Question 990538: When a certain number is divided by 2, there is a remainder of 1. When the same number is divided by 3, 4 and 5, there are remainders of 2, 3, and 4 respectively. What is the smallest positive number for which the conditions are satisfied?
Found 2 solutions by CubeyThePenguin, ikleyn:
Answer by CubeyThePenguin(3113)   (Show Source):
You can put this solution on YOUR website! Let the number be n.
n = 1 mod 2
n = 2 mod 3
n = 3 mod 4
n = 4 mod 5
The smallest value of n is 1 less than the LCM of 2, 3, 4, and 5, which is 59.
Answer by ikleyn(52788)  (Show Source):
You can put this solution on YOUR website! .
If you add 1 to the number, you will get the number multiple of 2, 3, 4, 5.
The smallest such a number is LCM(2,3,4,5) = 60.
THEREFORE, the number under the problem's question is 60-1 = 59. ANSWER
Solved.
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