15 of 100: EYE SEE

Consider the tens column $E + Y = E .$ It's given all digits are nonzero, so the only way for something to get added to $E$ to get the digit $E$ is by adding 10. This can only occur if there is a carry digit from the ones column and $Y = 9 .$ This also means there will be a carry digit into the hundreds column.

$\begin{array} {ccccc} \large & & & 1 & 1 & \\ \large & & & \color{#D61F06}S & \color{#20A900}E & \color{#20A900}E\\ \large + & & & \color{#20A900}E& \color{#3D99F6}9 & \color{#20A900}E \\ \hline \large & & & \color{#3D99F6}9 & \color{#20A900}E & \color{#D61F06}S \end{array}$

From the ones column we know that $E$ must be greater than 5 to have a carry digit into the tens column $(E$ can't be exactly 5 since that would make $S = 0 ).$ Also, $S$ must be an even digit since it comes about from doubling $E.$

Now consider the hundreds column, including the carry digit from the tens column: $1 + S + E = 9 ,$ which implies $S + E = 8 .$ Since $E$ must be greater than 5 and $S$ must be an even digit the only possible values that work are $E = 6$ and $S = 2 .$

2E=S+10

S+E+1=Y

E+Y+1=E+10

or, Y=9

so, S+E=8

2(8-S)=S+10

or, S=2, Answer!

the interest is the second digit: $E+Y\equiv E\mod{10}, or,E+Y+1\equiv E\mod{10} \to Y= 0,9$ ofcourse Y cant be 0 since obviously both $S,E$ must be positive(since all numbers are distinct) and hence have a sum greater than 0. so $Y=9$ . note that a 1 must carry over to the second digit so $E+E-10=S\to E=\dfrac{S+10}{2}$ . we have the first digit $S+E+1=9\to S+E=8\to S+\dfrac{S+10}{2}=8\to S=2$

Middle digit: $Y = 0$ if there were no carry; but this is not allowed. The alternative is that one was carried from the last digits; then $\boxed{Y =9}$ , and one is carried to the first digit.

The sum is now $\begin{array}{ccccc} & & 1 & 1 & \\ & & S & E & E \\ + & & E & 9 & E \\ \hline & & 9 & 1E & 1S \end{array}$ Combining expressions for first and last digits by substitution or adding the equations: $\begin{cases} 1 + S + E = 9 \\ E + E = 10+S \end{cases}\ \ \ \longrightarrow\ \ \ 3E = 18,$ so that $\boxed{E = 6}$ and $\boxed{S = 2}$ .

Let's denote the carry of $\color{#20A900}E\color{#333333}+\color{#20A900}E$ by $\color{#EC7300}C$ which is either $0$ or $1$ , because sum of two digits is always strictly between $0$ and $18$ .

Now, $\color{#20A900}E\color{#333333}+\color{#3D99F6}Y\color{#333333}+\color{#EC7300}C\color{#333333}\equiv\color{#20A900}E\color{#333333}\pmod{10}\implies\color{#3D99F6}Y\color{#333333}+\color{#EC7300}C\color{#333333}\equiv0\pmod{10}$ .

But, since $\color{#3D99F6}Y$ is non-zero digit and $\color{#EC7300}C$ is either $0$ or $1$ , we must have $\color{#3D99F6}Y\color{#333333}=9$ and $\color{#EC7300}C\color{#333333}=1$ .

Since, $\color{#EC7300}C\color{#333333}=1$ and $\color{#D61F06}S$ is non-zero, we must have $\color{#20A900}E\color{#333333}\geq6$ .

Also, $\color{#D61F06}S$ is even since, $2\color{#20A900}E\color{#333333}\equiv\color{#D61F06}S\color{#333333}\pmod{10}\implies2\color{#20A900}E\color{#333333}\equiv\color{#D61F06}S\color{#333333}\equiv0\pmod{2}$ .

But now since $\color{#D61F06}S$ is even non-zero digit, it must be equal to $2$ since $\color{#D61F06}S\color{#333333}+6\leq\color{#D61F06}S\color{#333333}+\color{#20A900}E\color{#333333}\leq\color{#3D99F6}Y\color{#333333}=9\implies\color{#D61F06}S\color{#333333}\leq3$ .

And indeed, $\color{#D61F06}S\color{#333333}=2, \color{#20A900}E\color{#333333}=6, \color{#3D99F6}Y\color{#333333}=9$ is the solution.

