Let x be a positive integer. How many x would you have to add into itself to get x x as the sum?
x x = x + x + x + . . . + x
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x x = x + x + x + x + . . .
Assume x = 4 (for easy explanation)
4 4 = 4 + 4 + 4 + 4 + . . .
4 4 = 4 1 + 4 1 + 4 1 + 4 1 + 4 1 + . . .
4 4 = 4 2 + 4 2 + 4 2 + 4 2 + 4 2 + . . .
4 4 = 4 3 + 4 3 + 4 3 + . . .
4 4 = 6 4 + 6 4 + 6 4 + 6 4
4 4 = 6 4 × 4
We need to add x x x − 1 times to make x x , which means:
x 1 × x x − 1 = x x
4 1 × 4 4 − 1 = 4 4
You did not show that it works for all x . You just checked 4 .
If you use x = 1 , the answer is x .
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But it is also equal to x 2 and x 1 0 0
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I know this @Hamza Anushath ! But I just wanted to tell that this solution would not be considered complete by most people! This solution would definitely reach you the correct answer, but in subjective, your solution should be like @Kaizen Cyrus .
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The y factor below indicates how many x we should add, as is the nature of multiplication.
x x = x x = x x x = x x − 1 = x + x + x + . . . + x x y y y