document.write( "Question 1172343: Find number of positive integers n such that n+2n^2+3n^3+...+2019n^2019
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Algebra.Com's Answer #797359 by ikleyn(52908) $\"\"$ $\"About$

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document.write( "It is well known, obvious and self-evident fact that every degree    gives the remainder 1, when is divided by (n-1).\r\n" );
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document.write( "Therefore, the sum  n + 2n^2 + 3n^3 + . . . + 2019n^2019  gives the remainder\r\n" );
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document.write( "    1 + 2 + 3 + . . . + 2019\r\n" );
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document.write( "when is divided by (n-1).\r\n" );
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document.write( "In turn, the sum  1 + 2 + 3 + . . . + 2019  is the sum of the first 2019 natural numbers, and, therefore, is equal to  \r\n" );
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document.write( "     =  = 1010*2019.\r\n" );
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document.write( "The number  1010*2019  has THIS DECOMPOSITION into the product of prime numbers\r\n" );
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document.write( "            1010*2019 = (2*5*101)*(3*673).\r\n" );
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document.write( "So, its decomposition is the product of 5 prime numbers with multiplicities 1 for each participating prime.\r\n" );
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document.write( "Therefore, the number of divisors of the number  1 + 2 + 3 + . . . 2019 = 1010*2019  is  \r\n" );
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document.write( "    (1+1)*(1+1)*(1+1)*(1+1)*1+1) = 2*2*2*2*2 =  = 32.\r\n" );
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document.write( "Each such divisor is the potential number (n-1).\r\n" );
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document.write( "THEREFORE, the answer to the problem's question is  32.\r\n" );
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