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

# Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.22.4" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px"]Find the 3-digit number whose ratio with the sum of its digits is minimal.

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.22.4" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.23.3"]Suppose that the number is $(abc)_{10}$. Then we would like to minimise $\frac{100a+10b+c}{a+b+c}$. Try to minimise for one variable at a time. [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.23.3"]Note that, $\frac{100a+10b+c}{a+b+c}=1+\frac{9(11a+b)}{a+b+c}$. In this expression, it is possible to minimise for $c$ independent of $a,b$. [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.23.3"]From the previous hint, the expression is minimised for $c=9$. Show that the ratio can now be rewritten as $1+\frac{10a-9}{a+b+9}$. Minimise for $b$. [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.23.3"]In the previous hint we see that the minimising value of $b$ is also 9. Finally, the ratio may be written as $10-\frac{189}{a+18}$. This is minimised for $a=1$. Hence the answer is 199. [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px"]

# Similar Problems

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