[Guest]

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

2(ELEVEN) + TEN+ ONE = THIRTY + THREE, where ELEVEN is divisible by 11.

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

[Guest]

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

2(ELEVEN) + TEN+ ONE = THIRTY + THREE, where ELEVEN is divisible by 11.

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

2 * (161018) + 218 + 781 = 293524 + 29511

E is 1,

L is 6,

V is 0,

N is 8,

T is 2,

O is 7,

H is 9,

I is 3,

R is 5,

Y is 4

[Pickett]

2(ELEVEN) + TEN+ ONE = THIRTY + THREE, where ELEVEN is divisible by 11.

First of all, I agree with huofo's answer, but since he didn't show how it was derived, I figured I would put my solution down. Also, note, that while my methodology would work with not too much effort, it was sort of luck that I came across the answer as quickly as I did. There may be a faster way, but this way works...

Wow, ok...let's start with some observations and use these as axioms:

Axiom 1. E, T, and O can't be 0

Axiom 2. N + N + N + E = Y + E...which means N + N + N = Y

Axiom 3. By Axiom 2 we know that if N is odd, Y is odd...and if N is even, Y is even and also N and Y can't be 0 or 5...

N = 1, Y = 3

N = 2, Y = 6

N = 3, Y = 9

N = 4, Y = 6

N = 6, Y = 8

N = 7, Y = 1

N = 8, Y = 4

N = 9, Y = 7

Axiom 4. E + E + E + N + X = T + E (where X <= 3), so that means E + E + N + X = #T (where X <= 3)

Axiom 5. Since ELEVEN is divisible by 11, we know that (EL + EV + EN) is also divisible by eleven.

Axiom 6. Since ELEVEN is divisible by 11, we know that (E + E + E) - (L + V + N) is a multiple of eleven...so that means 3E - L - V - N is a multiple of eleven...

Let's say E >= 5:

Then the left side becomes a 7 digit number. If this is the case, then T MUST be equal to 9 (in order to get the right side to be a 7 digit number)...But if E > 5, then the left side becomes too large to allow the right side to equal it. So we now know that 0 < E <= 5 and we know that T > E (by at least a factor of 2):

E = 1, T >= 2

E = 2, T >= 4

E = 3, T >= 6

E = 4, T >= 8

E = 5, T = 9

So T cannot be 1

If E = 1:

T = 2 or 3

Using Axiom 4, we can say that 2 + N + X = 9 (where X <= 3), so N must be 7 or 8 (so by axiom 3, Y must be 1 or 4)...but Y can't be 1, so N can't be 7...

So that means, T must be 2 or 3, and N must be 8, Y must be 4

if T = 3, when we add what we have so far, we end up with ####35 = ####45...so obviously that doesn't work...so T must be 2.

We now have 2(1L1V18) + 218+ O81 = 2HIR24 + 2HR11

We then have (by Axiom 6) 3 - (L + V + 8) is a multiple of eleven...so L + V must be 6...so the only combination of remaining numbers that allows for this is:

L = 0, V = 6 or L = 6, V = 0

We also know (by plugging the numbers we have) that 2L = H + 2 + X (where X <= 1), so 2L = H + 2 OR 2L = H + 3...so we have:

if L = 0, H = 7, but we can see right away that there is no way this will work, since 2 * 101### can never equal 27####...so L can't be 0.

so then L must be 6, and H must then be 9:

We are left with 2(161018) + 218+ O81 = 29IR24 + 29R11

Let's simplify this a little bit:

322254 + O81 = 29IR24 + 29R11

Going with that, we can then say that 2 + O + 1 = R + R, so O + 3 = 2R:

if O = 3, R = 3...which doesn't work

if O = 5, R = 6...which doesn't work

if O = 7, R = 5...it works, so we go with that.

322254 + 781 = 29I524 + 29511

It is now trivial to solve for I to see if our final solution satisifies the original:

323035 = 29I524 + 29511

293524 = 29I524...so I must be 3...and that does work.

So, we have our solution already, without needing to check if E >= 2. There MAY be other solutions out there that satisfy this, but this was the first one we came across...

