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

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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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