Categories
AMC 10 USA Math Olympiad

Enumerative Combinatorics- AMC 10A 2017 Problem 25 Sequential Hints

AMC 10A 2017, Problem 25 needed a clever trick of combinations and playing with numbers. See the solution with sequential hints.

[et_pb_section fb_built=”1″ _builder_version=”4.0″][et_pb_row _builder_version=”3.25″][et_pb_column type=”4_4″ _builder_version=”3.25″ custom_padding=”|||” custom_padding__hover=”|||”][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway|300|||||||” text_text_color=”#ffffff” header_font=”Raleway|300|||||||” header_text_color=”#e2e2e2″ background_color=”#0c71c3″ custom_padding=”20px|20px|20px|20px” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″]

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″]How many integers between $100$ and $999$, inclusive, have the property that some permutation of its digits is a multiple of $11$ between $100$ and $999?$ ( For example, both $121$ and $211$ have this property. )

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version=”3.25″][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=”4.0″ toggle_font=”||||||||” body_font=”Raleway||||||||” text_orientation=”center” custom_margin=”10px||10px”][et_pb_accordion_item title=”Source of the problem” open=”on” _builder_version=”4.0″]American Mathematics Competition [/et_pb_accordion_item][et_pb_accordion_item title=”Topic” _builder_version=”4.0″ open=”off”]Enumerative Combinatorics 

[/et_pb_accordion_item][et_pb_accordion_item title=”Difficulty Level” _builder_version=”4.0″ open=”off”]7/10

[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”3.29.2″ open=”off”]Introductory Combinatorics by Richard Brualdi

[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway|300|||||||” text_text_color=”#ffffff” header_font=”Raleway|300|||||||” header_text_color=”#e2e2e2″ background_color=”#0c71c3″ custom_margin=”48px||48px” custom_padding=”20px|20px|20px|20px” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color=”#0c71c3″ inactive_tab_background_color=”#000000″ _builder_version=”4.0″ tab_text_color=”#ffffff” tab_font=”||||||||” background_color=”#ffffff”][et_pb_tab title=”Hint 0″ _builder_version=”3.22.4″]So, well have a long look at the problem. With a little bit of thought, you might even crack this without proceeding any further !

[/et_pb_tab][et_pb_tab title=”Hint 1″ _builder_version=”4.0″]

Stuck ?…No worries. Try asking yourself – How many numbers between 100-999 are exactly divisible by 11 ? This is quite easy to figure out. 
What is the largest number in the range divisible by 11 ? Easy, 990.
How about the smallest such number ? Yeah, 110.
So, the number of integers in the range visible by 11 = ( ( 990110 ) / 11 ) + 1 = 81.
Now, how about you try taking things ahead from here onwards ?

[/et_pb_tab][et_pb_tab title=”Hint 2″ _builder_version=”4.0″]

Now see, 81 numbers are divisible by 11 in the range, as we just saw. All that’s left is to find out the permutations of these. Wait ! That’s simple, isn’t it ? Yes, 81 x 3 = 243. At this very juncture, ask yourself why. If you find out the answer to this “why”, you can might as well say you’ve gone far enough to solve the problem…  

[/et_pb_tab][et_pb_tab title=”Hint 3″ _builder_version=”4.0″]

    Let’s answer the “why” here. Say, we have a 3-digit number, abc. Now, ideally we would have had 6 permutations ( 3 ! simply ) for each such abc. But here’s a catch ! If abc is divisible by 11, so is cba.…!!! Yeah, that’s it. So, basically if we multiply by 6, we are accounting for same kind of permutations twice. So basically, each multiple of 11 in the range has it’s ( 6/2 ) = 3 permutations, that we are bothered about. This clearly justifies the fact that we can at maximum have 81 x 3 = 243 numbers in our desired solution set. But wait ? Why do I say, at maximum ? So…have we overcounted ? Yeah,we have. Why don’t you think about it…?          

[/et_pb_tab][et_pb_tab title=”Hint 4″ _builder_version=”4.0″]

Well, as you might have felt, we did overcount. We did not account for the numbers where 0 could be one of the digits. We overcounted cases where the middle digit of the number is 0 and the last digit is 0. So, what are these ? Let’s find them out. In how many of the numbers is the last digit 0 ? That’s easy…they have a pattern. It goes like 110, 220,….990. That makes it 9. Now, in how many of those 243 numbers that we are bothered about, does ‘0’ occur as a middle digit ? With a little bit of insight, you’d find out they are 803, 902, 704, 605. And well their permutations too. So that makes it 4 x 2 = 8.  So, in total, 9+8 = 17 elements have been overcounted. 

 Subtract that from 243, ( 24317 ) = 226, that’s your answer.

[/et_pb_tab][/et_pb_tabs][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section fb_built=”1″ _builder_version=”4.0″ global_module=”40396″ saved_tabs=”all”][et_pb_row _builder_version=”4.0″][et_pb_column type=”4_4″ _builder_version=”4.0″][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway|300|||||||” text_text_color=”#ffffff” header_font=”Raleway|300|||||||” header_text_color=”#e2e2e2″ background_color=”#0c71c3″ min_height=”12px” custom_margin=”50px||50px” custom_padding=”20px|20px|20px|20px” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″]

Connected Program at Cheenta

[/et_pb_text][et_pb_blurb title=”AMC Training Camp” url=”https://www.cheenta.com/amc-10-12-training-camp-american-mathematics-competition/” url_new_window=”on” image=”https://www.cheenta.com/wp-content/uploads/2018/03/matholympiad.png” _builder_version=”4.0″ header_font=”||||||||” header_text_color=”#e02b20″ header_font_size=”48px” link_option_url=”https://www.cheenta.com/amc-10-12-training-camp-american-mathematics-competition/” link_option_url_new_window=”on”]

One on One class for every student. Plus group sessions on advanced problem solving. 

A  special training program for American Mathematics Contest.

[/et_pb_blurb][et_pb_button button_url=”https://www.cheenta.com/amc-10-12-training-camp-american-mathematics-competition/” url_new_window=”on” button_text=”Learn More” button_alignment=”center” _builder_version=”4.0″ custom_button=”on” button_bg_color=”#0c71c3″ button_border_color=”#0c71c3″ button_border_radius=”0px” button_font=”Raleway||||||||” button_icon=”%%3%%” background_layout=”dark” button_text_shadow_style=”preset1″ box_shadow_style=”preset1″ box_shadow_color=”#0c71c3″][/et_pb_button][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway|300|||||||” text_text_color=”#ffffff” header_font=”Raleway|300|||||||” header_text_color=”#e2e2e2″ background_color=”#0c71c3″ custom_margin=”50px||50px” custom_padding=”20px|20px|20px|20px” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″]

Similar Problems

[/et_pb_text][et_pb_post_slider _builder_version=”4.0″ include_categories=”428″ image_placement=”left” hover_enabled=”0″][/et_pb_post_slider][/et_pb_column][/et_pb_row][/et_pb_section]

Leave a Reply

Your email address will not be published. Required fields are marked *