Hence, the answer is $\boxed{2}$ .

One basic principle in solving these sorts of things is that when you're adding n numbers, the most the sum will ever carry is n-1. This is because in order to carry n, a column must sum to at least 10n, and the greatest possible sum of a single column without carrying comes when every digit in the column is 9, which sums to 9n. That's n short, so in order to carry n, you'd need to carry n from the previous column. Which mean's you'd need to carry n from the column before that, and so forth, but eventually you'll hit the ones, which don't get to carry anything, and the whole thing breaks.

The point is, with just two addends, you can carry at most 1. Now, look at the middle column: if E + Y yields E, then Y plus whatever is carried from the ones column must end in 0. If we carry 0, Y must be 0, which violates the premise that all the letters are nonzero. So we must carry 1, which means Y must be 9.

This requires carrying a 1 from the right column, and results in carrying a 1 to the left column. The left column thus tells us S+E+1 = 9, and the right column tells us 2E = 10+S. You can do some simple algebra on that, or you can just iterate through the possible values for E -- since 2E is greater than 10, it must be 6, 7, 8, or 9, and the only value that works is 6. Either way, S comes out to 2.

$266+696= 962$

We have $2E = S$ , and I started trying with $S=2$ as a numerical digit for $S$ , $E$ can't be $1$ , therefore we take $E=6$ . We plug it in to get $S=12$ .

Then for the second column, we have ( $1+E+Y=E+10$ ),

solving for $Y$ , we get $Y=9$ .

Now, we have all the values, plug them in our equation to check.

Since Y is a non - zero digit, E+Y = E implies that Y=9 and also that we had carried over 1 in the previous step E+E = S (because we can carry at most 1 while summing any two digits). So, (E,S) = (6,2), (7,4), (8,6) or (9,8) (if E=5, then S is 0 which is not allowed). In the final step, S + E = Y, a single digit number. For the possible values of E and S, we notice that only 6 + 2 does not yield a double digit number. We try the problem with S = 2, E = 6 and Y = 9 and find that the solution is 266 + 696 = 962

1) E+E carries over into the tens column, therefore E= 6,7,8 or 9. (5+5= 10 and 0s aren’t allowed)

2) S+E has no carry over into the thousands column, therefore S= 1,2 or 3, as anything greater than 3 would equal 10 or more when added to any possible value of E.

Out of 1,2 and 3, 2 is the only viable answer.

see,,,,eye i saw my eyes 2 and than i saw yes and i press 2 how funny and coorrect .................

I'm coming in a bit late, but I approached this in a different way from all solutions I have seen so far.

Taking everything modulo 9, we see: $S+E+E+E+Y+E = Y+E+S \pmod 9$ ; cf. casting out nines. Hence, $3E=0\pmod 9$ . Thus, $E$ is a multiple of 3.

$E=9$ does not work, because the leftmost column would then give a carry (since $S>0$ ). Also $E=3$ does not work, because $0<Y<10$ . Therefore, $E=6$ , and consequently $\fbox{S=2}$ .

This might be rediculous (dont judge spelling) but I just was being dumb and seeing 'Eye See' as an 'I' and a 'C' and turning them into the numbers 9 and 3, adding them and 12 came up so I picked 2

In the one's column, E+E<18, thus E<9, and E is not equal to 5, since 5+5 =10 , thus S would equal to zero, which is not possible by the problem. Also, E cannot be less than or equal to 4, since that would create a problem in the ten's column ( since E+Y = E, if E is less than or equal to 4 here Y would have to be more than 9, not possible). Trying out remaining cases, we find E=6 hits. Thus, E+E=12, and hence S =2. Thus, this problem can be solved by some basic math.

I am not sure how valid the operations are but this method seems to make sense as well. I am curious to hear your thoughts.

First we'll find the value of Y, because that is the easiest approach.

If E+Y = E, then Y must be a number that is very close to 10, because when the carry is added, it becomes 10, thus E is essentially added to 0, and the 1 of 10 is carried over.

Now, is it possible for the carry to be more than 1?

No, because even if E is 9 (highest value possible), the maximum carry is 1.

Then Y has to be 1 less than 10, that is 9.

The rest is simple, try putting values of E that give 2, 4, 6, 8 as the unit's place digit.

