Elementary Number Theory MCQ Quiz in मल्याळम - Objective Question with Answer for Elementary Number Theory - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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നേടുക Elementary Number Theory ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Elementary Number Theory MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Elementary Number Theory MCQ Objective Questions

Top Elementary Number Theory MCQ Objective Questions

Elementary Number Theory Question 1:

Let k be the order of a mod n then ab ≡ 1(mod n) if and only if 

  1. k divides a
  2. k divides b
  3. k divides n
  4. k divides 1

Answer (Detailed Solution Below)

Option 2 : k divides b

Elementary Number Theory Question 1 Detailed Solution

Concept:

Let b ∈ Z such that a≡ 1(mod n).

Let us apply division algorithm to b and k then we have,

b = kq + r where 0 ≤ r ≤ k

Consider, ab = akq + r = (ak)q . ar

By hypothesis a≡ 1(mod n) and a≡ 1(mod n).

Hence, a≡ 1(mod n) where 0 ≤ r ≤ k

∴ r has to be equal to zero and otherwise the choice of k has the smallest positive integer such that a≡ 1(mod n) will be contradicted.

Hence, b = qk

⇒ k | b

⇒ k divides b

Hence, the correct answer is option 2)

Elementary Number Theory Question 2:

If 'p' is a prime, then the congruence x2 + 1 ≡ 0 (mod p) has a solution if and only if:

  1. p = 1 or p ≡ 1 (mod 3)
  2. p ≠ 1 or p ≡ 1 (mod 2)
  3. p = 2 or p ≡ 1 (mod 4)
  4. p = -2 or p ≡ 2 (mod 3)

Answer (Detailed Solution Below)

Option 3 : p = 2 or p ≡ 1 (mod 4)

Elementary Number Theory Question 2 Detailed Solution

Concept:

As we know that

\(\equiv\) r (mod b) means when a is divided by b  then r is the remainder or can be written as 

a = bq + r where q is any positive integer.  

Euler's criteria:

If p is a prime where p | a, then the quadratic congruence x2  \(\equiv\) (modmhas solutions or no solutions depending if \(a^\left({\frac{p-1}{2}}\right)=1. modp\)  or   \(a^\left({\frac{p-1}{2}}\right)=-1. modp\)

Calculation:

x2 + 1 ≡ 0 (mod p).

Case I: For x = 1

By definition of congruence and modulo we have 

p | x2 + 1 

Let, x = 1 then p | 12 + 1 \(\Rightarrow \) p | 2

As, p can divide 2 only when p = 2.

Case II: For x > 1

As by Euler's criteria, p | x2 + 1 if and only if \((-1)^\left({\frac{p-1}{2}}\right)=1. modp\).

Above condition can be only true if and only if the exponent of (-1) should be even.   ( As (-1)n = 1 if and only if n is even )

So, \(\frac{p-1}{2}\) should be even or is a multiple of 2.

\(\Rightarrow \frac{p-1}{2}= 2m\)   for m as some integer.

\(\Rightarrow p-1= 4m\)

p = 4m + 1 which is equivalent to p ≡ 1 (mod 4).   

Hence, If 'p' is a prime, then the congruence x2 + 1 ≡ 0 (mod p) has a solution if and only if p = 2 or p ≡ 1 (mod 4).

Elementary Number Theory Question 3:

What is the GCD of 4598 and 3211 ?

  1. 11
  2. 40
  3. 19
  4. 13

Answer (Detailed Solution Below)

Option 3 : 19

Elementary Number Theory Question 3 Detailed Solution

Concept:

Euclid division algorithm: If a and b are any integers and b > 0then there exists unique integers q and r such that a = bq + r where 0 ≤ r ≤ b.

Greatest common divisor: Let a and b be any two positive integers with atleast one of them different from zero then the greatest common divisor (GCD) of a and b is the positive integer d satisfying the following two conditions.

  1. d | a and d | b
  2. If c | a and c | b the c ≤ b then c ≤ d. Further c | d.

 

Calculations:

To find : gcd(4598, 3211) = ?

