rhallik wrote:
Hello!
Would it be possible to solve the task as follows:
I. We know that the reminder is 15. As 60 is divisible by 15 we can assume that 15.15 could be one solution.
II. We know that the reminder is 16 which is 2*8. As 60 can be divided by 2 we can assume that 18.16 could be one solution.
III. We know that the reminder is 17. As 60 is not divisible by 17 or vice versa, we can assume that 17.17 cannot be one of the solutions.
Also, please explain if my approach has a flaw.
Thank you in advance!
I think, it will not work every time. I will show you how.
Let's change the third option. Lets assume Remainder is 21 and quotient is 5.10
We know that the remainder is 0.10 As 21 is not divisible by 10 or vice versa, we can assume that 5.10 cannot be one of the solutions. Can this be true??
21/0.10 ------> 2100 / 10 ---------> well divisible by 10.
in the division 21/0.10 when you move the decimal sign of denominator to the right by two places, you add two zeros in the numerator to create the fraction 2100/10
When you are awarding two zeros to the numerator (i.e. to 21) remember you are actually giving 2 twos and 2 fives to the numerator. Why is so. That is because One zero can be obtained by multiplying 5 with 2
Now since the denominator (i.e. 10) contains 1 five and 1 two and since you just have given 2 fives and 2 twos, you can surely divide numerator by the denominator.