For S=2, E=6 (6+6=1'2')

And that is the answer.

There you go, first try.

Y can be worked out as 9 immediately as the only way E + Y could possibly equal E is if Y was equal to 9 as if we look at the units column it must equal something higher than 10 for a one to carryover making it equal to 10 + E is equal to E this means that looking at the 100's column we can figure out that E + S is equal to 8 as again a 1 is carried over

The last step is trial and error I got it on the first try 2 * 6 is 12 which gives us the number needed for the tens column which means S is equal to 2 266+ 696= 962 I'm not the best at explaining my methods sorry

Assume 2E < 10. That means E + Y = 10 + E or E + Y = E. Because Y cannot be zero or ten, 2E = 10 + S. E + Y + 1 = 10 + E

Y = 9

S + E + 1 = 9

3E - 10 + 1 = 9

E = 6

S = 2

I have already seen a solution very similar to this, though.

E having to serve as a adden and as a sum means y has to equal 9 and twice E has to be more than 10.

1 plus S plus E must equal 9. S has to be an even number (being produced by a doubles sum.)

E is either 6,7,or 8. Making S 2,4,or 6 Since 6+2+1= 9 E is 6 and S is 2.

S+E=Y, and there is no carry, then Y<=9. Since the numbers cannot repeat and are different digits Y>=3; 3<=Y<=9 Since (tens place) E-Y=E, and Y is not zero, then there must be a carry from the ones column. E+E=10 +S and S must be even because 2E must be even. E>=6 to create a carry Since Y<=9 and S+E=Y, S can be 1,2 or 3. Only 2 is even.

Going from right side to left and adding digits individually:

$E + E \equiv S$ ... (1)

$E + Y \equiv E$ ... (2)

$S + E \equiv Y$ ... (3)

$\because$ In (2), you can't get E after adding E and Y unless E is being added to a multiple of 10.

$\because$ Y is nonzero as given.

$\because$ Y can't be greater than 9 as it's single digit.

$\therefore$ Y is 9 and there must have been a carry of 1 from (1) to make the sum of these two as 10.

Thus, from (1) if carry is generated

$E + E = 10 + S$

And from (3) as carry is now generated from (2) as well

$S + E + 1 = Y$

$\Rightarrow S + E + 1 = 9$

$\Rightarrow S + E = 8$ ... (4)

Solving (1) and (4) $S = 2, E = 6$

Thus, $S = 2$

100S+11E+101E+10Y=100Y+10E+S,.
99S + 102E = 90Y,.
33S + 34E = 30Y,.
This implies S is even number,.
E is a multiple of 3, and Y > E,.
=> S = (Y - E) - (3Y + E)/33,.
3Y+E being less than 36, can only be equal to 33, so 3Y + E = 33 or Y + (E/3) = 11,.
E can not be 3 as Y < 10,.
E can not be 9 as Y > E,.
So E = 6, and consequently, Y = 9,.
and S = 9 - 6 - 1 = 2.

The key in such questions is to avoid brute force and look for a weakness in question which we can exploit to our advantage and solve the question easily. Lets look at all the information the question provides us. $E + E = S or 10+S$

$E + E \neq 20 + S$

because the maximum E can go is 9, Therefore max(E+E) = 18. The second piece of info. the question gives us is that:

$E + Y + (carried number)= 10 + E$

because since E is non zero , E+Y cant be equal to E itself. now ,

$Y + (carried number) = 10$

We can conclude that Y must be 9 and carried number would be 1 and not 0. Further, The second column gives us 10 + E, therefore it must also carry 1 forward to the third column. We have a pair of linear equation in two variables which can be solved easily.

$2E - 10 = S$

$S + E + 1 = 9$

Solving these two equations , we get E= 6 and $\boxed{ S = 2 }$

If E=6, then:

E+E=12

Carry the ten into the next column to get 1+6+?=something that ends in six.

This means that Y=9.

Carry the ten again to get: 1+2(from the first step)+6=9.

Therefore, S=2

Just brute force using Python

*Probably not the smartest way...

From the problem, we can derive the equation: $100S + 11E + 100E + E + 10Y = 100Y + 10E + S \therefore 99S + 102E=90Y$

Therefore, 99S + 102E must be divisible by 10 (because of the 90Y), so S has to be even.

Guessing and Checking from here then yields our answer, $\boxed{2}$ .