Imaginary Powerhouse ( i )
n = 1 2 3 + 4 5 6 7 8 × 9 1 0 1 1 . . . 1 4 1 5 + 1 6 1 7 1 8 . . . 2 4 × 2 5 2 6 . . . 3 4 3 5
For n as defined above, what is the value of i n ?
Clarification : i = − 1 denotes the imaginary unit .
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2 solutions
Aha. You managed to post a solution while i was writing mine. Nice :D
| i 1 = − 1 | i 5 = i 4 + 1 = 1 × i 1 = − 1 |
| i 2 = − 1 | i 6 = i 4 + 2 = 1 × i 2 = − 1 |
| i 3 = i 2 × i = − i | i 7 = i 4 + 3 = 1 × i 3 = − i |
| i 4 = 1 | i 8 = i 4 + 4 = 1 × i 4 = 1 |
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Solution:
n = 1 + [ 4 5 6 7 8 × 9 1 0 1 1 . . . 1 4 1 5 ] + [ 1 6 1 7 1 8 . . . 2 4 × 2 5 2 6 . . . 3 4 3 5 ]
Let n = 1 + a + b
a = 4 5 6 7 8 × 9 1 0 1 1 . . . 1 4 1 5 = large multiple of 4
b = 1 6 1 7 1 8 . . . 2 4 × 2 5 2 6 . . . 3 4 3 5 = large multiple of 4
a + b = large multiple of 4 + large multiple of 4 = larger multiple of 4
∴ i n = i 1 + a + b = i 1 + 4 = i 4 + 1 = i 5 = i 1 = − 1 = i
I actually redid the problem phrasing and format for you. Don't use too much huge characters and italic, too fancy and kiddy.
We note that i n = i n mod 4
Now, we have:
n ≡ ( 1 2 3 + 4 5 6 7 8 × 9 1 0 1 1 . . . 1 4 1 5 + 1 6 1 7 1 8 . . . 2 4 × 2 5 2 6 . . . 3 4 3 5 ) (mod 4) = 1 + 0 + 0 = 1 (mod 4) Note that 4 ∣ 4 and 4 ∣ 1 6
⟹ i n = i 1 = i