[Guest]

Substitute each of the capital letters in bold by a different base ten digit from 0 to 9 to satisfy the following set of cryptarithmetic relationships. None of the numbers contains leading zeroes.

(i) ABACDEBDECA is a perfect cube, and:

(ii) Each of AB, ACD and ECD is a perfect cube, and:

(iii) EBD is a perfect square.

Edited October 30, 2009 by K Sengupta

[EventHorizon]

ECA is a perfect cube, not ECD.

A=2

B=7

C=1

D=6

E=5

Method I used:

First I found the two digit cubes. (27 and 64)

I noticed that there is no three digit cube that starts with 6, so A=2 and B=7.

The only three digit cube is starting with 2 is 216, so C=1 and D=6.

I took the cube root of 27216111111 and rounded it up to 3008.

Cubing 3008 results in 27216576512, so E=5.

Verifying with the constraints...

AB = 27 = perfect cube

ACD = 216 = perfect cube

ECD = 516 = not a perfect cube (but ECA is the last part of the number so I assume you meant that and 512 is a perfect cube)

EBD = 576 = perfect square

[Guest]

It seams to me that here is no solution. All letters should represent different numbers. ACD and ECD should be perfect cubes and from available three digit perfect cubes 125, 216, 343, 512, 729 there is no pair ending with the same two digits CD. There is probably some typo in the question.

[Guest]

ECA is a perfect cube, not ECD.

A=2

B=7

C=1

D=6

E=5

Method I used:

First I found the two digit cubes. (27 and 64)

I noticed that there is no three digit cube that starts with 6, so A=2 and B=7.

The only three digit cube is starting with 2 is 216, so C=1 and D=6.

I took the cube root of 27216111111 and rounded it up to 3008.

Cubing 3008 results in 27216576512, so E=5.

Verifying with the constraints...

AB = 27 = perfect cube

ACD = 216 = perfect cube

ECD = 516 = not a perfect cube (but ECA is the last part of the number so I assume you meant that and 512 is a perfect cube)

EBD = 576 = perfect square

Nice methodolology and subsequent analytical derivation.

Yes, ECA is a perfect cube (and, not ECD in terms of the original post).

The typographical anomaly is sincerely regretted.

[Guest]

Substitute each of the capital letters in bold by a different base ten digit from 0 to 9 to satisfy the following set of cryptarithmetic relationships. None of the numbers contains leading zeroes.

(i) ABACDEBDECA is a perfect cube, and:

(ii) Each of AB, ACD and ECD is a perfect cube, and:

(iii) EBD is a perfect square.

27216576512

The only problem with this is that ECA (512) is a perfect cube and ECD (516) is not.