[Guest]

Substitute each of the capital letters in bold by a different base ten digit fro 0 to 9 to satisfy this alphametic equation. None of the numbers can contain any leading zero.

³√(ITS) + ³√(ANY) = ³√(SIGN)

Note: While a solution is trivial with the aid of a computer program, show how to derive it without one.

Edited June 22, 2009 by K Sengupta

[Pickett]

Boy, this is getting much tougher...You are just evil K Sengupta

. Let me first explain why this one is a lot harder. Obviously dealing with CUBE roots is no fun for anyone. Also, there are only 7 of the 10 digits used in this problem. Finally, when initially working it out, ITS and ANY can be interchangable...it isn't until you try to plug them into the SIGN part that you have to distinguish which 3-digit number goes with which word...so this one is VERY difficult. The most difficult part lies with the following:

We know that at least one of ³√(ITS) or ³√(ANY) can not be evaluated to whole numbers, since:

5^3=125

6^3=256

7^3=343

8^3=512

9^3=729

And we know that neither of them can be 7^3, since that has a 3 repeated in it (and neither of them have repeated numbers)...BUT that leaves us with all numbers that a "2" in them. Given that they do not share any common numbers, we know that at least one of the values must not be a whole number.

Finally, let me preface this with the statement that this is very difficult to explain well. My method does work, and isn't as much trial and error as it may sound in the below description...

Ok, so we know that the range for either ³√(ITS) or ³√(ANY) must be between 4.64 and 10 (exclusive)...which means that the ³√(SIGN) must be between 9.999... and 20 (exclusive).

Therefore, S cannot be 8, 9, or 0 because the highest number we can get from that range would be just under 8000.

Assuming you're not just REALLY mean, I would have to assume that the answer to this will be an equation that basically satisfies: x³√(a) + y³√(a) = z³√(a) which would leave us with the equation x + y = z...otherwise, there really wouldn't be much of a way to satisfy it.

If this is the case, we must find 3 digit numbers that contain perfect cubes as factors so that we can factor them out:

1³=1

2³=8

3³=27

4³=64

5³=125

6³=256

7³=343

8³=512

9³=729

x or y can't be 8 or 9, because anything multiplied by them would yield at least a 4 digit number.

So, we know that z must be between 3 and 13 (inclusive).

If either x or y is 1, we know that a must be a 3-digit number...so a would have to be between 100 and 125. If this were the case, then we would know that one of the 3-digit numbers would start with either 10 or 12. Which in turn makes it so that the other starts with 8 or 9...

So, we get these possibilites for the 3 digit number pairs:

102/816, 103/824, 104/832, 105/840, 106/848, 107/856, 123/984, 124/992

But we can rule out any that have duplicate numbers between them, so we are left with:

103/824, 104/832, 107/856, 123/984

That leaves us with only 4 possibilities:

³√103 + 2*³√103 = 3*³√103: ³√(103) + ³√(824) = ³√(2781) which doesn't work

³√104 + 2*³√104 = 3*³√104: ³√(104) + ³√(832) = ³√(2808) which doesn't work

³√107 + 2*³√107 = 3*³√107: ³√(107) + ³√(856) = ³√(2889) which doesn't work

³√123 + 2*³√123 = 3*³√123: ³√(123) + ³√(984) = ³√(3321) which doesn't work

So, we know that neither x or y is 1...which makes our life a bit more interesting, but we now know that z cannot be 3 or 4!

Let's start on the other side:

If x or y is 7, we know that a MUST be 2:

BUT that would mean one of the 3 digit numbers would be 686, which doesn't match, so x and y are not 7...

If x or y is 6, then we know that a MUST be 2 or 3, so we get these possibilities:

6*³√2 + 4*³√2 = 10*³√2: ³√(512) + ³√(128) = ³√(2000) which doesn't work

6*³√2 + 5*³√2 = 11*³√2: ³√(512) + ³√(250) = ³√(2662) which doesn't work

6*³√3 + 4*³√3 = 10*³√3: ³√(768) + ³√(192) = ³√(3000) which doesn't work

6*³√3 + 5*³√3 = 11*³√3: ³√(768) + ³√(375) = ³√(3993) which doesn't work

So, again, x or y is not 6...so x or y must be 2, 3, 4, or 5

If 5, the other must be 4 and then a must be 2, 3, 4, 5, 6, or 7

5*³√2 + 4*³√2 = 9*³√2: ³√(250) + ³√(128) = ³√(1458) which doesn't work

5*³√3 + 4*³√3 = 9*³√3: ³√(375) + ³√(192) = ³√(2187)...

HOLY COW, we have one that works!!

³√192 + ³√375 = ³√2187...which simplifies to 5*³√3 + 4*³√3 = 9*³√3

So we have:

I=1

T=9

S=2

A=3

N=7

Y=5

G=8

Now, maybe I can get some work done...Great puzzle...very difficult...

