What is the least number of square roots needed to express √1+√2+⋯+√100 if it must be expressed in the form a+b√c+d√e+⋯ where a,b,c,d,e,… are all integers?
Solution
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 1,4,9,16,25,36,49,64,81. Thus we have 1+25+11−2+3+2=40 such numbers by the principle of inclusion-exclusion. Thus, the answer is 100−40=60.
Question
How is it that the number being expressed in simplest form will give the least number of radicals? The solution seems to imply that but why is it true?
Answer
The square root of each square-free number less than 100
contributes a term to the sum with a surd (a square root that cannot be simplified out),
but all the square multiples of the number under the square root (4 times, 9 times, 16 times, etc.)
can be combined with the term for the original square-free number,
so they do not contribute any additional terms. For example,
√11+√44+√99=6√11.
One way to count the integers whose square roots do not contribute new surds to the sum is as follows:
- 10 perfect squares in the range 1 to 100, inclusive;
- 6 cases of twice a perfect square in the range 8 to 98, inclusive;
- 4 cases of 3 times a perfect square in the range 12 to 75, inclusive;
- 3 cases of 5 times a perfect square in the range 20 to 80, inclusive;
- 3 cases of 6 times a perfect square in the range 24 to 96, inclusive;
- 6 cases including 2 cases each of 7 times a perfect square, 10 times a perfect square, and 11 times a perfect square in the range 28 to 99, inclusive;
- 8 cases including 13×4, 14×4, 15×4, 17×4, 19×4, 21×4, 22×4, and 23×4.
These add up to 40 terms of the original sum that either are integers or
can be combined with other terms, leaving 60 unique surds.
This is the same as your result, of course.
Here is the sum completely worked out, confirming that 60 square
roots are needed:
√1+√2+⋯+√100=1+√2+√3+2+√5+√6+√7+2√2+3+√10+√11+2√3+√13+√14+√15+4+√17+3√2+√19+2√5+√21+√22+√23+2√6+5+√26+3√3+2√7+√29+√30+√31+4√2+√33+√34+√35+6+√37+√38+√39+2√10+√41+√42+√43+2√11+3√5+√46+√47+4√3+7+5√2+√51+2√13+√53+3√6+√55+2√14+√57+√58+√59+2√15+√61+√62+3√7+8+√65+√66+√67+2√17+√69+√70+√71+6√2+√73+√74+5√3+2√19+√77+√78+√79+4√5+9+√82+√83+2√21+√85+√86+√87+2√22+√89+3√10+√91+2√23+√93+√94+√95+4√6+√97+7√2+3√11+10=55+28√2+15√3+10√5+10√6+6√7+6√10+6√11+3√13+3√14+3√15+3√17+3√19+3√21+3√22+3√23+√26+√29+√30+√31+√33+√34+√35+√37+√38+√39+√41+√42+√43+√46+√47+√51+√53+√55+√57+√58+√59+√61+√62+√65+√66+√67+√69+√70+√71+√73+√74+√77+√78+√79+√82+√83+√85+√86+√87+√89+√91+√93+√94+√95+√97
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