I will be grateful if anyone provide me the idea to solve the following problem. The problem says, we have five digits a, b, c, d, e (0 <= a, b, c, d, e <= 9). Now for these five digits, we can convert each of 120 permutations(5!=120) of these five digits into a decimal number. if the sum of all the 120 numbers is x, then we can exclude a number N from the sum which results in a sum of 119 numbers. Now we will be given that sum of 119 numbers, we have to find the numbers N that we excluded. if multiple answers possible output them all in ascending order. Note : the initial five numbers will not be provided. Problem Link
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It's mainly math. Try to calculate the final sum. You know, that each digit will appear on each decimal position the same number of times.
Digit $$$a$$$ will appear $$$24$$$ times on decimal position $$$1$$$ through $$$5$$$.
Digit $$$a$$$ will contribute $$$24 \cdot a \cdot (10000+1000+100+10+1)$$$ to the sum.
The final sum of all $$$120$$$ permutations will be $$$24 \cdot (a+b+c+d+e) \cdot (10000+1000+100+10+1)$$$
So the sum of $$$120$$$ Permutations has to be divisible by $$$24 \cdot 11111=266664$$$. So first find, how much do you need to add to the given number, to make it divisible by $$$266664$$$.
we can also try out all possibilities, that will also fits in the time limit.
link to solution
Thanks a lot OleschY for your contribution.
TOO much Difficult for me , Can anyone tell me the expected rating of this problem.