To find the least number that meets these conditions, we can express the problem in terms of congruences. We have:
These congruences can be rewritten as:
Since all three congruences have the same form, we can solve for ( x ) using the least common multiple of the moduli (35, 45, and 55), which is 1035. Adding 10 to account for the negative remainder, we find that ( x = 1035 - 10 = 1025 ).
Thus, the least number is 1025.
810: quotient 1, remainder 1
1005
How about 14 because 14/9 = 1 with a remainder of 5
It is 38.
121
25
It is 1.1 = 0*12152128 r 1
301
127 is the least prime number greater than 25 that will have a remainder of 2 when divided by 25.
No answer is possible as any number divided by 9 must either be exactly divisible by 9 or leave a remainder less than 9.
810: quotient 1, remainder 1
1005
449 is the smallest number that satisfies those requirements.
How about 14 because 14/9 = 1 with a remainder of 5
It is 38.
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).
121