Latest TCS Aptitude Question SOLUTION: . ab313ab when divided by 12 leaves no remainder. What is the value of a+b?

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. ab313ab when divided by 12 leaves no remainder. What is the value of a+b?

Asked In TCS (10 years ago)

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Arindam Mukherjee  (10 years ago)

as it is divided by 12 then the digit should be divided by by 3& 4 respectively.now for divisible by 4 last 2 digit should be divisible by4 & sum of the no.s is divisible by 3..so sum is(3+1+3+2a+2b)/3 remainder should be 0.so(7+2a+2b)=should be 9,12,15..for 9..2(a+b)=2so (a+b)=1 so last 2 digit is 10..but it is not divisible by 4
so, we take 7+2a+2b=15,a+b=8/2=4..now ab may be 40 which is divisible by 4...so the solution is (a+b)=4

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sudipta sadhukhan  (10 years ago)

2(a+b)+7 mod 3 =0
a+b=16

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bhagyashree roy  (10 years ago)

ans is 4
for divisible by 12,the no is divisible by 3,4
now for 3
a+b+3+1+3+a+b=2(a+b)+7
and for 4
last 2 digit should be divided by 4
it means ab/4
so,a+b=4

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Pratyusha Bhadra  (10 years ago)

a=6,b=4.....a+b=10

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Soumit Ghosh  (10 years ago)

a=1,b=6
1631316 divided by 0 leaves no remainder
So a+b=1+6
=7

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ANUP KUMAR  (10 years ago)

a+b=4
since a+b+3+1+3+a+b= 2(a+b)+7: now let a+b=x
then on trial let x=4 then if a=4 b=0
then the required no is divisible by 3,4,and 6 so it divisble by 12 and leaves no reamainder

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Nitish kumar  (10 years ago)

for completely divisible by 12 this number have to divisible by 4 and 3 individually.
for divisible by 3
sum of the number is divisible by 3
(a+b+3+1+3+a+b)/3 gives rem=0;
for divisible to be 4-
last 2 digit is divisible by 4
ab=4
then a is either 1 or 3
and b is either 1 or 3
so a+b=4
4 is the ans

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Raushan shukla  (10 years ago)

ans -(a+b)=7, a=1 n b=6

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parvez sajjad  (10 years ago)

a+b=10
a=2
b=8

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Mita Pan  (10 years ago)

12=3*4 so,ab313ab has to be divided 3 and 4.(a+b+3+1+3+a+b)mod 3=0 and (a+b)mod 4=0. (a+b)=4 satisfy these conditions

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kanchanadevi  (7 years ago)

Which is the CRT ans

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TARUN  (6 years ago)

as it is divided by 12, it must be divided by 3 and 4.
hence last two digits are divisible by 4.
total sum of all digits are divisible by 3.
(a+b+3+1+3+a+b)%3=0
(7+2(a+b))%3=0
and
ab is divisible by 4.
let us take ab=12 which is divisible by 4.
==> a+b=3
==> (7+2(3))=13 which is not divisible by 3. so ab is not equal to 12.
let us take ab=16 which is divisible by 4.
==> a+b=7
==> 7+14 =21 which is divisible by 3.
so ab="16" ==> a+b=7.
let us take another example. ab=76 which is divisible by 4.
==> a+b=13
==> 7+2(13) = 33 which is divisible by 3.
so ab= "76" ==> a+b= 13

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TARUN  (6 years ago)

as it is divided by 12, it must be divided by 3 and 4.
hence last two digits are divisible by 4.
total sum of all digits are divisible by 3.
(a+b+3+1+3+a+b)%3=0
(7+2(a+b))%3=0
and
ab is divisible by 4.
let us take ab=12 which is divisible by 4.
==> a+b=3
==> (7+2(3))=13 which is not divisible by 3. so ab is not equal to 12.
let us take ab=16 which is divisible by 4.
==> a+b=7
==> 7+14 =21 which is divisible by 3.
so ab="16" ==> a+b=7.
let us take another example. ab=76 which is divisible by 4.
==> a+b=13
==> 7+2(13) = 33 which is divisible by 3.
so ab= "76" ==> a+b= 13

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