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! .
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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