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Consider all nine digit numbers which 1 , 2 , 3...9 1,2,3...9 digits appear at least only and one time, the sum of this numbers is N N .Find the greatest prime factor of N N .


The answer is 333667.

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1 solution

Paola Ramírez
Jan 7, 2015

First, we realized that digit sum of unities,tens,hundreds...columns it's the same. There are 9 ! 9! numbers of this kind, so each column had 8 ! 8! times the numbers 1 1 to 9 9 .

Sum of each colum is ( 9 ) ( 10 ) 2 ( 8 ! ) ( 1 0 k 1 ) \frac{(9)(10)}{2}(8!)(10^{k-1}) and k k is the column number.

The sum of all columns is N N and N = ( 9 ) ( 10 ) 2 ( 8 ! ) ( 1 0 0 ) N=\frac{(9)(10)}{2}(8!)(10^{0}) + ( 9 ) ( 10 ) 2 ( 8 ! ) ( 1 0 1 ) \frac{(9)(10)}{2}(8!)(10^{1}) + ( 9 ) ( 10 ) 2 ( 8 ! ) ( 1 0 3 ) \frac{(9)(10)}{2}(8!)(10^{3}) +...+ ( 9 ) ( 10 ) 2 ( 8 ! ) ( 1 0 8 ) \frac{(9)(10)}{2}(8!)(10^{8}) .

Factoring N = 8 ! × 45 × 111 , 111 , 111 N=8!\times 45\times111,111,111

Using divisibility for 111 , 111 , 111 111,111,111 , 111 , 111 , 111 = 9 × 37 × 333667 111,111,111=9\times 37\times 333667

N = 2 7 × 3 6 × 5 2 × 7 × 37 × 333667 333667 N=2^7×3^6×5^2×7×37×333667 \therefore \boxed{333667} is the greatest prime factor of N N .

what do you mean by the sum??

number from 1 to 9 has sum of 45

total sums will be 9! * 45 this will give 7 as the greatest prime factor

Ossama Ismail - 6 years, 5 months ago

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Sumatory of al permutations that can be formed with the number 1 to 9. For example with 1,2,3 , N will be 123+132+213+231+312+321=1332.

Paola Ramírez - 6 years, 5 months ago

how did you know that 333667 is a prime?

Andrea Virgillito - 4 years, 2 months ago

I did the same your method but at the end to determine whether that number was a prime I used a software.

Andrea Virgillito - 4 years, 2 months ago

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