Latest TCS Aptitude Question SOLUTION: Ques 23 : Choose the correct answer. P is an integer. P>883. If P-7 is a multiple of 11, then the largest number that will always divide (P+4) (P+15) is: Option 1 : 11 Option 2 :

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Ques 23 : Choose the correct answer.
P is an integer. P>883. If P-7 is a multiple of 11, then the largest number that will always divide (P+4) (P+15) is:

Option 1 : 11
Option 2 : 121
Option 3 : 242
Option 4 : None of these

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Let C be a positive integer such that C + 7 is divisible by 5. The smallest positive integer n (>2) such that C + n2 is divisible by 5 is:

Option 1 : 4
Option 2 : 5
Option 3 : 3
Option 4 : Does not exist

Asked In TCS ritabrata gautam (10 years ago)

Unsolved Read Solution (22)

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Read Solution (22) :

SOUBHIK BISWAS  (10 years ago)

p=11k+7
so (p+4)(p+15)=(11k+11)(11K+22)
=11(k+1)*11(k+2)
=121(k+1)(k+2)
here one of the terms of (k+1)and(k+2)is even and another is odd..so multiplication of odd and even no. produce a even no..so the no. is divisible by (121*2)=242

Is this solution Helpfull? Yes (78) | No (2)  

rubi saini  (10 years ago)

p=11k+7
so (p+4)(p+15)=(11k+11)(11K+22)
=11(k+1)*11(k+2)
=121(k+1)(k+2)
so 121 is the largest digit that will devide this term

Is this solution Helpfull? Yes (38) | No (24)  

srivani  (10 years ago)

P > 883 P − 7 > 876 The smallest number after 876 that is a multiple of 11 is 880 = 11*80 So P = 11k 7, for some integer k ≥ 80 (P 4)(P 15) = (11k 7 4) (11k 7 15) = (11k 11) (11k 22) = 11(k 1) * 11 (k 2) = 121 (k 1)(k 2) Now when k is even, then k 2 is even, so (k 1)(k 2) is even and when k is odd, then k 1 is even, so (k 1)(k 2) is even (k 1) (k 2) = 2n, for some integer n (P 4)(P 15) = 121 (k 1)(k 2) = 121 * 2n = 242n Answer: 3: 242

Is this solution Helpfull? Yes (19) | No (1)  

SUNNY KUMAR SHAW  (10 years ago)

GIVEN : C+7 is divisible by 5.
we have :- c+n^2
adding and subtracting seven we have

c+7+n^2-7........this should be divisible by 5
but in the question they have said that c+7 is divisible .....that means n^2-7 should be divisible.....by 5
NOW IN THE OPTION WE HAVE:
4
5
3
DOES NOT EXIST
NOW PLACING THE VALUES IN THE EQUATION N^2-7
WE SEE NEITHER 4 NOR 5 NOR 3 SATISFIES THAT SO THE ANSWER IS DOES NOT EXIST

Is this solution Helpfull? Yes (15) | No (2)  

Mantripragada Nikhil  (10 years ago)

23.given p-7 is a multiple of 11..so p-7=11n...and p=11n+7.
but p>883
11n+7>883...so 11n>876....n>79......
so(p+4)(p+15)==(11n+7+4)(11n+7+15)==(11n+11)(11n+22)=121(n+1)(n+2) which is divisible by 121

2.C+7=5n...as per question...so,c=5n-7
therefore,C+2n=5n-7+2n=7(n-1)...for this to be divisible by 5...n-1 is to be divisible by 5..smallest value of n(>2)...which satisfies the condition is n=6

Is this solution Helpfull? Yes (8) | No (6)  

Raju meka  (10 years ago)

The answer is does not exist.
Dividend = divisor * Quotient+ remainder.
c+7=5Q+0
==> c = 5Q-7
Now, C+n^2 = 5Q-7+n^2 which is not divisible any of the given "n" values.So, the answer is 4

Is this solution Helpfull? Yes (4) | No   

Hitesh Yererkar  (9 years ago)

p>883
p-7=876.

the smallest number after 876 is 880 which is multiple of 11=11*80.
so
p=11K+7. i.e k>80.
(p+4)(p+15)
=(11k+7+4)(11k+7+15)
=(11k+11)(11k+22)
=11(k+1)*11(k+2)
=121(k+1)(k+2).

here (k+1)(k+2)=2n.

