Question

Q. Let k = 1+ 11+ 111 + ........up to 50 terms. Find thousandth place digit in k.

please tell me the answer with explanation

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Answer 1

So we are given,

$\displaystyle\longrightarrow\sf{k=1+11+111+1111+\,\dots\dots50\ terms}$

Or,

$\displaystyle\longrightarrow\sf{k=1+11+111+1111+\,\dots\dots\,+\underbrace{\sf{111\dots111}}_{50}}$

I just divide and multiply 9 to it. Then I get,

$\displaystyle\longrightarrow\sf{k=\dfrac{1}{9}\times9[\,1+11+111+1111+\,\dots\dots\,+\underbrace{\sf{111\dots111}}_{50}\,]}$

$\displaystyle\longrightarrow\sf{k=\dfrac{1}{9}\,[\,9+99+999+9999+\,\dots\dots\,+\underbrace{\sf{999\dots999}}_{50}\,]}$

But each term can be written as,

$\displaystyle\longrightarrow\sf{k=\dfrac{1}{9}\,[\,(10-1)+(100-1)+(1000-1)+\,\dots\,+(1\underbrace{\sf{000\dots000}}_{50}-1)\,]}$

$\displaystyle\longrightarrow\sf{k=\dfrac{1}{9}\,[\,10-1+10^2-1+10^3-1+\,\dots\,+10^{50}-1\,]}$

$\displaystyle\longrightarrow\sf{k=\dfrac{1}{9}\,[\,10+10^2+10^3+\,\dots\,+10^{50}-(\underbrace{\sf{1+1+1+\,\dots\,+1}}_{50})\,]}$

$\displaystyle\longrightarrow\sf{k=\dfrac{1}{9}\,[\,10+10^2+10^3+\,\dots\,+10^{50}-50\,]\quad\quad\dots(1)}$

Consider the geometric series $\sf{10+10^2+10^3+\,\dots\,+10^{50}.}$

• $\sf{a=10\quad;\quad r=10\quad;\quad n=50}$

Then,

$\displaystyle\longrightarrow\sf{10+10^2+10^3+\,\dots\,+10^{50}=\dfrac{10\,\left(10^{50}-1\right)}{10-1}}$

$\displaystyle\longrightarrow\sf{10+10^2+10^3+\,\dots\,+10^{50}=\dfrac{10}{9}\,\left(10^{50}-1\right)}$

Then (1) becomes,

$\displaystyle\longrightarrow\sf{k=\dfrac{1}{9}\left[\dfrac{10}{9}\left(\,10^{50}-1\right)-50\right]}$

$\displaystyle\longrightarrow\sf{k=\dfrac{1}{9}\left[\dfrac{10^{51}-10}{9}-50\right]}$

$\displaystyle\longrightarrow\sf{k=\dfrac{10^{51}-10-450}{81}}$

$\displaystyle\longrightarrow\sf{k=\dfrac{10^{51}-460}{81}}$

But we see that,

$\displaystyle\longrightarrow\sf{k=\dfrac{1\overbrace{\sf{000\dots000}}^{51}-460}{81}}$

We separate last three zeroes in $\sf{10^{51}}$ from the other 48 zeroes.

$\displaystyle\longrightarrow\sf{k=\dfrac{1\overbrace{\sf{000\dots000}}^{48}000-460}{81}}$

So that, on subtracting $\sf{460}$ from $\sf{10^{51},}$ we get,

$\displaystyle\longrightarrow\sf{k=\dfrac{\overbrace{\sf{999\dots999}}^{48}540}{81}}$

$\displaystyle\longrightarrow\sf{k=\dfrac{\overbrace{\sf{111\dots111}}^{48}060}{9}}$

And from that 48 ones, let me separate 9 consecutive ones as each group from left, leaving the three right most ones alone. Then we see,

$\displaystyle\longrightarrow\sf{k=\dfrac{\overbrace{\sf{\overbrace{\sf{111\dots111}}^{9}\overbrace{\sf{111\dots111}}^{9}\dots\dots\overbrace{\sf{111\dots111}}^{9}}}^{5}111060}{9}\quad\quad\dots(2)}$

The reason for taking 9 consecutive ones as a group is due to the fact that,

• $\sf{111111111=9\times12345679}$

But I take $\sf{012345679,}$ since $\sf{111111111}$ has 9 digits.

And what about the remaining $\sf{111060\,?}$

• $\sf{111060=9\times12340}$

But I take $\sf{012340,}$ since $\sf{111060}$ has 6 digits.

Actually we have only to consider this $\sf{111060}$ only, which on division by 9 gives the thousands digit of k.

$\displaystyle\longrightarrow\sf{\dfrac{111060}{9}=12340}$

The thousands digit of $\sf{12340}$ is the same as that of k.

Hence 2 is the answer.

[Note:- Thousands place of a number is its fourth digit taken from right. If the number is a decimal number, then the third digit after the decimal point is the thousandths digit. ]

Well let me find the value of k !

After dividing the numerator by 9, (2) becomes,

$\displaystyle\longrightarrow\sf{k=\overbrace{\sf{012345679\,012345679\,\dots\,012345679}}^5012340}$

Or, undoubtedly,

$\displaystyle\longrightarrow\sf{\underline{\underline{k=12345679012345679012345679012345679012345679012340}}}$