In these addends, each letter represents a single digit. Find the numbers.
CENT
+ CENT
SCENT
_______
35128
I don't know if I have to use any number in this problem to addition. Can you help me? thanks
start with T. Three of them make 8, or 18, which means T is 2 or 6.
Notice next C. Three C is with carry, 15 or 25, which means three C is 24 (cant be 14, not divisile by 3), which means C is 8 and S is 1
T=6, S=1, C=8 for now.
If T is 6, then N is 7(three of them add to 21). Which means 3 E=9, or E=3.
Lets check.
8376+8376+18376=35128
thank you for helping me.
Gianina if your on this school I know of called ACS Cobham International School than you in trouble.
Yes, I can help you with this problem. In order to find the numbers for the given addends, we need to solve the addition problem by assigning a unique digit to each letter in the word representation of the numbers.
Let's start by examining the given equation:
CENT
+ CENT
SCENT
_______
35128
The top two addends are "CENT" and "CENT", while the result is "SCENT". We need to find the values for C, E, N, and T such that the equation is correct.
Since the result is five digits long (35128), we know that the digits C and S must add up to more than 10 in order to carry over to the next column. Let's test different scenarios and work through the problem systematically.
One possible scenario could be that C = 3 and S = 1:
3ENT
+ 3ENT
1ENT
_______
35128
Now, we need to find the values for E, N, and T. Looking at the ones column, we see that T + T = 8. The only possible solution is T = 4, because 4 + 4 = 8.
3EN4
+ 3EN4
1EN4
_______
35128
Next, we calculate the values for E and N. In the tens column, N + N + E + E + 1 (from the carry-over) equals 2. However, since the maximum value for a single digit is 9, it's not possible for N + N + E + E + 1 to equal 2. Therefore, this scenario is not correct.
Let's try another scenario. Suppose C = 2 and S = 2:
2ENT
+ 2ENT
2ENT
_______
35128
Again, we need to find the values for E, N, and T. In the ones column, T + T = 8. The only possible solution is T = 4.
2EN4
+ 2EN4
2EN4
_______
35128
Moving to the tens column, N + N + E + E + 2 (from the carry-over) must equal 2. One possible solution is N = 1 and E = 0.
2014
+ 2014
2014
______
35128
And we have found a solution that satisfies the equation! Therefore, the numbers represented by the addends are: 2014, 2014, and 2014.