Select Page

Understand the problem

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. )

Source of the problem
American Mathematics Competition
Topic
Enumerative Combinatorics

7/10

Suggested Book
Introductory Combinatorics by Richard Brualdi

So, well have a long look at the problem. With a little bit of thought, you might even crack this without proceeding any further !

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 ?

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…

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…?

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.

Connected Program at Cheenta

AMC Training Camp

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

A  special training program for American Mathematics Contest.

Similar Problems

Amplitude and Complex numbers | AIME I, 1996 Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1996 based on Amplitude and Complex numbers.

Roots of Equation and Vieta’s formula | AIME I, 1996 Problem 5

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1996 based on Roots of Equation and Vieta’s formula.

Triangle and integers | AIME I, 1995 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Triangle and integers.

Tetrahedron Problem | AIME I, 1992 | Question 6

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Tetrahedron Problem.

Sequence and greatest integer | AIME I, 2000 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and the greatest integer.

Arithmetic sequence | AMC 10A, 2015 | Problem 7

Try this beautiful problem from Algebra: Arithmetic sequence from AMC 10A, 2015, Problem. You may use sequential hints to solve the problem.

Series and sum | AIME I, 1999 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Series and sum.

Inscribed circle and perimeter | AIME I, 1999 | Question 12

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2011 based on Rectangles and sides.

Problem based on Cylinder | AMC 10A, 2015 | Question 9

Try this beautiful problem from Mensuration: Problem based on Cylinder from AMC 10A, 2015. You may use sequential hints to solve the problem.

Cubic Equation | AMC-10A, 2010 | Problem 21

Try this beautiful problem from Algebra, based on the Cubic Equation problem from AMC-10A, 2010. You may use sequential hints to solve the problem.