Find out the smallest number which leaves a remainder of 1 when divided by 2, 3, 4, 5, 6 but divided by 11 completely.

Dear student 

So smallest  number  satisfying  given  condition  is 121
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61 when divided by 2,3,4,5,6 leaves remainder but also not divisible by 11. But we can find the answer with multiples of 11.
So, it can be number itself means 11 because 11 is a prime number and when divided by 2,3,4,5,6 not gives exactly 1.
So. Now. 22 it's composite means completely divisible by 2 so it also cannot be the number.
33 is visible by 3 so this is also not the number.
44 it is completely divisible by 4 it is also not the number.
55 it is completely divisible by 5 it is also not the number.
66 is completely divisible by 6 so it is also not the number.
77 it is not divisible by any of the number but it do not leave 1 as remainder in each case.
88 is not divisible by other numbers but it is completely divisible by 4 and 2 it is also not the number.


Sorry I can't answer you full but you can try out this method may it helps or find the lcm by adding 1 in each case except 11.
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