Enumerative Combinatorics- AMC 10A 2017 Problem 25 Sequential Hints
Join Trial or Access Free ResourcesJoin Trial or Access Free Resources
and 
, inclusive, have the property that some permutation of its digits is a multiple of 
between and 
( For example, both 
and 
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"]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 = ( ( 990 - 110 ) / 11 ) + 1 = 81.
Now, how about you try taking things ahead from here onwards ?
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, ( 243 - 17 ) = 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"]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://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"]
Publishing original scientific research in a peer-reviewed journal is an achievement many scholars work toward for years. Achieving this feat early in one's academic journey requires exceptional intellectual rigor, creativity, and persistence. Aaditya Punatar and Muskan Sadana have reached this remarkable milestone, successfully publishing their collaborative research paper in the prestigious Journal of Emerging Investigators […]
Her project on quantum-accelerated Steiner trees for 15-minute cities earned a place among the finalists at the world's largest pre-college STEM competition. Prishaa Shrimali of Marymount School of New York, was selected as a Grand Finalist at the 2026 Regeneron International Science and Engineering Fair (ISEF) in Phoenix, Arizona. Her project was Optimizing Urban Accessibility: […]
In 2026, the following Cheenta students have been successful for Indian Statistical Institute's M.Stat Entrance. They ranked within the first 50 in the entire country in these entrances. I.S.I. M.Stat Entrance
In 2026, the following Cheenta students have been successful for Indian Statistical Institute's B.Stat Entrance and Chennai Mathematical Institute's B.Sc. Math Entrance. They ranked within the first 200 in the entire country in these entrances. Most of these students attended the problem solving workshops regularly, which happen 5 days every week. CMI B.Sc. Math Entrance […]