Degrees of the Same Radicand
If all variables below are positive integers more than 1 , how many digits does the smallest value of x have?
⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ x = a 3 x = b 4 x = c 5 x = d ⋮ 1 0 x = i
NOTE: The i is an algebraic variable, not an imaginary number ( − 1 ).
The answer is 759.
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2 solutions
Hi, can you please elaborate on the method of finding the number of digits, why does it work? Thanks!
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If a number n satisfies 1 0 1 ≤ n < 1 0 2 it has 2 digits. If 1 0 2 ≤ n < 1 0 3 it has 3 digits, and so on; in general, if 1 0 d − 1 ≤ n < 1 0 d then n has d digits.
Taking logs base 1 0 , this is d − 1 ≤ lo g 1 0 n < d . So to work out d (the number of digits) from n , we use d = ⌊ lo g 1 0 x ⌋ + 1 .
A question for you: why is it not ⌈ lo g 1 0 x ⌉ ?
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Oh ok, understood. Thanks!
For the question: (The floor function + 1) is equivalent of saying the (ceiling function)
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@Mahdi Raza – Ah - that's the point! It isn't always equivalent. You have to think about when it's not equivalent to work out why the particular form is used.
- x has to a number that has its power as the LCM of the radicals.
LCM ( 2 , 3 , … 1 0 ) = 2 5 2 0
- The smallest such value for x is
2 2 5 2 0
- Hence, we can find the length of x using a python code
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Since lcm ( 2 , 3 , ⋯ , 1 0 ) = 2 5 2 0 , x must be the 2 5 2 0 th power of an integer. The smallest possibility (since 1 isn't allowed) is x = 2 2 5 2 0 .
To find the number of digits, we need to know which two integer powers of 1 0 x lies between. The number of digits is given by ⌊ lo g 1 0 x ⌋ + 1 = ⌊ 2 5 2 0 lo g 1 0 2 ⌋ + 1 = 7 5 9 .