Latest TCS Aptitude Question SOLUTION: the sum of all possible two digit numbers formed from threee different one digit natural numbers when divided by the sum of original three no.s is equal to?

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the sum of all possible two digit numbers formed from threee different one digit natural numbers when divided by the sum of original three no.s is equal to?

Asked In TCS ratna (11 years ago)

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Anurag  (11 years ago)

Let the three nos. be x,y,z
So,the sum of all two digit nos formed by them is 10x+y+10x+z+10y+x+10y+z+10z+x+10z+y=22(x+y+z)
So22(x+y+z)/(x+y+z)=22

Is this solution Helpfull? Yes (25) | No (3)  

BRIJESH KUMAR GUPTA  (11 years ago)

sorry th correct answer is =33
i misunderstood the question
take p,q,r now 3*3=9 no will be formed
sum of it (10*p+p)+(10*q+q)+(10*r+r)+(10*p+q)+(10*p+r)+(10*q+r)+(10*q+p)+(10*r+p)+(10*r+q)=33(p+q+r) now divide it by p+q+r u get 33

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Jetipiti Babu  (11 years ago)

The maximum sum of three different single digit numbers can not be greater than 24(9+8+7).
Hence 36 is rejected.
22 has factors, 1, 2 and 11only, whose sum is only 14; hence this also is not possible.
The single digit factors of 18 are, 1, 2, 3, 6, 9. Of these sum of 3 + 6 + 9 = 18; Hence '18' is the required number for the given question.
Source(s):
My knowledge

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Raja Polavarapu  (11 years ago)

ANS: 22
(10X+Y)+(10X+Z)+(10Y+X)+(10Y+Z)+(10Z+X)+(10Z+Y)=[20(x+y+z)+2(X+Y+Z)]/(X+Y+Z)
=>22

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BRIJESH KUMAR GUPTA  (11 years ago)

ans=6
take any three digit let p,q,r
total numbers formed of 2 digits =3*3=9
total digits used in 9 numbers=9*2=18
out of which 6 of p 6 of q and 6 of r now adding
we get 6*(p+q+r).divide it by p+q+r u get 6

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