Use Those Radicals Sparingly!
1 + 2 + ⋯ + 1 0 0
If the sum above can be expressed as
a 0 + a 1 b 1 + a 2 b 2 + ⋯ + a n b n
where a 0 , a 1 , a 2 , … , a n , b 1 , b 2 , … , b n are integers, what is the minimum value of n ?
The answer is 60.
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1 solution
All d nos can be expressed in power os primes so d minimum radical should be equal to d no. Of primes +1. 1 is equivalent to sum of all perfect squares
In order to find the least number of square roots, we must express this number in simplest form. Thus, the numbers in the radicand that are in simplest form that will get counted are numbers with at most one of each prime factor. Then the question simplifies to "How many numbers are there between 1-100 with at most one of each prime factor?"
There are a few ways to proceed from here. I think the best way would be to complementary count. Then we are looking for numbers with at least 2 factors of each prime, thus multiples of perfect squares. Our perfect squares are 1 , 4 , 9 , 1 6 , 2 5 , 3 6 , 4 9 , 6 4 , 8 1 . Thus we have 1 + 2 5 + 1 1 − 2 + 3 + 2 = 4 0 such numbers by the principle of inclusion and exclusion . Thus, the answer is 1 0 0 − 4 0 = 6 0 .