Q: Explain why 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 are composite numbers.

Given 7 \times 11\times 13 +13 =13 \times (7 \times 11 +1)=3 \times 78 This number is multiple of two integers.Hence it has more than two factors.Hence it is a composite number.similarly in 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 +5 = 5 ( 7 \times 6 \times 4 \times 3 \times 2 \times 1 +1 )=5 \times 1009 This number is multiple of two integers.Hence it has more than two factors.Hence it is a composite number.

Q: If two positive integers a and b are written as a=x^2y^2 and b=xy^2 ; x , y are prime numbers then the HCF(a,\,b) is

HCF by using the prime factorization method.Step1: \text{Find the prime factorization of each of the number} Step2: \text{The product of all common prime factor is the HCF} Here, \Rightarrow a=x\times x\times y\times y \Rightarrow b=x\times y\times y common Factors are x\times y\times y Thus , HCF(a,b)= xy^2

Q: If two positive integers p and q can be expressed as p = ab^2 and q = a^3b ; a, b being prime numbers, then LCM (p, q) is

Given, p=a{ b }^{ 2 }=a\times b\times b q=a^{ 3 }{ b }=a\times a\times a\times b L.C.M of p,q is a^{ 3 }{ b }^{ 2 }

Q: A merchant has 120 litres of oil of one kind, 180 litres of another kind and 240 litres of third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity. What should be the greatest capacity of such a tin?

\text{We need to find the HCF or GCD that is Greatest Common Divisor} {120 = 2\times2\times2 \times3\times5} {180 = 2\times2\times3\times3 \times5} {240 = 2\times2\times2\times2\times3\times5} {GCD = 2\times2\times3\times5= 60} \text{The greatest capacity = 60 liters} \text{So the merchant needs to fill 60 liters of all types of oils } \text{Hence option C is correct}

Question 1

The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is?

Accepted Answer

The least number divisible by all the number from  1  to  10  will be the L.C.M of following number, 1=1  2=2	imes 1  3=3	imes 1  4=2	imes 2  5=5	imes 1  6=2	imes 3  7=7	imes 1  8=2	imes 2	imes 2  9=3	imes 3 10=2	imes 5 So, the L.C.M. of these number is  1	imes 2	imes 2	imes 2	imes 3	imes 3	imes 5	imes 7=2520 Hence, the least number divisible by all the numbers from  1  to  10  is  2520

Question 2

Explain why 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 are composite numbers.

Accepted Answer

Given   7 	imes 11	imes 13 +13 =13 	imes (7 	imes 11 +1)=3 	imes 78  This number is multiple of two integers.Hence it has more than two factors.Hence it is a composite number.similarly in 7 	imes 6   	imes 5   	imes 4    	imes 3     	imes 2   	imes 1 +5     = 5 ( 7 	imes 6   	imes 4    	imes 3     	imes 2   	imes 1 +1 )=5 	imes 1009    This number is multiple of two integers.Hence it has more than two factors.Hence it is a composite number.

Question 3

Find  HCF  and  LCM  of  404  and  96  and verify that  HCF 	imes LCM =  Product of the two given numbers.

Accepted Answer

Let  a = 404  and   b = 96 Expressing the number as a product of prime number 404=2	imes 2	imes 101 96=2	imes 2	imes 2	imes 2	imes 2	imes 3 	herefore  LCM of  404  and  96  is  2	imes 2	imes 2	imes 2	imes 2	imes 101	imes 3=9696 	herefore  HCF of  404  and  96  is  2	imes 2=4 Now  HCF 	imes LCM=4	imes 9696=38784 Product of two given numbers  404 	imes 96=38784 Hence  HCM 	imes LCM=  product of two given numbers.

Question 4

The LCM of two numbers is 14 times their HCF. The sum of LCM and HCF is 600. If one number is 280, then the other number is

Accepted Answer

Let the other number be  'x'  Given: LCM+HCF=600   \Rightarrow 14HCF+HCF=600   \Rightarrow 15HCF=600   \Rightarrow HCF=40   \Rightarrow LCM=14	imes 40=560  Since HCFxLCM=Product of 2 numbers=> 40 	imes560  =  280	imes x  \Rightarrow x=\dfrac { { 40	imes 560 } }{ { 280 } }    \Rightarrow x=40	imes 2   \Rightarrow x=80  hence the other number is  80 .

Question 5

Check whether  6^n  can end with the digit  0  for any natural number  n .

Accepted Answer

If any digit has the last digit  10  that means it divisible by  10. The factor of  10 = 2 	imes 5,  So value of  6^n  should be divisible by  2  and  5. Both  6^n  is divisible by  2  but not divisible by  5. So, it can not end with  0.

Question 6

If two positive integers  a  and  b  are written as  a=x^2y^2  and  b=xy^2 ;  x ,  y  are prime numbers then the  HCF(a,\,b)  is

Accepted Answer

HCF by using the prime factorization method.Step1: 	ext{Find the prime factorization of each of the number} Step2: 	ext{The product of all common prime factor  is the HCF} Here, \Rightarrow a=x	imes x	imes y	imes y \Rightarrow b=x	imes y	imes y common Factors are  x	imes y	imes y Thus , HCF(a,b)= xy^2

Question 7

If two positive integers  p  and  q  can be expressed as  p = ab^2  and  q = a^3b ;  a, b  being prime numbers, then LCM  (p, q)  is

Accepted Answer

Given, p=a{ b }^{ 2 }=a	imes b	imes b q=a^{ 3 }{ b }=a	imes a	imes a	imes b L.C.M of  p,q   is  a^{ 3 }{ b }^{ 2 }

Question 8

A merchant has  120  litres of oil of one kind,  180  litres  of another kind and  240  litres of third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity. What should be the greatest capacity of such a tin?

Accepted Answer

ext{We need to find the HCF or GCD that is Greatest Common Divisor}   {120 = 2	imes2	imes2 	imes3	imes5} {180 = 2	imes2	imes3	imes3 	imes5} {240 = 2	imes2	imes2	imes2	imes3	imes5} {GCD = 2	imes2	imes3	imes5= 60} 	ext{The greatest capacity = 60 liters} 	ext{So the merchant needs to fill 60 liters of all types of oils } 	ext{Hence option C is correct}

Question 9

What is the  HCF  of smallest prime number and the smallest composite number?

Accepted Answer

Smallest prime number is  2  and smallest composite number is  4 . Clearly,  2  is a factor of  4 , so their H.C.F. is  2 . Thus, the H.C.F. of the smallest prime number and the smallest composite number is  2 .

Question 10

For any Positive integer  n , prove that  {n}^{3}-n  divisible by  6 .

Accepted Answer

The condition for any number to be divisible by  6  is
that the number must be individually divisible by  3  and  2.  Check whether   n^{3}-n   is divisible by  3.  n^{3}-n=n\left ( n+1 \right )\left ( n-1 \right )
  When a number is divided by  3  then by the remainder
theorem, the remainder obtained is either  0  or  1  or  2.  n=3p   or   n=3p+1   or   n=3p+2  , where  
p   is some
integer.If   n=3p  , then the number is divisible by  3. If   n=3p+1  , then   n-1=3p+1-1=3p  . The
number is divisible by  3. If   n=3p+2  , then   n+1=3p+2+1=3\left ( p+1
\right )  . The number is divisible by  3.  So, any number in the form of   n^{3}-n=n\left (
n+1 \right )\left ( n-1 \right )   is divisible by  3.  Check whether   n^{3}-n   is divisible by  2. When a number is divided by  2 , the remainder
obtained is either  0  or  1  by the remainder theorem.  n=2p   or   n=2p+1  , where  
p    is some
integer.If   n=2p  , then the number is divisible by  2. If   n=2p+1   then   n-1=2p+1-1=2p  . The number
is divisible by  2.  So, any number in the form of   n^{3}-n=n\left (
n+1 \right )\left ( n-1 \right )   is divisible by  2.  Since, the given number   n^{3}-n=n\left ( n+1
\right )\left ( n-1 \right )   is divisible by both  3  and  2.  Therefore, according
to the divisibility rule of 6, the given number is divisible by  6 .

Hence,   n^{3}-n=n\left ( n+1 \right )\left ( n-1
\right )   is divisible by  6.