Vanuatu
Assume that the symbol ^@\ddagger x \ddagger^@ denotes the largest integer not exceeding ^@x^@. For example, ^@\ddagger 3 \ddagger = 3^@, and ^@\ddagger 4.9 \ddagger = 4^@.
What is the value of
^@\ddagger \sqrt{1}\ddagger + \ddagger\sqrt{2}\ddagger + \ddagger\sqrt{3}\ddagger \space + ...... + \ddagger\sqrt { 16 }\ddagger.^@
Answer:
^@38^@
- Given, ^@\ddagger x \ddagger^@ is the largest integer not exceeding ^@x^@.
Let ^@x^@ be a number which lies between two square numbers ^@a^@ and ^@b^@, i.e. ^@ a < x < b^@ then ^@\sqrt{x}^@ will lie between ^@\sqrt{a}^@ and ^@\sqrt{b}^@, i.e. ^@\sqrt{a} < \sqrt{x} < \sqrt{b}^@ - Here, square numbers from ^@1^@ to ^@16^@ are ^@ 1, 4, 9, \text{ and } 16 ^@.
Because, the root of the numbers that are greater than or equal to 1 and less than ^@4, i.e. 1 \le x < 4^@ will be greater than or equal to 1 and less than ^@2, i.e. 1 \le \sqrt{x} < 2^@
Therefore, ^@\ddagger \sqrt{x} \ddagger = 1^@ for ^@ 1 \le x < 4 ^@
^@\implies ^@^@\ddagger \sqrt{ 1 } \ddagger^@ ^@+^@ ^@\ddagger \sqrt{ 2 } \ddagger^@ ^@+^@ ^@\ddagger \sqrt{ 3 } \ddagger^@ ^@=^@ ^@1^@ ^@+^@ ^@1^@ ^@+^@ ^@1^@ ^@ = 3 \times 1 = 3 ^@ - The root of the numbers that are greater than or equal to 4 and less than ^@9, i.e. 2 \le x < 9^@ will be greater than or equal to 2 and less than ^@3, i.e. 2 \le \sqrt{x} < 3^@
Therefore, ^@\ddagger \sqrt{x} \ddagger = 2^@ for ^@ 4 \le x < 9 ^@
^@\implies ^@^@\ddagger \sqrt{ 4 } \ddagger^@ ^@+^@ ^@\ddagger \sqrt{ 5 } \ddagger^@ ^@+^@ ^@\ddagger \sqrt{ 6 } \ddagger^@ ^@+^@ ^@\ddagger \sqrt{ 7 } \ddagger^@ ^@+^@ ^@\ddagger \sqrt{ 8 } \ddagger^@ ^@=^@ ^@2^@ ^@+^@ ^@2^@ ^@+^@ ^@2^@ ^@+^@ ^@2^@ ^@+^@ ^@2^@ ^@ = 5 \times 2 = 10 ^@ - Similarly, for ^@ 9 \le x < 16, \sqrt{x} ^@ will be ^@3 \le \sqrt{x} < 4^@
Therefore, ^@\ddagger \sqrt{x} \ddagger = 3 ^@ for ^@ 9 \le x < 16^@
^@\implies ^@^@\ddagger \sqrt{ 9 } \ddagger^@ ^@+^@ ^@\ddagger \sqrt{ 10 } \ddagger^@ ^@+^@ ^@\ddagger \sqrt{ 11 } \ddagger^@ ^@+^@ ^@\ddagger \sqrt{ 12 } \ddagger^@ ^@+^@ ^@\ddagger \sqrt{ 13 } \ddagger^@ ^@+^@ ^@\ddagger \sqrt{ 14 } \ddagger^@ ^@+^@ ^@\ddagger \sqrt{ 15 } \ddagger^@ ^@=^@ ^@3^@ ^@+^@ ^@3^@ ^@+^@ ^@3^@ ^@+^@ ^@3^@ ^@+^@ ^@3^@ ^@+^@ ^@3^@ ^@+^@ ^@3^@ ^@ = 7 \times 3 = 21 ^@
And ^@\sqrt{ 16 } = 4^@
^@\implies\ddagger\sqrt{ 16 }\ddagger = 4^@ - ^@\ddagger \sqrt{1}\ddagger + \ddagger\sqrt{2}\ddagger + \ddagger\sqrt{3}\ddagger \space + ...... + \ddagger\sqrt { 16 }\ddagger = 3 + 10 + 21 + 4^@
Hence, the value of ^@\ddagger \sqrt{1}\ddagger + \ddagger\sqrt{2}\ddagger + \ddagger\sqrt{3}\ddagger \space + ...... + \ddagger\sqrt { 16 }\ddagger ^@ is ^@38^@
