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Olympiad workbook
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IOQM
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Basics of Mathematics

Olympiad workbook · IOQM · Basics of Mathematics

An integer n is such that [(n)/(9)] is a three digit number with...

An integer nn is such that \left[\frac{n}{9}\right]isathreedigitnumberwithequaldigits,and[n−1724]is a three digit number with equal digits, and \left[\frac{n-172}{4}\right] is a 4 digit number with the digits 2,0,2,42,0,2,4 in some order. What is the remainder when nn is divided by 100 ?

Correct answer

91

Step-by-step solution

 Let [n−1724]=k,k∈I⇒n−1724∈[k,k+1)n−172∈[4k,4k+4)n∈[4k+172,4k+176)n9∈[4k+1729,4k+1769) Now, [n9]∈{111,222,333,…,999}⇒4k+1769>aaa‾>4k+1729, for a∈{1,…,9} For a=9k∈(2203.75,2204.75)⇒k=2204\begin{aligned} & \text { Let }\left[\frac{n-172}{4}\right]=k, k \in I \\ & \begin{array}{l}\Rightarrow \quad \frac{n-172}{4} \in[k, k+1) \\ \quad n-172 \in[4 k, 4 k+4) \\ \quad n \in[4 k+172,4 k+176) \\ \quad \frac{n}{9} \in\left[\frac{4 k+172}{9}, \frac{4 k+176}{9}\right)\end{array} \\ & \text { Now, }\left[\frac{n}{9}\right] \in\{111,222,333, \ldots, 999\} \\ & \Rightarrow \quad \frac{4 k+176}{9}>\overline{a a a}>\frac{4 k+172}{9}, \text { for } a \in\{1, \ldots, 9\} \\ & \text { For } a=9 \\ & k \in(2203.75,2204.75) \\ & \Rightarrow \quad k=2204\end{aligned} [n−1724]=2204\left[\frac{n-172}{4}\right]=2204......(i)And[n9]=999And \left[\frac{n}{9}\right]=999......(ii) From (i): n−1724∈[2204,2205)⇒n∈[8988,8992)\begin{aligned} & \frac{n-172}{4} \in[2204,2205) \\ & \Rightarrow \quad n \in[8988,8992) \end{aligned} From (ii): n9∈[999,1000)⇒n∈[8991,9000)⇒n=8991⇒91\begin{aligned} & \frac{n}{9} \in[999,1000) \\ & \Rightarrow \quad n \in[8991,9000) \\ & \Rightarrow \quad n=8991 \\ & \Rightarrow \quad 91 \end{aligned}
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