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mortimer
 Senior Boarder
Posts: 48 
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what number that has a remainder of 1 and is divisable by 2,3,4,5, and 6 but has no leftover when divided by 7?
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NGR
 Senior Boarder
Posts: 64
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Probably 'What number leaves a remainder of 1 on division by 2,3,4,5, or 6, but no remainder on division by 7?'
Of course, there's more than one solution, but it's easy to see that the smallest is
25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 301
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Lambdalana
Senior Boarder
Posts: 72
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721.
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quest_marsman
Senior Boarder
Posts: 72
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SBE> what number that has a remainder of 1 and is divisable by SBE> 2,3,4,5, and 6 but has no leftover when divided by 7?
spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler.... spoiler....
The necessary and sufficient condition for a number to be divisible by 2, 2, 3, 5, and 6 is that it should be divisible by 60. Hence, the puzzle is to solve the integer equation
7a - 60b = 1
where a and b are integers.
Number theory (theory of congruences) says that there is exactly one solution in the range of (0...6) for b and (0...59) for a. Since checking 7 is much easier than 60, we consider 1, 61, 121, ..., 361 for divisibility by 7, and find 301 = 43 * 7. So, (43, 5) is a particular solution.
Since (a, b) = (43, 5) is a solution, any (a, b) in the form (60k+43, 7k+5) is the general solution. (60*7*k gets cancelled in the expression.). So, the number is in the form 7*(60k+43) = 420k + 301.
Answer:
Any number in the form (420k + 301) is a solution.
The smallest positive solution is 301, followed by 721, 1141, ....
The largest negative solution is -119, followed by -539, -959,....
Happy puzzling!
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ciproantib
Senior Boarder
Posts: 71
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SN> The necessary and sufficient condition for a number to be SN> divisible by 2, 2, 3, 5, and 6 is that it should be divisible by SN> 60.
Should be '... to be divisible by 2, 3, 4, 5 and 6 is ......'
Sorry for the typo.
Happy puzzling!
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quest2006
Senior Boarder
Posts: 60
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The everyday concept of dividing two integers to get a remainder and quotient applies in a well-defined way only to positive numbers or when the remainder is zero. When dividing -119 by 6, it makes sense to get either a quotient of -20 and a remainder of 1, or a quotient of -19 and a remainder of -5. There is a convention in some computer-programming environments to give the remainder the same sign as the divisor, but it is only a convention, and not universal.
I consider that the problem is only about positive numbers and the negative solutions don't apply.
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