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A is a set of positiv integer such than when divide by 2,3,4,5,6 leaves a remainder 1,2,3,4,5 respectively.how many integer between 0-100 lies in the set?
First figure out the least positive number which is exactly divisible by 2, 3, 4, 5 & 6 i.e. LCM (2,3,4,5,6) = 60.So any number in the form of 60x k where k belongs to n will be divisible by 2 ,3 ,4, 5 & 6Now we need a remainder 1 when divided by 2. So we can add 1 to 60 and say that 61 satisfies this condition . But can we do that?We also need a remainder 2 when divided by 3. But 61 does not fit into this.So let’s go back to the initial requirement.In order to get a reminder 1 when a number is divided by 2, there are two ways - one is to add 1 to 60 and other is to subtract 1 from 60Similarly to get a remainder 2 when divided by 3, one way to achieve it is to add 2 to 60 and the other is to subtract 1 (which is 3 - 2) from 60. Now you can find a pattern here. Though the remainders are different, the difference between the divisor and the respective remainders is constant which is 1 in this case.So the required numbers will be in the form of 60 k – 1 , where k is any natural number.And in the given range of 0 to 100, there is only number which is 59. (for k = 1)
First figure out the least positive number which is exactly divisible by 2, 3, 4, 5 & 6 i.e. LCM (2,3,4,5,6) = 60.
So any number in the form of 60x k where k belongs to n will be divisible by 2 ,3 ,4, 5 & 6
Now we need a remainder 1 when divided by 2. So we can add 1 to 60 and say that 61 satisfies this condition . But can we do that?
We also need a remainder 2 when divided by 3. But 61 does not fit into this.
So let’s go back to the initial requirement.
In order to get a reminder 1 when a number is divided by 2, there are two ways - one is to add 1 to 60 and other is to subtract 1 from 60
Similarly to get a remainder 2 when divided by 3, one way to achieve it is to add 2 to 60 and the other is to subtract 1 (which is 3 - 2) from 60. Now you can find a pattern here. Though the remainders are different, the difference between the divisor and the respective remainders is constant which is 1 in this case.
So the required numbers will be in the form of 60 k – 1 , where k is any natural number.
And in the given range of 0 to 100, there is only number which is 59. (for k = 1)
number should be in form of2k+1,3l+2,4h+3,5m+4,6t+5so it will be 59 only
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