Outline Mathematics
Number Theory

When 3AA1 is divisible by 11?

Here's a problem to tackle:

3AA1 is divisible by 11. Find A.

Solution


|Up| |Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2017 Alexander Bogomolny

Solution

3AA1 is divisible by 11. Find A.

A is a digit of a 4-digit number divisible by 11. A appears twice in the decimal representation of the number, with two other digits being 3,3,5,7,9 and 1,0,1,2,4,6. There is a well known criterion for divisibility by 11: a number is divisible by 11 iff the alternating sum of its digits is divisible by 11. The sum in question (for number 3AA1) equals

3 - A + A - 1 = 2

and is never divisible by 11. We are forced to conclude that the problem has no solution.

May we modify the problem a little to make sure a solution exists? Does the following problem have a solution: find a decimal digit B such that number B4B3 is divisble by 11.

Since it is still a question of divisibility by 11, we are going to compute the alternating sum of the digits, which, in this case, is given by

B - 4,1,2,3,4 + B - 3 = 2B - 7,4,5,6,7,8.

Let's estimate the possible values of the alternating sum:

-7,-4,-5,-6,-7,-8| ≤ 2B - 7 ≤ 11,10,11,12,13,14.

There are only two numbers divisible by 11 in this range: 0 and 11. The former leads us nowhere:

2B = 7,6,7,8,9,

with no integer solutions. We now check the latter:

2B - 7 = 11,9,10,11,12,13, or
2B = 18.

In other words, B = 9,5,6,7,8,9.

Answer: B = 9 and the number is 9493,9494,4939,9493,9394.

Check the answer: 9493 / 11 = 863,833,638,386,863,336. Very good.

You may want to try your hand at a similar problem of divisibility by 9.


Related material
Read more...

  • Divisibility Criteria
  • Fermat's Little Theorem
  • Divisibility by 7, 11, and 13
  • Divisibility Criteria (Further Examples)
  • Criteria of divisibility by 9 and 11
  • Division by 81
  • 5109094x171709440000 = 21!, find x

  • |Up| |Contact| |Front page| |Contents| |Algebra|

    Copyright © 1996-2017 Alexander Bogomolny

     62647697

    Search by google: