(x1000 - x500 - 1)/(x - 1)
(x1000 + 0x999 + ... + 0x501 - x500 + ... + 0x - 1)/(x - 1)
Now you can find the remainder, if you like...
6.5
more info: An unknown polynomial f(x) of degree million yields a remainder of 1 when divided by x - 1, a remainder of 3 when divided by x - 3, a remainder of 21 when divided by x - 5. Find the remainder when f(x) is divided by (x - 1)(x - 3)(x - 5).
12*1 + 3 = 15 12*2 + 3 = 27 12*4 + 3 = 51 when 51 divided by 15 gives the remainder as 6.
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.
What you have to do here is find the least common multiple of 46 and 8, then add 3.
35
11.0503
127.6
To find the remainder you can do the 'chunking' method. This would leave you with a remainder of 2 because 4x9=36 38-36=2
6.5
506
more info: An unknown polynomial f(x) of degree million yields a remainder of 1 when divided by x - 1, a remainder of 3 when divided by x - 3, a remainder of 21 when divided by x - 5. Find the remainder when f(x) is divided by (x - 1)(x - 3)(x - 5).
If it leaves no remainder when divided by 2 then it is an even number.
12*1 + 3 = 15 12*2 + 3 = 27 12*4 + 3 = 51 when 51 divided by 15 gives the remainder as 6.
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.
An even number can be divided by 2 evenly. An odd number will have a remainder of 1 when divided by 2.
The remainder is 8. To find the remainder when a number is divided by 9, add the digits together; if this sum has more than 1 digit, repeat until one digit remains. This digit is the remainder, unless it is 9 in which case the remainder is 0. examples: remainder 53 → 5 + 3 = 8 → remainder is 8 when 53 is divided by 9. remainder 126 → 1 + 2 + 6 = 9 → remainder is 0, ie no remainder, 126 is divisible by 9. remainder 258 → 2 + 5 + 8 = 15 → 1 + 5 = 6 → remainder is 6 when 258 is divided by 9.