By using Euclid´s division algorithm, we have

\(4598 = 3211 ​\times 1 + 1387\)

\(3211= 1387 \times 2 + 437\)

⇒ \(1387 = 437 \times 3 + 76 \)

⇒ \(437 = 76 \times 5 + 57\)

⇒ \(76= 57 \times 1 + 19\)

⇒ \(57 = 19 \times 3 + 0\)

∴ gcd(4598, 3211) = 19

Hence, the correct answer is option 3)

Elementary Number Theory Question 4:

Find the series of all positive integers, when divided by 5 leaves the remainder 3.

  1. 1, 6, 11, 16, 21, .......
  2. 3, 8, 13, 18, 23, .......
  3. 5, 8, 11, 14, 17, .......
  4. None of these

Answer (Detailed Solution Below)

Option 2 : 3, 8, 13, 18, 23, .......

Elementary Number Theory Question 4 Detailed Solution

Concept:

Euclid's division lemma: Given positive integers a and b there exist whole numbers q and r satisfying,

a = bq + r, where, 0 ≤ r ≤ b         ----(1)

  • The remainder is always less than the divisor
  • If r = 0 then a = bq so b divides a.
  • Similarly, if b divides a then a = bq.

Calculation:

Using the division lemma, we can easily get the answer. 

To get the first number, let q = 0, r = 3 (given) and b = 5 (given)

⇒ a = 5(0) + 3         [using (1)]

⇒ a = 3

Similarly, To get the second number, let q = 1, r = 3 (given) and b = 5 (given)

⇒ a = 5(1) + 3

⇒ a = 8

Now, to get the third number, let q = 2, r = 3 (given) and b = 5 (given)

⇒ a = 5(2) + 3

⇒ a = 13

Thus, The next number will be 5 more than the preceding number.

So, the series would be 3, 8, 13, 18, 23, ...

Hence, the series of all positive integers, when divided by 5 leaves the remainder 3 is 3, 8, 13, 18, 23, ...

Elementary Number Theory Question 5:

Find x if x ≡ 1(mod 3) ;  x ≡ 2(mod 5) ;  x ≡ 3(mod 7) is the simultaneous system of linear congruence using chinese remainder theorem.

  1. 60
  2. 123
  3. 357
  4. 52

Answer (Detailed Solution Below)

Option 4 : 52

Elementary Number Theory Question 5 Detailed Solution

Concept: 

Chinese remainder theorem: Let n1, n2, ... ,nr be positive integers such that (ni, nj) = 1 for i ≠ j.Then the system of linear congruence x ≡ ai (mod ni) 1 ≤ i ≤ r, has a simultaneous solution which is unique modulo n1, n2, ... ,nr.

And the solution for the above system of linear congruence is given by x = a1N1x1 + a2N2x2 + ... + akNkxk

Calculations:

Given, x ≡ 1(mod 3) ;  x ≡ 2(mod 5) ;  x ≡ 3(mod 7)

we have n1 = 3 , n2 = 5, n3 = 7 and a1 = 1, a2 = 2, a3 = 3

And, gcd(3, 5) =  gcd(5, 7) = gcd(3, 7) = 1

Let n = n1. n2. n3 = 3 . 5 . 7 = 105

n = 105 Then, N1 = n/n1 = 105/3 = 35

N2 = n/n2 = 105/5 = 21 and N3 = n/n3 = 105/7 = 15

Now to find x1, xand x3 consider the congruences

N1x≡ 1(mod n1) ⇒ 35x1 ≡ 1(mod 3) ⇒ x1 = -1

N2x≡ 1(mod n2) ⇒ 21x2 ≡ 1(mod 5) ⇒ x2 = 1

similarly, N3x≡ 1(mod n3) ⇒15x3 ≡ 1(mod 7) ⇒ x3 = 1

So, on substituting all the values in  x = a1N1x1 + a2N2x2 + a3N3x3

x = 1 × 35 × (-1) + 2 × 21 × 1 + 3 × 15 × 1 ⇒ x = 87 - 35 x = 52

Hence, the correct answer is option 4)

Elementary Number Theory Question 6:

Which of the following statement is not among the Peano's axioms?

  1. If x is a number, the successor of x is a number.
  2. Zero is a number
  3. If P and Q are two numbers that are the same, and R and S are also the same, P + R is same as Q + S.
  4. If a set S of numbers contains zero and also the successor of every number in S, then every number is in S.

Answer (Detailed Solution Below)

Option 3 : If P and Q are two numbers that are the same, and R and S are also the same, P + R is same as Q + S.

