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2011 Square Start (Posted on 2011-09-16) Difficulty: 3 of 5
Determine the smallest perfect square whose base ten representation begins with 2011 (reading left to right). How about the smallest positive perfect cube whose base ten representation begins with 2011 (reading left to right)?

*** Calculators are allowed, but no computer programs.

No Solution Yet Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

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Solution solution | Comment 3 of 5 |

sqrt(2011) = 44.84417464955732
sqrt(2012) = 44.85532298401161

so if the answer has an even number of digits, its square root's digits must lie, lexicographically, above the first sequence of digits but below the second: 4485. The result is 4485^2 = 20115225.

But the answer might have an odd number of digits:

sqrt(20110) = 141.8097316829843
sqrt(20120) = 141.8449858119772

14181^2 = 201100761

The even-digit (8-digit) one is smaller, so the answer to part 1 is 20115225.

For cube roots we have to try numbers of digits that are congruent to 1, 2 or 0, mod 3:

2011^(1/3) = 12.6222668331058
2012^(1/3) = 12.6243586904251

20110^(1/3) = 27.1938495320821
20120^(1/3) = 27.1983563020572

201100^(1/3) = 58.5873727874249
201200^(1/3) = 58.5970823289993

The three candidates are therefore 12623^3, 27194^3 and 5859^3. Clearly this last is the smallest: 5859^3 = 201127054779, which is thus the answer to part 2.

Edited on September 16, 2011, 1:54 pm
  Posted by Charlie on 2011-09-16 13:48:37

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