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Number Theory
Level
4
Given five different non-zero digits, prove that it is possible to form 120 different five-digit numbers containing each of these five digits.
Submit your answer as the largest prime factor of the sum of the 120 numbers.
The answer is 271.
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1 solution
For each position we can fix one of the numbers and with this fixed position we can have 4! arrangements. Now we can consider that the number in the units digit is fixed and now we can have 4! arrangements with this fixed digit. If we consider only the total sums of the digits in the units place, we have 4! arrangements with each digit fixed. Therefore the total sum of the digits in the units place is: 4! x s where the s is the sum of the 5 given digits. Similarly, if we fix the digit in the tens place and find the total sum of the digits in the tens place, we get: 4! x 10 x s (multiplying by 10 because we're considering the tens place. We multiply by 100 for the next place , 1000 for the next place, and so on) If we proceed like this for the next three places, and add all of our results, we get the total sum of our arrangements which is: 4! x s x (1+10+100+1000+10000)= =4!x s x (11111). Prime factorising, we get: =2^3 x 3 x 5 x s x 41 x 271 Since the maximum value of s is 35 , the largest possible prime factor of s is 7 Therefore our answer is 271