Quiz

Question: If n ∈ N, then 11n+2 + 122n+1 is divisible by
Options:
A.113
B.123
C.133
D.None of these
Answer: Option C
: C

Putting n = 1 in 11n+2+122n+1 We get, 111+2+122×1+1 = 113+123 = 3059, which is divisible by 133.

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Question 1. For all natural numbers n, 23n7n1 is divisible by 
  1.    64
  2.    36
  3.    49
  4.    25
Answer: Option C
: C

Substitute n=1 in23n7n1, we get
2371=0divisible by all positive integers Substitute n=2 in23n7n1, we get
26141=49
LetP(n):23n7n1is divisible by 49
P(2) is true.
Assume P(k) is true
23k7k1=49m
Substituting k+1 in place of n, we get
23k+37(k+1)1=8.23k7k8=8.(23k7k1)+7.7k=49(8m+k)divisible by 49
P(k+1) is true

Hence, P(n) is true.
Question 2. Which of the following illustrates the inductive step to prove a statement P(n) about natural numbers n by mathematical induction, where k is an arbitrary natural number?
  1.    P(1) is true
  2.    P(k) is true
  3.    P(k) is true ⇒ P(k+1) is true
  4.    none of these
Answer: Option C
: C

Suppose there is a given statement P(n) involvingnatural numbers n such that i) The statement is true for a specific natural number m i.e. P(m) is true. This is known as the base case. ii) If the statement is true for n = k (where k is some positive integer), then the statement is also true for n=k+1, i.e., the truth of P(k) implies the truth of P(k+1).

The second step is called the inductive step.
Question 3.  n ϵ N, P(n):4.7n+20 is divisible by 24. State true or false.
  1.    True
  2.    False
  3.    17
  4.    24
Answer: Option A
: A

P(n):4.7n+20isdivisibleby24
P(1):28+20=48divisibleby24P(k):4.7k+20isdivisibleby24assumetrue4.7k+20=24qNow,4.7k+1+20=7.(4.7k)+20=6.4.7k+(4.7k+20)=24.7k+24qP(k+1)istrue

P(n)istrue4.7n+20isdivisibleby24
Question 4. For all natural numbers n, 102n1+1 is divisible by
  1.    13
  2.    11
  3.    9
  4.    none of these
Answer: Option B
: B

Substituting n=1, we have
101+1=11LetP(n):102n1+1is divisible by 11

P(1) is true.
Assume P(k) is true
102k1+1=11mNow,P(k+1):102k+1+1=100.102k1+1=99.102k1+102k1+111.(9.102k1+m)divisible by 11

P(k+1) is also true.
Hence, P(n) is true.
Question 5. 7n3n is always divisible by 
  1.    6
  2.    10
  3.    4
  4.    8
Answer: Option C
: C

Substituting n=1, we get
7131=4
Let P(n):7n3nis divisible by 4
P(1) is true
Assume P(k) is true
7k3kis divisible by 47k3k=4mNow,7k+13k+1=7.(7k3k)+4.3k=4.(7m+3k)divisible by 4
P(k+1) is true

Hence, P(n) is true.

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