ABC+CBA=DEDD Where A, B, C, D, E are distinct digits satisfying this addition fact, then E is?
A) 3
B) 5
C) 2
D) 4
Answers
Answered by
7
Answer:
C. 2
Step-by-step explanation:
4 5 7
+ 7 5 4
-----------------
1 2 1 1
Therefore , A = 4
B = 5
C = 7
D = 1
E = 2
Answered by
1
Answer:
Answer is C) 2
Step-by-step explanation:
- So let us take the maximum value of each solve so:
- Maximum number = 9
- But A B C D and E are different digits so we need another number for adding:
- 2nd maximum number 8
- But in the hundred's place you cannot give the D as any other number except 1 (9 + 8 = 17)
- So:
- D = 1
- => A B C
+ C B A
_______
1 E 1 1
- Now there is no non-decimal numbers that add up to 1 but we need 1 at 1000nds place because we made the D as 1.
- But 11 is the only number that has 1 at all the places. So we could take any two digits to add up to 11: like 8 + 3, 9 + 2, 7 + 4 etc...
- So for example we have taken:
- C = 9
- A = 2
- => 2 B 9
+ 9 B 2
_______
1 E 1 1
- Now there is 1 carry over to tens place and we need same digits to add up to 11 but there is no non-decimal numbers that add up to 11.
1
- => 2 B 9
+ 9 B 2
_______
1 E 1 1
- But, As there is a carry over, it will automatically been added so 11 - 1 = 10
- Now we could only take 5 and add itself to 10
- So:
- B = 5
1
- => 2 5 9
+ 9 5 2
_______
1 E 1 1
- Then again a carry over of 1 to hundreds place.
- Now by taking the values of C and A we could add up it to 11 and carry over of 1 which is 11+1 = 12
1 1
- => 2 5 9
+ 9 5 2
_______
1 2 1 1
There You Go !!
Your answer is:
A B C 2 5 9
+ C B A to + 9 5 2
______ ______
D E D D 1 2 1 1
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