Question 1064777: Find the smallest number that when divided by 2, 3, 4, 5, or 6, it leaves a remainder of one but when divided by 7 has no remainder.
my work:
So I first I found the LCM of 2, 3, 4, 5, 6 which is 60.
Then I added 1 to 60 which is 61.
Then I multiplied 61 by 7 repeatedly till it fulfilled the condition that remainder should be 1. I got the answer 146461 which seems to correct but not completely sure.
Answer by ikleyn(52930)   (Show Source):
You can put this solution on YOUR website! .
Let N be our unknown number under the question.
Consider the number N-1.
According to the condition, it is multiple of 2, 3, 4, 5 and 6.
Hence, N-1 is the multiple of 60.
It means that our N is somewhere among these numbers:
60+1, 2*60+1, 3*60+1, 4*60+1, 5*60+1, 6*60+1
i.e, among the numbers 61, 121, 181, 241, 301.
Check, which of these numbers is multiple of 7.
Answer. 301.
See the lesson
- The number that leaves a remainder 1 when divided by 2, by 3, by 4, by 5 and so on until 9
in this site.
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