TIFR CSE 2011 | Part A | Question: 20

Question

Let $n>1$ be an odd integer. The number of zeros at the end of the number $99^{n}+1$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$
• E. None of the above

Answer 1

For odd $n,$

$\begin{align} 99^n &= (100-1)^n\\[1em] \quad &= \small 100^n - {n \choose 1}100^{n-1} +\quad \cdots \quad - {n \choose n-2}100^2 + {n \choose n-1}100^1\normalsize\,\color{blue}{-1}\\[2em] \hline 99^n \color{red}{+1} &= \small 100^n - {n \choose 1}100^{n-1} +\quad \cdots \quad - {n \choose n-2}100^2 + {n \choose n-1}100^1\normalsize\,\color{blue}{-1}\,\color{red}{+1}\\[1em] &= 100  \left(\small 100^{n-1} - {n \choose 1}100^{n-2} + \cdots - {n \choose n-2}100 \normalsize + \color{red}{n} \right) \end{align}$

Since $n$ is odd, it cannot end in a $0$

Thus, $99^n+1 = 100 \left( \ldots \text{ doesn't end with } 0 \right)$

which means $99^n+1$ ends with exactly $2$ zeros

Hence, option b) is correct.

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Alternative way:

$\begin{array}{llrlr} 99 &\times &01 &= &\color{blue}{99}\\ 99 &\times &\color{#f0f}{0}99 &= &98\color{red}{01}\\ 99 &\times &(\ldots)801 &= &(\ldots)2\color{blue}{99}\\ 99 &\times &(\ldots)\color{#f0f}{2}99 &= &(\ldots)6\color{red}{01}\\ 99 &\times &(\ldots)601 &= &(\ldots)4\color{blue}{99}\\ 99 &\times &(\ldots)\color{#f0f}{4}99 &= &(\ldots)4\color{red}{01}\\ 99 &\times &(\ldots)401 &= &(\ldots)6\color{blue}{99}\\ 99 &\times &(\ldots)\color{#f0f}{6}99 &= &(\ldots)2\color{red}{01}\\ 99 &\times &(\ldots)201 &= &(\ldots)8\color{blue}{99}\\ 99 &\times &(\ldots)\color{#f0f}{8}99 &= &(\ldots)\color{#f0f}{0}\color{red}{01} \end{array}$

Thus, $99^n$ always ends in a $99$ when $n$ is odd, but never in a $999$.

Hence, $99^n+1$ will always end with exactly $2$ zeros.

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Note: We couldn't just say that $99^3+1$ ends with exactly $2$ zeros, so b must be correct. This is because we also have an option e which says None of the above. Had it not been there, we could have marked b without having to prove that the pattern will continue.

Correct Answer: $B$

Comments

can u please elaborate  on "Since nn is odd, it cannot end in a 0" ??

Answer 2

Number of zeros at the end of $99^n + 1$ is decided by the number of $9’s$ at the end of $99^n .$

For example, if there are $2,$ $9’s$ at the end of $99^n$ then $99^n + 1 = *****99 + 1 = *******00$

and if there are $3,$ $9’s$ at the end of $99^n$ then $99^n + 1 = *****999 + 1 = *******000$

Now, we have to decide how many $9’s$ are at the end of $99^n$ when $n$ is odd and $>1.$

$99^n$ has unit digit as $9$ when $n$ is odd and $>1$ because when we multiply $99$ even number of times, we get $1$ as unit digit and again multiply by $99$ will give $9$ as unit digit.

So, it confirms that $99^n$ ends with $9.$

Now, to get last $2$ digits, we have to do $mod\; 100$ operation on $99^n$.

$99^n = (100-1)^n = 100^n - n*(100)^{n-1}+.......-(n*(n-1)/2)*100^2+n*100-1$

Here, in the binomial expansion of $99^n$, all terms have $100$ as a multiple except $“-1”.$ So, all terms will give remainder as $0$ except $-1$ when divided by $100.$

So, $99^n\; mod\; 100= -1 \; mod\; 100 = (100-1) = 99$

So, it confirms that $99^n$ ends with $99.$

Now, to get last $3$ digits, we have to do $mod\; 1000$ operation on $99^n$.

when $n$ is odd and $>1,$ all the terms in the binomial expansion of $99^n$ have $1000$ as multiple except $n*100-1$ and binomial coefficients are always integers. So, all terms will give remainder as $0$ except $100n-1$ when divided by $1000.$

So, $99^n\; mod\; 1000= 100n-1$

Since, $100n-1\; mod\;1000$ will give $999$ when $n=10,20,30,…$ i.e. multiple of $10$ but here $n$ is odd, it means we can’t get last $3$ digits as $999.$

So, $99^n$ always ends with $99$ but not $999$ when $n$ is odd and $>1.$ So, number $99^n + 1$ must be ended with $2$ zeros.