A number when divided by 2, 3, 4, 5 and 6 leaves remainder 1, 2, 3, 4 and 5, it is divisible by 7, then the least possible number is
Answers
Answer:
119
Hope this helps.
Step-by-step explanation:
Let x be such a number.
We can ignore the first two bits of information since leaving a remainder of 5 when divided by 6 automatically means there'll be a remainder of 1 when divided by 2 and a remainder of 2 when divided by 3.
So we just need: remainders 3, 4, 5, 0 when divided by 4, 5, 6, 7.
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Remainder of 5 when divided by 6 => remainder of 5 or 11 when divided by 12.
Remainder of 3 when divided by 4 => remainder of 3, 7 or 11 when divided by 12.
Therefore x must leave a remainder of 11 when divided by 12.
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Remainder of 11 when divided by 12 => remainder of 11, 23, 35, 47 or 59 when divided by 60.
Remainder of 4 when divided by 5 => remainder of 4, 9, 14, 19, ... , 54 or 59 when divided by 60.
Therefore x must leave a remainder of 59 when divided by 60.
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So the form of x is
x = 60k + 59
= 56(k+1) + 4k + 3.
Since 7 divides evenly into 56(k+1), for x to give a remainder of 0 when divided by 7, we must have 7 dividing evenly in 4k+3.
The smallest (non-negative) k that makes this true is k = 1, as that make 4k+3 = 7, which is certainly divided evenly by 7 !!
Therefore, the number we are looking for is
x = 60 + 59 = 119
Answer:
119
Hope this helps.
Step-by-step explanation:
Let x be such a number.
We can ignore the first two bits of information since leaving a remainder of 5 when divided by 6 automatically means there'll be a remainder of 1 when divided by 2 and a remainder of 2 when divided by 3.
So we just need: remainders 3, 4, 5, 0 when divided by 4, 5, 6, 7.
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Remainder of 5 when divided by 6 => remainder of 5 or 11 when divided by 12.
Remainder of 3 when divided by 4 => remainder of 3, 7 or 11 when divided by 12.
Therefore x must leave a remainder of 11 when divided by 12.
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Remainder of 11 when divided by 12 => remainder of 11, 23, 35, 47 or 59 when divided by 60.
Remainder of 4 when divided by 5 => remainder of 4, 9, 14, 19, ... , 54 or 59 when divided by 60.
Therefore x must leave a remainder of 59 when divided by 60.
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So the form of x is
x = 60k + 59
= 56(k+1) + 4k + 3.
Since 7 divides evenly into 56(k+1), for x to give a remainder of 0 when divided by 7, we must have 7 dividing evenly in 4k+3.
The smallest (non-negative) k that makes this true is k = 1, as that make 4k+3 = 7, which is certainly divided evenly by 7 !!
Therefore, the number we are looking for is
x = 60 + 59 = 119
Step-by-step explanation: