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# Divisibility AMC 8, 2016 Problem 5

## Competency in Focus: Divisibility.

This problem from American Mathematics contest (AMC 8, 2016) is based on the concept of divisibility .

## Next understand the problem

[/et_pb_text][et_pb_text _builder_version="4.2.2" text_font="Raleway||||||||" text_font_size="20px" text_letter_spacing="1px" text_line_height="1.5em" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]The number $N$ is a two-digit number. • When $N$ is divided by $9$, the remainder is $1$. • When $N$ is divided by $10$, the remainder is $3$. What is the remainder when $N$ is divided by $11$?   $\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7$[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="4.0"][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="4.2.2" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" _builder_version="4.2.2" hover_enabled="0" open="on"]

American Mathematical Contest 2016, AMC 8  Problem 5 [/et_pb_accordion_item][et_pb_accordion_item title="Key Competency" open="off" _builder_version="4.2.2"]

### Divisibility

[/et_pb_text][et_pb_tabs _builder_version="4.2.2"][et_pb_tab title="HINT 0" _builder_version="4.0.9"]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title="HINT 1" _builder_version="4.2.2"]When $N$ is divided by $10$ it leaves remainder $3$ i.e., $N=10\times P+3 \textbf{ ,where } P$ is an integer. i.e., The unit digit of $N$ must be $3$ because unit digit of $10P$ is zero.[/et_pb_tab][et_pb_tab title="HINT 2" _builder_version="4.2.2"]$N$ leaves remainder $1$ when divided by $9$. i.e., $N=9\times Q+1$ where $Q$ is an integer. Since $10\times P+3=9\times Q+1$ the unit digit of $9\times Q +1$ must be $3$.        [/et_pb_tab][et_pb_tab title="HINT 3" _builder_version="4.2.2"]Since $N$ is a two digit number then the only possibility is $Q=8$ i.e., $N=9\times 8+1=73$[/et_pb_tab][/et_pb_tabs][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section fb_built="1" fullwidth="on" _builder_version="4.2.2" global_module="50833"][et_pb_fullwidth_header title="AMC - AIME Program" button_one_text="Learn More" button_one_url="https://www.cheenta.com/amc-aime-usamo-math-olympiad-program/" header_image_url="https://www.cheenta.com/wp-content/uploads/2018/03/matholympiad.png" _builder_version="4.2.2" title_level="h2" background_color="#00457a" custom_button_one="on" button_one_text_color="#44580e" button_one_bg_color="#ffffff" button_one_border_color="#ffffff" button_one_border_radius="5px"]

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