SOLUTION: If we factor the number 2012 into primes, we get 2012 = 2×2×503. If instead of multiplying these numbers we add them, the result is 507. It turns out this is a rather large-ish res
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-> SOLUTION: If we factor the number 2012 into primes, we get 2012 = 2×2×503. If instead of multiplying these numbers we add them, the result is 507. It turns out this is a rather large-ish res
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Question 557666: If we factor the number 2012 into primes, we get 2012 = 2×2×503. If instead of multiplying these numbers we add them, the result is 507. It turns out this is a rather large-ish result for numbers near 2012. If we did it for 2010 = 2×3×5×67, the total is 77, while the total for 2013 = 3×11×61 is 75. On the other hand, last year (2011) was prime, so had a total of…2011.
If we go back to 1944 = 2^33^5, the total is only 21. But what was the most recent year when the total was only 20? Answer by richard1234(7193) (Show Source):
You can put this solution on YOUR website! Note that 2^10 = 2*2*...*2 = 1024, sum of the prime factors is 20. We can prove that 1024 is the largest such number that satisfies the constraints (and is less than 2012).
We want to maximize given that
where p_j's are (not necessarily distinct) primes. However, by AM-GM inequality,
. If i is fixed, the LHS is a constant, and equality occurs if and only if p_1 = p_2 = ... = p_i. p_1 = 2 results in the largest possible N, 2^10 or 1024.
