I really can't check them all!
Add any number of parentheses to the expression above such that the resultant number is an integer and is minimized. Submit your answer as this minimum number.
For instance, we could add 1 pair of parentheses like , the resultant number of which is an integer but is not necessarily minimized.
The answer is 3003.
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1 solution
The expected answer is in the form of (1x a 1 x...x a n)/(2 x a (n+1) x.... x a m) with a1, a2, ..., am represents permutation of 3,4,...,12. By adjusting the parentheses, we can decide which one becomes numerator and which one becomes denumerator, with exception of 1 and 2, which are already fixed. The multiplication result of numbers 1 until 13 is k= 2^10 . 3^5. 5^2. 7. 11.13. To get an integer, for each prime, there should be more or equal power in numerator compared to the one in denominator. Ideally, for each prime with odd power in k, to get minimum value, we should place the number such that the power of that prime in numerator only differs by 1 compared to it's power in denominator. Also, ideally, the allocation lets each of the prime with even power in k has equal power in numerator compared with denominator. If such ideal is reachable, the minimum integer result is 3.7.11.13 = 3003. It can be proven to be possible by setting the parentheses this way: (1:(2:3:4:5)):(6:7:8:9):(10:11):(12:13)