Solve the following question and mark the most appropriate option.
In the addition problem shown below, distinct letters of the alphabet are used to represent the digits from 0 to 9. E and D are both less than 5 with D > E so that the sum of D and E is the average of W and T.
W | O | R | L | D | |
+ | T | R | A | D | E |
C | E | N | T | E | R |
Which of the following letters of the alphabet represents the average of the first nine natural numbers?
Options
E
E
E
T
T
T
W
W
W
L
L
L
Solution 1
Since W + T = (CE), C must be 1.
Since D and E are both less than 5, they can only take values 0, 2, 3 or 4 (1 has already been used)
If either of D or E is 0, then D + E must yield a D or an E.
But in the above example D + E = R, neither D nor E can be 0.
So (D, E) = (3, 2), (4, 2) or (4, 3) and the corresponding values of R will be 5, 6 or 7.
Since R is the average of W and T, the corresponding values of W + T will be 10, 12 or 14.
Since E can only take values 2 or 3, and the sum W + T ends in 0, 2 or 4, E must be 2.
The first 2 digits of the answer are 1 and 2 respectively.
Which means W + T = 12, and therefore D + E = R, must be 6.
We can conclude that (D, E) has to be (4, 2).
W | O | 6 | L | 4 | |
+ | T | 6 | A | 4 | 2 |
1 | 2 | N | T | 2 | 6 |
Now L + 4 must be 12 so that L is 8 and 1 is carried over.
W | O | 6 | 8 | 4 | |
+ | T | 6 | A | 4 | 2 |
1 | 2 | N | T | 2 | 6 |
The digits remaining are 0, 3, 5, 7 and 9.
A total of 12 can be obtained as 5 + 7 or 3 + 9
Suppose (W, T) is (3, 9) then 6 + 1 + A = 9 yields A = 2 which is not possible
Suppose (W, T) is (9, 3) then 6 + 1 + A = 13 yields A = 6 which is not possible
So (W, T) is (7, 5) or (5, 7)
Suppose (W, T) is (7, 5) then 6 + 1 + A = 15 yields A = 8 which is not possible
Therefore (W, T) must be (5, 7) and 6 +1 + A = 7, yields A = 0.
The only 2 digits now remaining are 3 and 9.
Hence O has to be 3 and N has to be 9.
5 | 3 | 6 | 8 | 4 | |
+ | 7 | 6 | 0 | 4 | 2 |
1 | 2 | 9 | 7 | 2 | 6 |
W = 5, is the average of 1 to 9
Solution 2
Since W + T = (CE), C must be 1.
Since D and E are both less than 5, they can only take values 0, 2, 3 or 4 (1 has already been used)
If either of D or E is 0, then D + E must yield a D or an E.
But in the above example D + E = R, neither D nor E can be 0.
So (D, E) = (3, 2), (4, 2) or (4, 3) and the corresponding values of R will be 5, 6 or 7.
Since R is the average of W and T, the corresponding values of W + T will be 10, 12 or 14.
Since E can only take values 2 or 3, and the sum W + T ends in 0, 2 or 4, E must be 2.
The first 2 digits of the answer are 1 and 2 respectively.
Which means W + T = 12, and therefore D + E = R, must be 6.
We can conclude that (D, E) has to be (4, 2).
W | O | 6 | L | 4 | |
+ | T | 6 | A | 4 | 2 |
1 | 2 | N | T | 2 | 6 |
Now L + 4 must be 12 so that L is 8 and 1 is carried over.
W | O | 6 | 8 | 4 | |
+ | T | 6 | A | 4 | 2 |
1 | 2 | N | T | 2 | 6 |
The digits remaining are 0, 3, 5, 7 and 9.
A total of 12 can be obtained as 5 + 7 or 3 + 9
Suppose (W, T) is (3, 9) then 6 + 1 + A = 9 yields A = 2 which is not possible
Suppose (W, T) is (9, 3) then 6 + 1 + A = 13 yields A = 6 which is not possible
So (W, T) is (7, 5) or (5, 7)
Suppose (W, T) is (7, 5) then 6 + 1 + A = 15 yields A = 8 which is not possible
Therefore (W, T) must be (5, 7) and 6 +1 + A = 7, yields A = 0.
The only 2 digits now remaining are 3 and 9.
Hence O has to be 3 and N has to be 9.
5 | 3 | 6 | 8 | 4 | |
+ | 7 | 6 | 0 | 4 | 2 |
1 | 2 | 9 | 7 | 2 | 6 |
W = 5, is the average of 1 to 9
Solution 3
Since W + T = (CE), C must be 1.
Since D and E are both less than 5, they can only take values 0, 2, 3 or 4 (1 has already been used)
If either of D or E is 0, then D + E must yield a D or an E.
But in the above example D + E = R, neither D nor E can be 0.
So (D, E) = (3, 2), (4, 2) or (4, 3) and the corresponding values of R will be 5, 6 or 7.
Since R is the average of W and T, the corresponding values of W + T will be 10, 12 or 14.
Since E can only take values 2 or 3, and the sum W + T ends in 0, 2 or 4, E must be 2.
The first 2 digits of the answer are 1 and 2 respectively.
Which means W + T = 12, and therefore D + E = R, must be 6.
We can conclude that (D, E) has to be (4, 2).
W | O | 6 | L | 4 | |
+ | T | 6 | A | 4 | 2 |
1 | 2 | N | T | 2 | 6 |
Now L + 4 must be 12 so that L is 8 and 1 is carried over.
W | O | 6 | 8 | 4 | |
+ | T | 6 | A | 4 | 2 |
1 | 2 | N | T | 2 | 6 |
The digits remaining are 0, 3, 5, 7 and 9.
A total of 12 can be obtained as 5 + 7 or 3 + 9
Suppose (W, T) is (3, 9) then 6 + 1 + A = 9 yields A = 2 which is not possible
Suppose (W, T) is (9, 3) then 6 + 1 + A = 13 yields A = 6 which is not possible
So (W, T) is (7, 5) or (5, 7)
Suppose (W, T) is (7, 5) then 6 + 1 + A = 15 yields A = 8 which is not possible
Therefore (W, T) must be (5, 7) and 6 +1 + A = 7, yields A = 0.
The only 2 digits now remaining are 3 and 9.
Hence O has to be 3 and N has to be 9.
5 | 3 | 6 | 8 | 4 | |
+ | 7 | 6 | 0 | 4 | 2 |
1 | 2 | 9 | 7 | 2 | 6 |
W = 5, is the average of 1 to 9