A combinatorics problem by Naitik Sanghavi -1
Probability
Level
4
Four digit numbers are formed using digits 1,2,3,4(repetition among the digits is allowed) Find the number of such four digit numbers divisible by 11.
Note: Repetition is allowed.
The answer is 44.
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1 solution
There may be a more elegant solution but I was able to list out multiples of 11 by knowing the divisibility by 11 rule. That is that when you add up the digits and alternate the signs, the sum is 0. For instance, 1243 is divisible by 11 because 1-2+4-3=0
Here's an attempt at a way without listing out the multiples directly: Let's fix the subtraction of the first two digits. The only possible values for this subtraction are -3,-2,-1,0,1,2,3.
So the other two digits must subtract and then add to the first two digit's subtraction and be 0.
For purposes of symmetry we will only look at 0 and the positive.
How many ways can we subtract x-y and get 0 when limited to 1 thru 4? We can see that there are 4 ways to do this.
What about 1? There are 3 ways to do this. What about 2? There are 2 ways to do this. What about 3? There is 1 way to do this.
The logic would be the same for the negatives.
So if we fix the first digit to 1, the possible subtraction of the first two digits are 0,1,2,3
So this would result in 4+3+2+1 = 10 ways
For 2, we have -1,0,1,2 which results in 3+4+3+2=12 ways
For 3 we have -2,-1,0,1 which results in 2+3+4+3=12 ways
And for 4 we have -3,-2,-1,0 which looks a lot like the 1 case. But this would result in 1+2+3+4 =10 ways.
10+12+12+10=44 numbers divisible by 11