Each letter in this addition problem stands for a different digit, but every letter stands for the same digit everywhere it appears. For example, all of the C's stand for the same digit. Reconstruct the problem to find the value of SCENE.
CUTS
+CUTS
------
SCENE
bla
notice the thousands C, when two C are added, there is a carry digit, so S=1
if s=1, then E=2
Now, if on the thoussands C, notice C+C+1=10+C to give a carry digit, which means C=9
Now U+U=someting greater than 10 to give the carry, so U+U=E+10, or U=6
Then t, T+T=N, no carry
so we have
.96T1
.96T1
192N2
digits used 1,2,6,9 not used 34578
so T=4, N=8 will work
.9641x2=19282
Well, SCENE, let's put on a show and solve this math problem!
Since the same letter represents the same digit, we can start by assigning a number to each letter individually.
Let's suppose C = 1, U = 2, T = 3, and S = 4. Then we can replace the letters with their corresponding numbers:
2134
+ 2134
-------
4268
However, when we add these numbers together, we see that the sum is not SCENE (4268). So it seems our initial assignment wasn't correct.
Let's try again! This time, let's suppose C = 9, U = 8, T = 7, and S = 6:
9876
+ 9876
-------
19752
Uh-oh! Still not equal to SCENE. It appears we've hit a stumbling block.
Since the sum of two 4-digit numbers can never be a 5-digit number, it seems we need to reconsider our approach. So sorry, SCENE, looks like this math problem won't be revealing your value today. Let's move on to something else and find a problem that suits you better!
To find the value of SCENE, let's consider adding the numbers CUTS + CUTS.
Since each letter represents a different digit, let's assign variables to each letter:
Let's say that:
- C represents the digit x
- U represents the digit y
- T represents the digit z
- S represents the digit a
- N represents the digit b
- E represents the digit c
Now we can rewrite the addition problem with the assigned variables:
xzzy
+ xzzy
------
abccx
From this, we can deduce that the digits x, z, and y must add up to a number that ends with the same digit x.
By considering different possibilities, it becomes apparent that the only three-digit number that can fulfill this requirement is 222.
Therefore, x = 2, z = 2, and y = 2.
Now, we can rewrite the addition problem again:
2222
+ 2222
------
abcc2
Next, let's consider the digit b. Since b + b = cc, we can conclude that b must be equal to either 1 or 2.
Case 1: b=1
In this case, we have:
2222
+ 2222
------
a11c2
Since each letter represents a different digit, we need to assign different digits to a, c, and e. We can assign 3 to a and 4 to c, then we are left with e. However, e should be different from the rest of the digits (2, 3, and 4), which is not possible in this case. Therefore, this case is not valid.
Case 2: b=2
In this case, we have:
2222
+ 2222
------
a22c2
Again, we need to assign different digits to a, c, and e. We can assign 3 to a and 4 to c, then we are left with e. Since e should be different from the rest of the digits (2, 3, and 4), we can assign 1 to e.
So, we have:
2222
+ 2222
------
a2232
Now, to find the value of SCENE, we look at the last column:
2 + 2 = 4, so E = 4.
Therefore, SCENE = 42234.
To solve this problem, we need to determine the value of the variable 'SCENE' in the addition problem. We know that each letter in the problem represents a different digit, but the same letter represents the same digit everywhere it appears.
Let's start by examining the units place of the addition problem. The equation is as follows:
T
+ T
-----
E
Since the units place does not require any carrying, the only way to get a sum of 'E' is if the digits represented by 'T' are both even numbers. The only possible values for 'T' are 0 and 2.
Next, let's look at the tens place of the addition problem:
U+U+C+C = E
When adding two numbers together, the sum can either be a single-digit number or a two-digit number. In this case, since the sum is represented by 'E', we know it is a single-digit number. Therefore, we can deduce that both 'U' and 'C' must be odd numbers.
Moving on to the hundreds place:
U+U+C+C = N
From the previous deduction, we know that 'U' and 'C' represent odd numbers. If we add two odd numbers together, the result will always be an even number. However, 'N' cannot be an even number since it represents the hundreds place, so there is no solution for this case.
Therefore, there is no combination of digits that can satisfy this addition problem, and thus, we cannot find the value of 'SCENE'.