If we exchange the positions of just two digits from the lowest common multiple of all positive numbers up to n, we will get the square of the sum of all positive numbers up to n. What is the largest prime factor of the sum between that lowest common multiple and that square?
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Using a calculator, I found that the lowest number satisfying the condition of the problem is n = 9 :
lcm ( 1 , 2 , . . . , 9 ) = 2 0 2 5
( 1 + 2 + . . . + 9 ) 2 = 4 5 2 = 2 5 2 0
2 0 2 5 + 2 5 2 0 = 4 5 4 5 = 3 2 ⋅ 5 ⋅ 1 0 1
The answer, therefore, is 1 0 1 .