SOLUTION: What is the lowest number n that when divided by 3 leaves a remainder of 1, when divided by 4 leaves a remainder of 3, and when divided by 5 leaves a remainder of 3?

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Question 1097708: What is the lowest number n that when divided by 3 leaves a remainder of 1, when divided by 4 leaves a remainder of 3, and when divided by 5 leaves a remainder of 3?
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
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Answer. The number under he question is 43.

Let us consider this part of the condition first:


    "the number . . . when divided by 4 leaves a remainder of 3, and when divided by 5 leaves a remainder of 3."



Let N be our number.  Then that part of the condition implies that the number N-3 is divided by 4 and by 5.

Hence, N-3 is divided by 20.



It means that our number N has the form  N = 20k + 3.


Check the first numbers of this form at k = 1, 2, 3, . . . :

k            1    2    3
N = 20k+3    23   43   63


Which of them gives the remainder of 1 when divided by 3 ?


But of course, N = 43.


It is your answer.


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