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Question 3
M is a 4-digit number. If the left most digit is removed, then the resulting three digit number is $$\frac{1}{9}^{th}$$ of M. How many such M's are possible?
Solution
According to question,
Number AXYZ/9 = XYZ
It can be written as
A000/9 + XYZ/9 = XYZ
A000/9 + XYZ/9 = 9XYZ/9
A000/9 = 8XYZ/9
A000 = 8XYZ
We have to find the maximum value of A so that when A000 is divided by 8 gives three digit numbers.
7000/8= 875 and 8000/8=1000.
So, answer is 7.
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