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

• 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
• 10 years agoHelpfull: Yes(78) No(2)
• 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
• 10 years agoHelpfull: Yes(38) No(24)
• 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
• 10 years agoHelpfull: Yes(19) No(1)
• 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
• 10 years agoHelpfull: Yes(15) No(2)
• 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
• 10 years agoHelpfull: Yes(8) No(6)
• 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
• 10 years agoHelpfull: Yes(4) No(0)
• 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.
  
• 10 years agoHelpfull: Yes(4) No(0)
• 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
• 8 years agoHelpfull: Yes(4) No(0)
• sorry in second question it will be C+N^2
• 10 years agoHelpfull: Yes(2) No(0)
• 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
• 10 years agoHelpfull: Yes(2) No(1)
• ans :option 1 that is 4 ,c =8 then 8+4^2=25 which can divisible by 5
• 10 years agoHelpfull: Yes(1) No(7)
• 242 is the correct answer
• 10 years agoHelpfull: Yes(1) No(1)
• 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
• 10 years agoHelpfull: Yes(0) No(7)
• 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
  
• 10 years agoHelpfull: Yes(0) No(2)
• 1st questn-121
  2nd questn-3
• 10 years agoHelpfull: Yes(0) No(6)
• 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.
• 10 years agoHelpfull: Yes(0) No(0)
• 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.
• 10 years agoHelpfull: Yes(0) No(0)
• 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
• 10 years agoHelpfull: Yes(0) No(4)
• 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..
• 10 years agoHelpfull: Yes(0) No(0)
• for first its 11
  
• 10 years agoHelpfull: Yes(0) No(2)
• 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
• 9 years agoHelpfull: Yes(0) No(0)
• 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
• 8 years agoHelpfull: Yes(0) No(0)