Latest TCS Aptitude Question SOLUTION: A four digit no. equals the sum of twice of its reversed and half of it.What is the sum of digit of such a no..?

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A four digit no. equals the sum of twice of its reversed and half of it.What is the sum of digit of such a no..?

Asked In TCS Arpit Namdev (12 years ago)

Solved Aman Read Solution (5)

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Aman  (12 years ago)

If number is abcd, then
1000a+100b+10c+d = 2*(a+10b+100c+1000d) + (1000a+100b+10c+d)/2= 502a+70b+205c+(4001/2)d
498a+30b= 195c+ (3999/2)d
min possible value of d=2, c=1,
then 498a+30b= 4194
possible value of a so that 498a has 4 at unit place.
then a=8, b=7

so number will be 8712.

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Aman  (12 years ago)

18

Such number is 8712.
sum of digits = 8+7+1+2= 18
CHECK
8712 = 2*2178 + 8712/2

Is this solution Helpfull? Yes (6) | No (2)  

trish  (12 years ago)

can you plz explain the process

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Karan  (12 years ago)

Let the number be (abcd)

Then, (abcd)=2*(dcba) +0.5*(abcd)
or, 0.5*(abcd)=2*(dcba)
or, (abcd)=4*(dcba)

Then, dcba
* 4
------
abcd
------

Now, 'd' needs to be an even number (because 4*a = even number only)
Next 'd' cannot be 0 or 1.
So,d=2
Next,a=8;

Then we have the following two equations ==>>
4*b+3=c+10*x (where x= carry over)
and, 4*c=b (No carry,because 4*d=a is already 4*2=8)

Solving we get, b=7 and c=1 ...

So, ANSWER = a+b+c+d =8+7+1+2=18
so, ANSWER=18

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

chand basha  (10 years ago)

given that 1st digit is 4 times of last digit and 2nd digit is 7 times of third digit... so first number will be the multiple of 4 and 2nd digit is the multiple of 7..
by checking 3rd digit should be 0 or 1 and last digit is 1 or 2 as when multiplied by 4 or 7 we should not get any double number

2 minutes enough for solving rather than writing equations

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