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# 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

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[/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_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 !

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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 ?

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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…

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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…?

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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.

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# Connected Program at Cheenta

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

A  special training program for American Mathematics Contest.

# Similar Problems

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## Does there exist a Magic Rectangle?

Magic Squares are infamous; so famous that even the number of letters on its Wikipedia Page is more than that of Mathematics itself. People hardly talk about Magic Rectangles.

Ya, Magic Rectangles! Have you heard of it? No, right? Not me either!

So, I set off to discover the math behind it.

#### Does there exist a Magic Rectangle?

First, we have to write the condition mathematically.

Take a table of dimension $m$ x $n$. Now fill in the tables with positive integers so that the sum of the rows, columns, and diagonals are equal. Does there exist such a rectangle?

Let’s start building it from scratch.

Now let’s check something else. Let’s calculate the sum of the elements of the table in two different ways.

Let’s say the column, row and diagonal sum be $S$. There are $m$ rows and $n$ columns.

#### Row – wala Viewpoint

The Rows say the sum of the elements of the table is $S.m$. See the picture below.

#### Column – wala Viewpoint

The Rows say the sum of the elements of the table is $S.n$. See the picture below.

Now, magically it comes that the $S.m = S.n$. Therefore the number of rows and columns must be equal.

Whoa! That was cute!

Visit this post to know about Magic Square more.

Edit 1: Look into the comments for a nice observation that if we allowed integers, and the common sum is 0, then we may not have got the result. Also we need to define the sum of the entries of a diagonal of a rectangle.

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## Big idea – stabilization

In a counting problem, you may need to stabilize the number of cases. This means, by letting some of the variables change, one may fix the remaining cases.

This is the central idea in the following problem from American Mathematical Contest 8 (AMC 8, 2018, Problem 9).

### Problem

In a sign pyramid a cell gets a “+” if the two cells below it have the same sign, and it gets a “-” if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a “+” at the top of the pyramid?

### Sequential Hints

(How to use this discussion: Do not read the entire solution at one go. First, read more on the Key Idea, then give the problem a try. Next, look into Step 1 and give it another try and so on.)

#### Hint 1

Start from the top! Fix the + sign at the top of the pyramid and try filling in the second row from left. Do you see any pattern?

(What happens if you plugin a + sign in the left most block of second row from top?)

#### Hint 2

If we fix a + sign in the left most block of second row from top, the other box must contain a + sign. (Why?)

On the other hand fixing a – sign in the left most block of second row from top will force the remaining block to contain – sign.

Thus by varying the left most block of the second row, we can fix the entire row.

Will this work for the third row?

#### Hint 3

In fact the same trick will work for third row.

Try this. Fix a plus sign in the first row. Then fix second row as well ( plus, plus or minus, minus).

Then try to plugin some sign in the third row’s first block. And check that the remaining two blocks get automatically fixed. This is precisely known as stabilization.

#### Final Hint

Thus, we should only worry about the first block of second, third and fourth. Fixing them fixes the entire row.

There are 2 choices for the first block of second row, 2 choices for first block of third row and 2 choices for first block of third row. Hence in total we have $2 \times 2 \times 2 = 8$ cases.

Recommended book

Principles and Techniques in Combinatorics Kindle Edition by Chen Chuan-Chong

Try chapter 1.