How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?

# Product of Digits, ISI Entrance 2017

## Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]  Let $g : \mathbb{N} \to \mathbb{N}$ with $g(n)$ being the product of digits of $n$.        (a) Prove that $g(n)\le n$ for all $n \in \mathbb{N}$ .        (b) Find all $n \in \mathbb{N}$ , for which $n^2-12n+36=g(n)$. [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" link_option_url="https://www.cheenta.com/isicmientrance/" link_option_url_new_window="on"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.3.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.3.4" hover_enabled="0"]
I.S.I. (Indian Statistical Institute, B.Stat, B.Math) Entrance. Subjective Problem 5 from 2017
[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.22.4" open="off"]Inequality

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.22.4" open="off"]9 out of 10

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Do you really need a hint? Try it first!

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Let $n$ be a $k$ digit(s) number number , then $n$ can be written as                             $n=a_0+10a_1+10^2a_2+\cdots+10^{k-1}a_{k-1}$ Where ,$a_0,a_1,...,a_{k-1}$ are digits of $n$.

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$a_0,a_1,...,a_{k-1} \in [1,9]$ as the range of the function $g$ is $\mathbb{N}$ $\Rightarrow a_0,a_1,...,a_{k-1}\neq 0$ . Now $g(n)=a_0a_1a_2\cdots a_{k-1}\le \underbrace{10\cdot10\cdot10\cdots 10}_{(k-1) times} \cdot a_{k-1}$   [Since $a_0,a_1,...,a_{k-1} \le 10$]  Equality holds when $k=1$ .

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$\Rightarrow g(n)\le 10^{k-1}\cdot a_{k-1}$ $\Rightarrow g(n)\le 10^{k-1}\cdot a_{k-1}+\cdots+10^2a_2+10a_1+a_0$    [Since $a_0,a_1,...,a_{k-1} >0$]  $\Rightarrow g(n)\le n$        (Proved) .

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$n^2-12n+36=g(n)$ $\Rightarrow n^2-12n+36 \le n$     [Since $g(n) \le n$ ] $\Rightarrow n^2-13n+36 \le 0$ $\Rightarrow (n-9)(n-4) \le 0$ $\Rightarrow 4\le n\le 9$ $\Rightarrow n={4,5,6,7,8,9}$ (Ans.)  .

## Connected Program at Cheenta

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

## Similar Problems

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