37 is the number.
The largest remainder would be 8, because if it were 9 you could divide the number once more. The largest remainder you can have is always one less than what you're dividing by. So if you're dividing by 10, your largest remainder is 9. If you're dividing by 100, it's 99. And so on.
An even number can be divided by 2 evenly. An odd number will have a remainder of 1 when divided by 2. There s the same amount of each.
The largest [integer] remainder is 10. If the remainder was any more you would get one (or more) lots of 11.
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.
Restate the question to, "What multiple of 7 is one more than a multiple of 15?" The answer is 91.
If the remainder is greater than the divisor then you can divide it once more and get one more whole number and then have less remainders.
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, ...
5. To leave a remainder of 1 when divided by 2, the number must be odd. To leave a remainder of 2 when divided by 3, the number must also be two more than a multiple of 3. The multiples of 3 which are odd are 1 x 3, 3 x 3, 5 x 3, etc. The smallest odd multiple of three is 1 x 3 = 3 ⇒ required number is 3 + 2 = 5.
11, 31, 51 and more at intervals of 20
you need to tell the number that you divided by. no one is able to figure it out until you give more information.
A number that can be divided by 6 with a remainder of 1 is one more than a multiple of 6. In mathematical terms, it can be represented as 6n + 1, where n is any integer. For example, if n = 2, then the number would be 6(2) + 1 = 13, which when divided by 6 leaves a remainder of 1.
Since 17 is a factor of 119, the remiander will be the same modulo 17. Thus, the remainder, on division by 17 would be 19 except that 19 contains one more 17 - leaving a final remainder of 2.