# Understand the problem

##### Source of the problem

##### Topic

##### Difficulty Level

##### Suggested Book

# Start with hints

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…*

Let’s answer the “why” here. Say, we have a 3-digit number, * abc*. Now, ideally we would have had

**6**permutations (

*simply ) for each such*

**3 !***.*

**abc***But here’s a catch*

*!*If

*is divisible by*

**abc****11**, so is

*…!!! Yeah, that’s it. So, basically if we multiply by*

**cba.****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**, ( **243** – **17** ) = **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.