Question

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 is true?

विकल्प

• (EO)² = (WTR)

• (ED)² + (OE)² = (DA)²

• (EA) x (CW) = (OA) x (CE)

• (LD) = (EO) x (D)

Answer 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

Option 1 is 23² = 576, which is false.
Option 2 is 24² + 32² = 402 , which is true
Option 3 is 20 x 15 = 30 x 12, which is false
Option 4 is 84 = 23 x 4, which is false.

Hence the second option 24² + 32² = 402 is the answer.