[Guest]

Solve this alphametic equation, where each of the capital letters represents a different decimal digit from 0 to 9. None of T and B is zero.

√(THE) + √(TOP) = √(BEE)

Edited April 21, 2009 by K Sengupta

[Guest]

Solve this alphametic equation, where each of the capital letters represents a different decimal digit from 0 to 9. None of T and B is zero.

√(THE) + √(TOP) = √(BEE)

um do the 2 E's have the same value?

[Guest]

um do the 2 E's have the same value?

Any given letter represents a decimal digit. For example, E=1 implies that E has the value of 1 throughout the given equation. However, two different letters cannot assume the same value. For example, if E =1, then none of the other letters can be equal to 1.

Edited April 21, 2009 by K Sengupta

[Guest]

I can't figure this one out. I figure that the numbers have to be perfect squares, so only 10^2-31^2 can possible work for any of these numbers because 32^2 is a 4 digit number and 9^2 is a 2 digit number. 10^2, 11^2, and 12^2 can work for either of the first numbers because they have repeating numbers in them. Therefore, the smallest possible numbers are 13^2 and 14^2. Because you have 13+14 or higher the last number has to be more than 27^2. The only number that has the last two digits repeating is 30^2 which is 900. You can't possible have 0 as the last digit for the first number so I don't see how this is possible. But maybe I'm WAYYY off target with this. And I most likely am.

[Guest]

I can't figure this one out. I figure that the numbers have to be perfect squares, so only 10^2-31^2 can possible work for any of these numbers because 32^2 is a 4 digit number and 9^2 is a 2 digit number. 10^2, 11^2, and 12^2 can work for either of the first numbers because they have repeating numbers in them. Therefore, the smallest possible numbers are 13^2 and 14^2. Because you have 13+14 or higher the last number has to be more than 27^2. The only number that has the last two digits repeating is 30^2 which is 900. You can't possible have 0 as the last digit for the first number so I don't see how this is possible. But maybe I'm WAYYY off target with this. And I most likely am.

I agree with you completely.

They might not be perfect squares. Which makes it much more difficult

[Guest]

I can't figure this one out. I figure that the numbers have to be perfect squares, so only 10^2-31^2 can possible work for any of these numbers because 32^2 is a 4 digit number and 9^2 is a 2 digit number. 10^2, 11^2, and 12^2 can work for either of the first numbers because they have repeating numbers in them. Therefore, the smallest possible numbers are 13^2 and 14^2. Because you have 13+14 or higher the last number has to be more than 27^2. The only number that has the last two digits repeating is 30^2 which is 900. You can't possible have 0 as the last digit for the first number so I don't see how this is possible. But maybe I'm WAYYY off target with this. And I most likely am.

The numbers may not necessarily correspond to perfect squares.

As an example, V(12) + V(27) = V(75), but none of 12, 27 and 75 are perfect squares.

Edited April 21, 2009 by K Sengupta

[Guest]

The numbers may not necessarily correspond to perfect squares.

As an example, V(12) + V(27) = V(75), but none of 12, 27 and 75 are perfect squares.

THE = 108

TOP = 192

BEE = 588 (E = 8)

let y = Sqrt(THE) + SQRT(TOP) = Sqrt(108) + SQRT(192)

y^2 = 108 + 192 + 2sqrt (108 x 192)

y^2 = 300 + 2sqrt (9x12x16x12)

y^2 = 300 + 2sqrt (9 x 16 x 144)

y^2 = 300 + 2(3 x 4 x 12)

y^2 = 300 + 288

y = sqrt (588)

Done!