It is 38.
810: quotient 1, remainder 1
1005
How about 14 because 14/9 = 1 with a remainder of 5
121
The numbers divided by 5 gives 1 remainder means it should end either 1 or 6. But in case of number ends with 6 will be divided by 2. So only number ending with 1 have to consider. Also the number should be divisble by 7. So when number ending with 3 is multiplied by 7 will give result of number ending 1. So, possibilities are 7*3, 7*13, 7*23, etc In that least number which solves this problem 7*43 = 301. So the least possible number is 301.
127 is the least prime number greater than 25 that will have a remainder of 2 when divided by 25.
810: quotient 1, remainder 1
No answer is possible as any number divided by 9 must either be exactly divisible by 9 or leave a remainder less than 9.
1005
How about 14 because 14/9 = 1 with a remainder of 5
The smallest number which can be divided by both 4 and 5 without a remainder is 20. This is also known as the Least Common Multiple (LCM).
25
It is 1.1 = 0*12152128 r 1
121
301
The least common denominator requires two values. It is the largest number that evenly divides (that is, without a remainder) two numbers. Since 59 is prime, the least common divisor between it and another number will either be 1 or 59. It would be 59, if 59 divided the other number. Otherwise it will be 1.
The numbers divided by 5 gives 1 remainder means it should end either 1 or 6. But in case of number ends with 6 will be divided by 2. So only number ending with 1 have to consider. Also the number should be divisble by 7. So when number ending with 3 is multiplied by 7 will give result of number ending 1. So, possibilities are 7*3, 7*13, 7*23, etc In that least number which solves this problem 7*43 = 301. So the least possible number is 301.