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

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\times 1\$\$ 3=3\times 1\$\$ 4=2\times 2\$\$ 5=5\times 1\$\$ 6=2\times 3\$\$ 7=7\times 1\$\$ 8=2\times 2\times 2\$\$ 9=3\times 3\$\$10=2\times 5\$So, the L.C.M. of these number is \$1\times 2\times 2\times 2\times 3\times 3\times 5\times 7=2520\$Hence, the least number divisible by all the numbers from \$1\$ to \$10\$ is \$2520\$

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: 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

Let the other number be \$'x'\$ Given:\$LCM+HCF=600 \$\$ \Rightarrow 14HCF+HCF=600 \$\$ \Rightarrow 15HCF=600 \$\$ \Rightarrow HCF=40 \$\$ \Rightarrow LCM=14\times 40=560 \$Since HCFxLCM=Product of 2 numbers=>\$40 \times560\$ = \$280\times x\$\$ \Rightarrow x=\dfrac { { 40\times 560 } }{ { 280 } } \$\$ \Rightarrow x=40\times 2 \$\$ \Rightarrow x=80 \$hence the other number is \$80\$.

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

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.\$

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

If any digit has the last digit \$10\$ that means it divisible by \$10.\$The factor of \$10 = 2 \times 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.\$

Q: Find \$HCF\$ and \$LCM\$ of \$404\$ and \$96\$ and verify that \$HCF \times LCM =\$ Product of the two given numbers.

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

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}\$

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: What is the \$HCF\$ of smallest prime number and the smallest composite number?

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 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$$ 2=2\times 1$$ 3=3\times 1$$ 4=2\times 2$$ 5=5\times 1$$ 6=2\times 3$$ 7=7\times 1$$ 8=2\times 2\times 2$$ 9=3\times 3$$10=2\times 5$So, the L.C.M. of these number is $1\times 2\times 2\times 2\times 3\times 3\times 5\times 7=2520$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 \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.

Question 3

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\times 40=560 $Since HCFxLCM=Product of 2 numbers=>$40 \times560$ = $280\times x$$ \Rightarrow x=\dfrac { { 40\times 560 } }{ { 280 } }  $$ \Rightarrow x=40\times 2 $$ \Rightarrow x=80 $hence the other number is $80$.

Question 4

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

Question 5

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&#10;that the number must be individually divisible by $3$ and $2.$ Check whether $ n^{3}-n $ is divisible by $3.$$ n^{3}-n=n\left ( n+1 \right )\left ( n-1 \right )&#10;$ When a number is divided by $3$ then by the remainder&#10;theorem, the remainder obtained is either $0$ or $1$ or $2.$$ n=3p $ or $ n=3p+1 $ or $ n=3p+2 $, where $&#10;p $ is some&#10;integer.If $ n=3p $, then the number is divisible by $3.$If $ n=3p+1 $, then $ n-1=3p+1-1=3p $. The&#10;number is divisible by $3.$If $ n=3p+2 $, then $ n+1=3p+2+1=3\left ( p+1&#10;\right ) $. The number is divisible by $3.$ So, any number in the form of $ n^{3}-n=n\left (&#10;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&#10;obtained is either $0$ or $1$ by the remainder theorem.$ n=2p $ or $ n=2p+1 $, where $&#10;p $  is some&#10;integer.If $ n=2p $, then the number is divisible by $2.$If $ n=2p+1 $ then $ n-1=2p+1-1=2p $. The number&#10;is divisible by $2.$ So, any number in the form of $ n^{3}-n=n\left (&#10;n+1 \right )\left ( n-1 \right ) $ is divisible by $2.$ Since, the given number $ n^{3}-n=n\left ( n+1&#10;\right )\left ( n-1 \right ) $ is divisible by both $3$ and $2.$ Therefore, according&#10;to the divisibility rule of 6, the given number is divisible by $6$. &#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;Hence, $ n^{3}-n=n\left ( n+1 \right )\left ( n-1&#10;\right ) $ is divisible by $6.$

Question 6

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 $10 = 2 \times 5,$ 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 7

Find $HCF$ and $LCM$ of $404$ and $96$ and verify that $HCF \times LCM =$ 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\times 2\times 101$$96=2\times 2\times 2\times 2\times 2\times 3$$\therefore$ LCM of $404$ and $96$ is $2\times 2\times 2\times 2\times 2\times 101\times 3=9696$$\therefore$ HCF of $404$ and $96$ is $2\times 2=4$Now $HCF \times LCM=4\times 9696=38784$Product of two given numbers $404 \times 96=38784$Hence $HCM \times LCM=$ product of two given numbers.

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

$\text{We need to find the HCF or GCD that is Greatest Common Divisor}$ ${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 9

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:$\text{Find the prime factorization of each of the number}$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 ,$HCF(a,b)= xy^2$

Question 10

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 $2$.