=(p+4)(p+15)
=121(k+1)(k+2)
=121*2n.
=242n.

Is this solution Helpfull? Yes (4) | No   

Raghav Goyal  (7 years ago)

Ans=242

P-7 is a multiple of 11.
P-7=11k for some k.
P=11k+7
substituting (P+4)(P+15)=(11k+7+4)(11k+7+15)
=(11k+11)(11k+22)
=11(k+1)x11(k+2)
=121(k+1)(k+2)
Now since k+1 and k+2 are consectuive no.
(k+1)(k+2) is always divisible by 2.
hence ans=121*2=242

Is this solution Helpfull? Yes (4) | No   

ritabrata gautam  (10 years ago)

sorry in second question it will be C+N^2

Is this solution Helpfull? Yes (2) | No   

anushri  (10 years ago)

p=11k+7
so (p+4)(p+15)=(11k+11)(11K+22)
=11(k+1)*11(k+2)
=121(k+1)(k+2)
so 121 is the largest digit that will devide this term
but i thnk the ans is 242 ...
becz p> 883
and p-7 is divisibl by 11...
after 883 ..no. wch is divisible by 11 is 891
so p= 891+7
9 mins

so
4 mins

(898+4)(898+15)= 823526
823526/242 =3403
hence ans is 242

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

syambabu  (10 years ago)

ans :option 1 that is 4 ,c =8 then 8+4^2=25 which can divisible by 5

Is this solution Helpfull? Yes (1) | No (7)  

Priya Singh  (9 years ago)

242 is the correct answer

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

yamana santhosh kumar  (10 years ago)

ans:option 4,because the least no is (c=3).If n=6 then its divisible by 5.
sol:3+6*2=15. it can divisible by 5

Is this solution Helpfull? Yes | No (7)  

vinothkumar  (10 years ago)

option3 :3
The value of c is 3 and the integer is 6
for (c+7) the value of C must be 3, 8.. inorder to divide by 5 and the least value is 3, and if n>2, then 6 is the next choice other than 1

Is this solution Helpfull? Yes | No (2)  

sharmila paul  (10 years ago)

1st questn-121
2nd questn-3

Is this solution Helpfull? Yes | No (6)  

selva  (10 years ago)

p is larger than 883 means i will chse 898. then p-7 is the multiple of 11 so 898-7=891 which is the multiple of 11(11*811=891). (p+4)(p+15)=902*913=823526.
given options r 11,121,242. the 11 is the only one which always divide the 823526.

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selva  (10 years ago)

p is larger than 883 means i will chse 898. then p-7 is the multiple of 11 so 898-7=891 which is the multiple of 11(11*81=891). (p+4)(p+15)=902*913=823526.
given options r 11,121,242. the 11 is the only one which always divide the 823526.

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Akanksha Yadav  (10 years ago)

since C+7 is divisible by 5
C+7=5(x)
C=5x-7
now
C+n^2=5x-7+n^2
C+n^2 will only be divisible by 5 when n=3
therefore ans 3

Is this solution Helpfull? Yes | No (4)  

sindhu  (10 years ago)

if c=4 then how c+7=divisible by 5 its not satisfying the first condition and how can we go for 2nd c+n^2.
plz give a clear explaination..

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Amol Rathod  (9 years ago)

for first its 11

Is this solution Helpfull? Yes | No (2)  

ramesh  (8 years ago)

P-7 is a multiple of 11.
P-7=11k for some k.
P=11k+7
substituting (P+4)(P+15)=(11k+7+4)(11k+7+15)
=(11k+11)(11k+22)
=11(k+1)x11(k+2)
=121(k+1)(k+2)
Now since k+1 and k+2 are consectuive no.
(k+1)(k+2) is always divisible by 2.
hence ans=121*2=242

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sweta  (7 years ago)

P-7 is a multiple of 11.
P-7=11k for some k.
P=11k+7
substituting (P+4)(P+15)=(11k+7+4)(11k+7+15)
=(11k+11)(11k+22)
=11(k+1)x11(k+2)
=121(k+1)(k+2)
Now since k+1 and k+2 are consectuive no.
(k+1)(k+2) is always divisible by 2.
hence ans=121*2=242

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