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$N$ is a $4$ digit number If the left most digit is removed the resulting number is $ \frac{1}{9}{th} \ of \ N$ How many such numbers are possible?
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Let $N=abcd,$ meaning $N = 1000a+100b+10c+d.$

Now, given that $100b+10c+d = N/9$

So, $1000a = 8N/9 \implies 1125a = N.$

So, possible values of $a = \{1,2,3,4,5,6,7,8\}.$

(When $a=9, N=10000$ and this is no longer a 4 digit number)

Thus, no. of possible values of $N = 8.$
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I was confused with  $a=8,N=9000$ Please report if my reasoning is right or not!
Then other terms will sum $1000$ and complete number will be $9000+1000=10000$ thats what you have mentioned right?
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Thanks. Corrected that.
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