Latest infosys Aptitude Question SOLUTION: If abcde is a five digit number and abcde*4=edcba, then what is the value of b+c+d?

• Placement Papers
• /
• infosys
• /
• Numerical Ability
• /
• Number System

infosys Question Papers: infosys Interview Experience(10) INFOSYS Company Profile

(#M40153997) INFOSYS Numerical Ability question number Keep an EYE

If abcde is a five digit number and abcde*4=edcba, then what is the value of b+c+d?

Asked In infosys rasagnya (8 years ago)

Unsolved Read Solution (7)

Is this Puzzle helpful?   (12)   (14) Submit Your Solution Number System

Read Solution (7) :

arun gupta  (8 years ago)

hy .. can you tell me that these questions are of infosys recently ?

Is this solution Helpfull? Yes (15) | No (5)  

ulli sairam  (8 years ago)

how can we find that number 21978???

Is this solution Helpfull? Yes (9) | No (2)  

mettu naveen  (8 years ago)

A must be 1 or 2 since any other number would means that the answer would have 6 digit. A also have to be even since 4 time any number is even. thus A is 2. E would have to be 3 or 8. but E cannot be 3 since the last digit have to be 8 or 9. thus E is 8. now 4xB must be less than 10 thus B is one or 2. but A is already 2 thus B is one. 4 x D +3 must end with 1 two possible digit 2 and 7 but 2 already used thus D is 7. 4xC+2 must end with C. two possible numbers 0 and 9. now verify the options

Is this solution Helpfull? Yes (9) | No (1)  

Dheeraj Kumar Rao  (8 years ago)

21978*4=87912 then b+c+d=1+9+7=17

Is this solution Helpfull? Yes (6) | No (3)  

  (6 years ago)

Since EDCBA is a 5-digit number, we know ABCDE is < 1/4 × 100000.
ABCDE < 25000, so A is 1 or 2.

But EBCDA is a multiple of 4, so it is even, so A = 2.
4 * ....E = ....2, so E is either 3 or 8.
But, 4 × 2nnnn cannot be 3nnnn so E is 8.

We have 2BCD8 * 4 = 8DCB2.
and therefore 4 × BCD + 3 = DCB.
Let's look at it case by case, for values of D:
D = 1, B = 7
D = 2, B = 1
D = 3, B = 5
D = 4, B = 9
D = 5, B = 3
D = 6, B = 7
D = 7, B = 1
D = 8, B = 5
D = 9, B = 9
D = 0, B = 3

But B cannot be greater than 2, or there would be a carry, and B cannot
be equal to 2, since A = 2 (and we assume each letter represents a
different number).
So, B = 1 and D = 2 or 7.
But again, A is already 2, so D = 7.

So we have 21C78 * 4 = 87C12
We now have 4 * C + 3 = C (mod 10)
3*C = 7 mod 10
and therefore C = 9

So the answer is 21978 * 4 = 87912

Is this solution Helpfull? Yes (2) | No   

priya kumari  (8 years ago)

A-2,B -1,C-9,D-7,E-8
Therefore B+C+D= 17

Is this solution Helpfull? Yes (1) | No (2)  

Pratyush Gangwar  (8 years ago)

21978
*4
________
87912

so b+c+d=17

Is this solution Helpfull? Yes (1) | No (4)  

Please Login to Read Solution.