go into ANY two digit number - its smallest positive multiple is 171. Therefore the remainder must be the original 2 digit number. However, the question states that the remainder is 6. That is a contradiction. Consequently, there cannot be any solution to the question as stated.
Unless they are divided by themselves or one, they will always have a remainder.
49 can be divided by these numbers without leaving a remainder: 1 7 and 49.
57 can be divided by 3 and 19 with no remainder. 51 can be divided by 3 and 17 with no remainder.
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.
In mathematics, numbers that can be divided without a remainder are referred to as divisible numbers. For example, any integer can be divided by its factors, such as 2, 3, or 5, if those numbers evenly divide it. Additionally, whole numbers can be divided by themselves and 1 without a remainder. Essentially, any number can be divided evenly by its divisors.
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.
Unless they are divided by themselves or one, they will always have a remainder.
1 can be divided into both numbers with no remainder
49 can be divided by these numbers without leaving a remainder: 1 7 and 49.
57 can be divided by 3 and 19 with no remainder. 51 can be divided by 3 and 17 with no remainder.
It is an Odd number. Even numbers are those which have no remainder when divided by 2.
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.
0.25
546
-3
0.5