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5040 distinct seven digit base ten positive integers X₁, X₂, ...., X₅₀₄₀ are such that each X_i is formed by using each of the digits from 1 to 7 exactly once.

Determine the remainder when X₁ + X₂+ .......+ X₅₀₄₀ is divided by 7.

Edited March 23, 2010 by K Sengupta

[Guest]

5040 distinct seven digit base ten positive integers X₁, X₂, ...., X₅₀₄₀ are such that each X_i is formed by using each of the digits from 1 to 7 exactly once.

Determine the remainder when X₁ + X₂+ .......+ X₅₀₄₀ is divided by 7.

If a number is abcdefg

Then it is simply a*10^6 + b*10^5 + c*10^4 + ... + g*10^0

Consider the one's place.

Taking into account all permutations, the ones place will be occupied by each digit 1 to 7 with uniform frequency.

Because there are 7! permutations, each digit will have 6! instances when it will appear in the one's place.

Summing together all of the one's places for all 5040 numbers, you can factor out 6! to get 6!*(1+2+3+4+5+6+7)

But 1+2+3+4+5+6+7 = 28 which is congruent to 0 mod 7.

Thus the sum of all one's place digits is divisible by 7.

For 10's place digits, the final sum is 10*6!*28 which is divisible by 7 also

For 100's place, the sum is 100*6!*28 which is divisible by 7

...

...

Each of these seven sums is divisible by 7, so in turn, their sum is divisible by 7.

Thus the remainder is 0.