N is a four digit number. If the leftmost digit is removed, the resulting three digit number is 1/9th of N. How many such N are possible?

Question

N is a four digit number. If the leftmost digit is removed, the resulting three digit number is 1/9th of N. How many such N are possible with solution?

Answer 1

Suppose we consider the four digit number $N$ as "$abcd$", i.e.,

\begin{eqnarray} N=1000a+100b+10c+d. \end{eqnarray}

We are given

\begin{align} \frac{1000a+100b+10c+d}{9}&=(100b+10c+d)\\ \Rightarrow 125a-100b-10c-d=0. \end{align}

This allows $a$ to vary from $1-7$, and the resulting numbers are $1125,2250,3375,4500,5625,6750,$ and $7875$.

Answer 2

Let $x\in[1,9]$ denote the leftmost digit.

Let $y\in[000,999]$ denote the remaining digits.

$1000x+y=9y\implies1000x=8y\implies125x=y$.

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Hence there are $7$ possible values of $N$:

• $x=\color\red{1} \implies y=\color\green{ 125} \implies N=\color\red{1}\color\green{ 125}$
• $x=\color\red{2} \implies y=\color\green{ 250} \implies N=\color\red{2}\color\green{ 250}$
• $x=\color\red{3} \implies y=\color\green{ 375} \implies N=\color\red{3}\color\green{ 375}$
• $x=\color\red{4} \implies y=\color\green{ 500} \implies N=\color\red{4}\color\green{ 500}$
• $x=\color\red{5} \implies y=\color\green{ 625} \implies N=\color\red{5}\color\green{ 625}$
• $x=\color\red{6} \implies y=\color\green{ 750} \implies N=\color\red{6}\color\green{ 750}$
• $x=\color\red{7} \implies y=\color\green{ 875} \implies N=\color\red{7}\color\green{ 875}$

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In the remaining cases, $y\not\in[000,999]$:

• $x=\color\red{8} \implies y=\color\green{1000}$
• $x=\color\red{9} \implies y=\color\green{1125}$

Answer 3

You have:

$$ \begin{align*} 1000a+100b+10c+d=9(100b+10c+d) &\iff 1000a=800b+80c+8d \\ &\iff125a=100b+10c+d=bcd \end{align*} $$

The answer is clearly given by all the values of the digit $a$ such that $125a$ have three digits, then there are seven possibilities ($a=1,2,3,4,5,6,7$ are good; $a=0$ $a=8,9$ must be discarded).