[Guest]

First of all, I agree with huofo's answer, but since he didn't show how it was derived, I figured I would put my solution down. Also, note, that while my methodology would work with not too much effort, it was sort of luck that I came across the answer as quickly as I did. There may be a faster way, but this way works...

Wow, ok...let's start with some observations and use these as axioms:

Axiom 1. E, T, and O can't be 0

Axiom 2. N + N + N + E = Y + E...which means N + N + N = Y

Axiom 3. By Axiom 2 we know that if N is odd, Y is odd...and if N is even, Y is even and also N and Y can't be 0 or 5...

N = 1, Y = 3

N = 2, Y = 6

N = 3, Y = 9

N = 4, Y = 6

N = 6, Y = 8

N = 7, Y = 1

N = 8, Y = 4

N = 9, Y = 7

Axiom 4. E + E + E + N + X = T + E (where X <= 3), so that means E + E + N + X = #T (where X <= 3)

Axiom 5. Since ELEVEN is divisible by 11, we know that (EL + EV + EN) is also divisible by eleven.

Axiom 6. Since ELEVEN is divisible by 11, we know that (E + E + E) - (L + V + N) is a multiple of eleven...so that means 3E - L - V - N is a multiple of eleven...

Let's say E >= 5:

Then the left side becomes a 7 digit number. If this is the case, then T MUST be equal to 9 (in order to get the right side to be a 7 digit number)...But if E > 5, then the left side becomes too large to allow the right side to equal it. So we now know that 0 < E <= 5 and we know that T > E (by at least a factor of 2):

E = 1, T >= 2

E = 2, T >= 4

E = 3, T >= 6

E = 4, T >= 8

E = 5, T = 9

So T cannot be 1

If E = 1:

T = 2 or 3

Using Axiom 4, we can say that 2 + N + X = 9 (where X <= 3), so N must be 7 or 8 (so by axiom 3, Y must be 1 or 4)...but Y can't be 1, so N can't be 7...

So that means, T must be 2 or 3, and N must be 8, Y must be 4

if T = 3, when we add what we have so far, we end up with ####35 = ####45...so obviously that doesn't work...so T must be 2.

We now have 2(1L1V18) + 218+ O81 = 2HIR24 + 2HR11

We then have (by Axiom 6) 3 - (L + V + 8) is a multiple of eleven...so L + V must be 6...so the only combination of remaining numbers that allows for this is:

L = 0, V = 6 or L = 6, V = 0

We also know (by plugging the numbers we have) that 2L = H + 2 + X (where X <= 1), so 2L = H + 2 OR 2L = H + 3...so we have:

if L = 0, H = 7, but we can see right away that there is no way this will work, since 2 * 101### can never equal 27####...so L can't be 0.

so then L must be 6, and H must then be 9:

We are left with 2(161018) + 218+ O81 = 29IR24 + 29R11

Let's simplify this a little bit:

322254 + O81 = 29IR24 + 29R11

Going with that, we can then say that 2 + O + 1 = R + R, so O + 3 = 2R:

if O = 3, R = 3...which doesn't work

if O = 5, R = 6...which doesn't work

if O = 7, R = 5...it works, so we go with that.

322254 + 781 = 29I524 + 29511

It is now trivial to solve for I to see if our final solution satisifies the original:

323035 = 29I524 + 29511

293524 = 29I524...so I must be 3...and that does work.

So, we have our solution already, without needing to check if E >= 2. There MAY be other solutions out there that satisfy this, but this was the first one we came across...

Yes, basically, my solution is the same.

1. since 2* E <= T => 1<= E <=4

2. Consider the last 2 digital, 3*EN + NE = TY + EE or 3*EN + NE = 100 + TY + EE . Convert it to the following equation:

30*E + 3*N + 10*N + E = 10*T + Y + 11*E or 30*E + 3*N + 10*N + E = 10*T + Y + 11*E + 100

=> 20*E + 13*N = 10*T + Y or 20*E + 13*N = 10*T + Y +100 (equation 1)

If E =1, T = 2, then 13N = Y (which is not work) or 13N = 100+ Y(which only N =8 , Y =4 is the answer) .

Balabala......

Sorry that I did not have a lot of time to explain more detail.