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What is the least possible number, the square of which, when added to the sum of the squares of 10 and 11, results in a perfect square?

Sagot :

To solve the problem of finding the least possible number [tex]\(n\)[/tex] such that the square of [tex]\(n\)[/tex] added to the sum of the squares of 10 and 11 results in a perfect square, we can follow these detailed steps:

1. Calculate the squares of the given numbers 10 and 11:
[tex]\[ 10^2 = 100 \quad \text{and} \quad 11^2 = 121 \][/tex]

2. Find the sum of these squares:
[tex]\[ 100 + 121 = 221 \][/tex]

3. Determine the least possible number [tex]\(n\)[/tex] such that [tex]\( n^2 \)[/tex] added to 221 results in a perfect square. Let's denote the sum of 221 and [tex]\( n^2 \)[/tex] as [tex]\( \text{total\_sum} \)[/tex]. Thus, we want [tex]\( \text{total\_sum} \)[/tex] to be a perfect square.

4. Initialize [tex]\( n \)[/tex] and start checking from the smallest possible values:

- For [tex]\( n = 1 \)[/tex]:
[tex]\[ n^2 = 1^2 = 1 \][/tex]
[tex]\[ \text{total\_sum} = 221 + 1 = 222 \quad (\text{not a perfect square}) \][/tex]

- For [tex]\( n = 2 \)[/tex]:
[tex]\[ n^2 = 2^2 = 4 \][/tex]
[tex]\[ \text{total\_sum} = 221 + 4 = 225 \quad (\text{perfect square as } 15^2 = 225) \][/tex]

5. Verify the result:
- To ensure that 225 is indeed a perfect square, check:
[tex]\[ \sqrt{225} = 15 \quad \text{which is an integer.} \][/tex]

So, the least possible number [tex]\( n \)[/tex] is indeed 2.

Finally, summing up,

- The squares of 10 and 11 are 100 and 121, respectively.
- Their sum is 221.
- The least possible number [tex]\( n \)[/tex] whose square added to 221 results in a perfect square is [tex]\( \boxed{2} \)[/tex].

Thus, the least possible number [tex]\( n \)[/tex] is 2.