Latest TCS Aptitude Question SOLUTION: A number ab3133ab is divisible by 12. find minimum value of (a+b) ?

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A number ab3133ab is divisible by 12.
find minimum value of (a+b) ?

Asked In TCS aanchal kushwaha (12 years ago)

Unsolved Read Solution (14)

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Read Solution (14) :

rahul  (12 years ago)

answer is 4..
for divisible by 12;
the no. is devided by 3,4
now, for 3
a+b+3+1+3+3+a+b= 2(a+b)+10............(1)
for 4;
last two digit devided by 4
it means; ab/4.........................(2)
from equestion 1 and 2
suitable value is of (a+b)=4;
a=0, b=4;

Is this solution Helpfull? Yes (19) | No (5)  

aanchal kushwaha  (12 years ago)

i think answer vil b 4. bcz a=4 b=0. so min value of a+b=4.

but in some site its given 7.

plz tell wich vil b correct.

Is this solution Helpfull? Yes (18) | No (9)  

Rajesh Nimmala  (12 years ago)

the answers are depends on given condition.if a and b are non zero values,then u will get 7 otherwise answr is 4

Is this solution Helpfull? Yes (13) | No (1)  

prai  (12 years ago)

a+b=4(minimum)
as for 12 divisibility.. multiple of 3 and 4..
multiple of 3:-
10+2(a+b) its sum should be divisible of 3..
i.e. 4+0;3+6... like this so leat value which is also divisible by 4 will be
4+0
so ans 4

Is this solution Helpfull? Yes (4) | No   

Ritesh Tiwari  (12 years ago)

in dis question ask's the minimum value so it should be 4.

Is this solution Helpfull? Yes (4) | No   

deepasha guru  (12 years ago)

Ans is 7.
a number to b divisible by 12 must be divisible by 3 &4
For 3 sum of digits must b divisible by 3.
For 4 last two digits must b divisible by 4.
a & b can be 1 & 6

Is this solution Helpfull? Yes (3) | No (4)  

venkatrao  (12 years ago)

In this question the answer is always 4. if a and b are non-zero values then a+b will be 4 i.e. 0 and 4.

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

priyankaa tamta  (12 years ago)

the condtn fr divisibily by 12 z it shold be divide by 3,4...
so the sum of a+b=4 ans

Is this solution Helpfull? Yes (2) | No   

Sanchit Khaneja  (12 years ago)

answer is 4 bcz it has to be divisble by both 2 and 3 as per by divisbility rule of 12

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chandrika  (12 years ago)

for question ab313ab ans is 7

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

Rajesh Kumar Sethi  (12 years ago)

In this question the answer is always 4. if a and b are non-zero values then a+b will be 4 i.e. 0 and 4.
otherwise a = 1 and b =3

Is this solution Helpfull? Yes (1) | No (6)  

preeti arya  (12 years ago)

a=4 and b=0

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

gaurav kumar  (10 years ago)

ans is 4 we know that a no. is divisible by 12 only when it is divisible by 3 and 4 so for satisfying these conditions we have sum of digits as a+b+3+1+3+3+a+b/3 which gives 2(a+b)+10/3 and also for no. divisible by 4 last 2 digit must be divisible by 4 so let a=4 and b=0 so for div by 3 we have 2*4+10/3=6 so this criteria is satisfied and for div by 4 last digit is 40 which is divisible by 4 and one case is also when we have a=5 and b=2 so 2*7+10/3=8 and 52/4=13 but minimum value is 4+0 ie ans would be 4

Is this solution Helpfull? Yes (1) | No   

snehil  (12 years ago)

this number can be divisible by 12 if last digits would be
12,24,36,48,60,72,84,96........
here only two palces given ie ab so max valu of ab will be 96
so add 9+6=15
ans

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