Question 550112
The natural number n is the smallest number satisfying the following properties: when divided by 3 remainder is 1,
when divided by 4 remainder is 2,
when divided by 5 remainder is 3,
when divided by 6 remainder is 4,
what is the remainder in case of division by 7?
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I can't think of an algebra equation that can solve this but we can use some logic here.
Since one of the requirements is divide by 5, remainder 3, we know the number has a unit value of 3 or 8
Also if it meets the requirement of divide by 6, remainder 4, it will automatically satisfy the requirement of divide 3, remainder 1
This narrows it down, I wrote numbers;
18, 23, 28, 33, 38, 43, 48, 53, 58, 63
easy to throw out 18, 28, 33, 48, 63
tried, 23, 38, 43, 53, then came the winner 58!!
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"what is the remainder in case of division by 7"
58/7 = 8 remainder 2
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I am thinking there may be a more efficient way to do this, but just can't come up with it this Christmas Eve.  Merry Christmas!!