Who am I?
a) i am divisible by 8 and 10 but not by 6.
b) if you divide me by 11, i leave a remainder of 5.
c) I am divisible by 3 and 4 and leave a remainder of 2 when divided by 5
Answers
Answer:
a. 80
b. 16, 27, 38..
c. 12
Hope it helps you....
a) We have to find the numbers that are divisible by both 8 and 10 but not by 6.
For a number to be divisible by 10 it should end with 0.
Therefore, the numbers that was are divisible by 10 are 10, 20, 30, 40, 50, 60, 70, 80, .............. and so on
Out of the above the numbers that are divisible by 8 are 40, 80, ...... and so on.
Also, the above numbers 40, 80 are not divisible by 6.
Hence, our above condition is satisfied.
The numbers that are divisible by 8 and 10 but not by 6 are 40, 80, 120, ....... and so on.
The pattern observed here is that the difference between two consecutive terms is 40.
b) We need to find such numbers that on division by 11 leaves a remainder.
We will solve this by adding 5 to the numbers that are divisible by 11.
For example, the terms that are divisible by 11 are 11, 22, 33, 44,...............
We will add 5 to all the numbers in the series above and we will get
16, 27, 38, 49, ................... and so on.
The numbers thus obtained that leave a remainder of 5 on division by 11 are 16, 27, 38, 49, .................... and so on.
The pattern observed here is that the difference between two consecutive terms is 11.
c) According to the question we have to find a number that is divisible by 3 and 4 both but leaves a remainder 2 when divided by 5.
Now the smallest common number that is divisible by both 3 and 4 is 12.
Also 12 on division by 5 leaves a remainder 2.
Therefore, our condition is satisfied.
The number that is divisible by both 3 and 4 and leaves a remainder of 2 when divisible by 5 is 12.
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