[Guest]

Boy, this is getting much tougher...You are just evil K Sengupta

. Let me first explain why this one is a lot harder. Obviously dealing with CUBE roots is no fun for anyone. Also, there are only 7 of the 10 digits used in this problem. Finally, when initially working it out, ITS and ANY can be interchangable...it isn't until you try to plug them into the SIGN part that you have to distinguish which 3-digit number goes with which word...so this one is VERY difficult. The most difficult part lies with the following:

We know that at least one of ³√(ITS) or ³√(ANY) can not be evaluated to whole numbers, since:

5^3=125

6^3=256

7^3=343

8^3=512

9^3=729

And we know that neither of them can be 7^3, since that has a 3 repeated in it (and neither of them have repeated numbers)...BUT that leaves us with all numbers that a "2" in them. Given that they do not share any common numbers, we know that at least one of the values must not be a whole number.

Finally, let me preface this with the statement that this is very difficult to explain well. My method does work, and isn't as much trial and error as it may sound in the below description...

Ok, so we know that the range for either ³√(ITS) or ³√(ANY) must be between 4.64 and 10 (exclusive)...which means that the ³√(SIGN) must be between 9.999... and 20 (exclusive).

Therefore, S cannot be 8, 9, or 0 because the highest number we can get from that range would be just under 8000.

Assuming you're not just REALLY mean, I would have to assume that the answer to this will be an equation that basically satisfies: x³√(a) + y³√(a) = z³√(a) which would leave us with the equation x + y = z...otherwise, there really wouldn't be much of a way to satisfy it.

If this is the case, we must find 3 digit numbers that contain perfect cubes as factors so that we can factor them out:

1³=1

2³=8

3³=27

4³=64

5³=125

6³=256

7³=343

8³=512

9³=729

x or y can't be 8 or 9, because anything multiplied by them would yield at least a 4 digit number.

So, we know that z must be between 3 and 13 (inclusive).

If either x or y is 1, we know that a must be a 3-digit number...so a would have to be between 100 and 125. If this were the case, then we would know that one of the 3-digit numbers would start with either 10 or 12. Which in turn makes it so that the other starts with 8 or 9...

So, we get these possibilites for the 3 digit number pairs:

102/816, 103/824, 104/832, 105/840, 106/848, 107/856, 123/984, 124/992

But we can rule out any that have duplicate numbers between them, so we are left with:

103/824, 104/832, 107/856, 123/984

That leaves us with only 4 possibilities:

³√103 + 2*³√103 = 3*³√103: ³√(103) + ³√(824) = ³√(2781) which doesn't work

³√104 + 2*³√104 = 3*³√104: ³√(104) + ³√(832) = ³√(2808) which doesn't work

³√107 + 2*³√107 = 3*³√107: ³√(107) + ³√(856) = ³√(2889) which doesn't work

³√123 + 2*³√123 = 3*³√123: ³√(123) + ³√(984) = ³√(3321) which doesn't work

So, we know that neither x or y is 1...which makes our life a bit more interesting, but we now know that z cannot be 3 or 4!

Let's start on the other side:

If x or y is 7, we know that a MUST be 2:

BUT that would mean one of the 3 digit numbers would be 686, which doesn't match, so x and y are not 7...

If x or y is 6, then we know that a MUST be 2 or 3, so we get these possibilities:

6*³√2 + 4*³√2 = 10*³√2: ³√(512) + ³√(128) = ³√(2000) which doesn't work

6*³√2 + 5*³√2 = 11*³√2: ³√(512) + ³√(250) = ³√(2662) which doesn't work

6*³√3 + 4*³√3 = 10*³√3: ³√(768) + ³√(192) = ³√(3000) which doesn't work

6*³√3 + 5*³√3 = 11*³√3: ³√(768) + ³√(375) = ³√(3993) which doesn't work

So, again, x or y is not 6...so x or y must be 2, 3, 4, or 5

If 5, the other must be 4 and then a must be 2, 3, 4, 5, 6, or 7

5*³√2 + 4*³√2 = 9*³√2: ³√(250) + ³√(128) = ³√(1458) which doesn't work

5*³√3 + 4*³√3 = 9*³√3: ³√(375) + ³√(192) = ³√(2187)...

HOLY COW, we have one that works!!

³√192 + ³√375 = ³√2187...which simplifies to 5*³√3 + 4*³√3 = 9*³√3

So we have:

I=1

T=9

S=2

A=3

N=7

Y=5

G=8

Well done !!!

This is indeed the answer which I also arrived at by way of precisely the same procedure as quoted above.

I was unsure about the answer until that was reinforced by the foregoing methodology