Elementary Number Theory Question 6 Detailed Solution

Concept:

Peano's axioms

1. Zero is a number.

2. For every natural number n, the successor of n is also a natural number. We denote the successor of n by S(n).

3. Zero is not the successor of any natural number n i.e.,S(n) = 0 is false.

4. Two numbers of which the successors are equal are themselves equal i.e.,for all natural numbers m and n, S(m) = S(n) if and only if m = n.

5. If a set K of numbers contains zero and also the successor of every number in K, then every number is in K (induction axiom).

(or) ) If K is a set of natural numbers such that

  • 0 is in K, and
  • for every natural number n, if n ∈ K then S(n) ∈ K. then K contains every natural number.

But option 3) is Euclid's axiom but not Peano's axiom.

Hence, the correct answer is option 3)

Elementary Number Theory Question 7:

Euclid's division algorithm can be applied to:  

  1. All integers
  2. All integers except zero
  3. Only positive integers
  4. More than one of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 3 : Only positive integers

Elementary Number Theory Question 7 Detailed Solution

Concept:

Euclid's division lemma: If a and b are two positive integers, then there exists unique positive integers q and r such that 

 \(a=bq+r\) where \(0\leq{r}\ <\ {b}\) .

If b ∣ a then r = 0 otherwise r must satisfy stronger inequality 0 < r < b.

Explanation:

From the above discussion, we can conclude that Euclide division lemma can be applied in all positive integers

Hence, option 3 is correct.

Additional Information

  • Euclid’s division algorithm is a technique to compute the Highest Common Factor (HCF) of two given positive integers.
  • The HCF of two positive integers a and b is the largest positive integer d that divides both a and b.

 

Elementary Number Theory Question 8:

If 26! = n8k, where k and n are positive integers, then what is the maximum value of k?

  1. 6
  2. 7
  3. 8
  4. 9

Answer (Detailed Solution Below)

Option 2 : 7

Elementary Number Theory Question 8 Detailed Solution

Explanation:

\([{26\over2}]+[{26\over2^2}]+[{26\over2^3}]+[{26\over2^4}]\)

= 13 + 6 + 3 + 1 ([.] is greatest integer function)

= 23

then

26! = a223

⇒ 26! = a(23)7× 22

⇒ 26! = 4a × 87

Comparing with 26! = n8k we get maximum value of k is 7.

Hence option (2) is true.

Elementary Number Theory Question 9:

For any positive integers 'p' and 9, there exist unique integers 'a' and 'b' such that p = 9a + b, where b must satisfy:

  1. 0 ≤  b < 3
  2. 0 <  b < 9  
  3. 0 <  b < 3
  4. 0 ≤  b < 9  

Answer (Detailed Solution Below)

Option 4 : 0 ≤  b < 9  

Elementary Number Theory Question 9 Detailed Solution

Concept:

Euclid's division lemma: It states that if 'p' and 'q' are positive integers then there exists 'a' and 'b' such that

p = aq + b    where,  0  b < q

Calculation:

Here in the given problem, we have 'p', and 'q' = 9 as any positive integers, then by applying the above lemma there exists 'a' and 'b' such that

p = 9q + b where,

0 ≤ b < 9 ( As q = 9)

Hence, the correct answer is 0 ≤  b < 9.  

Elementary Number Theory Question 10:

The last three digits of 2563 × 6325 are________

  1. 225
  2. 245
  3. 275
  4. 375

Answer (Detailed Solution Below)

Option 4 : 375

Elementary Number Theory Question 10 Detailed Solution

Concept:

To find the last three digits of a number, you can simply take the number modulo 1000. This operation gives you the remainder when dividing the number by 1000, which effectively keeps only the last three digits.

Explanation:

Firstly, find 2563 mod(1000)

252 ≡ 625 (mod1000)

⇒ 2562 ≡ 62531 (mod1000)

⇒ 62531 ≡ 625 (mod1000)

⇒ 2562 ≡ 625 (mod1000)

⇒ 2563 ≡ 625 (mod1000) ....(1) 

To find 

625 x 6325 (mod1000)

Now, 625 x 63 ≡ 375 (mod1000)

625 x 632 ≡ 625 (mod1000)

625 x 633 ≡ 375 (mod1000)

........................................

⇒ 625 x 6325 ≡ 375 (mod 1000) 

⇒ 2563 x 6325 ≡ 375 (mod1000)

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