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Find the value of \frac{4}{\left ( 216 \right )^{\frac{-2}{3}}}+\frac{1}{\left ( 256 \right )^{\frac{-3}{4}}}+\frac{2}{\left ( 243 \right )^{\frac{-1}{5}}}.

Answer. 214
Solution. We have, \frac{4}{\left ( 216 \right )^{\frac{-2}{3}}}+\frac{1}{\left ( 256 \right )^{\frac{-3}{4}}}+\frac{2}{\left ( 243 \right )^{\frac{-1}{5}}}
We know that
216 = 6.6.6 = 63
256 = 4.4.4.4 = 44
243 = 3.3.3.3.3 = 35
So, \frac{4}{\left ( 216 \right )^{\frac{-2}{3}}}+\frac{1}{\left ( 256 \right )^{\frac{-3}{4}}}+\frac{2}{\left ( 243 \right )^{\frac{-1}{5}}}

= \frac{4}{\left ( 6^{3} \right )^{\frac{-2}{3}}}+\frac{1}{\left ( 4^{4} \right )^{\frac{-3}{4}}}+\frac{2}{\left ( 3 ^{5}\right )^{-\frac{1}{5}}}
= \frac{4}{\left ( 6 \right )^{3\times \frac{-2}{3}}}+\frac{1}{\left ( 4 \right )^{4\times \frac{-3}{4}}}+\frac{2}{\left ( 3 \right )^{5\times \frac{-1}{5}}}
\because \left ( \left ( a \right )^{m} \right )^{n}= \left ( a \right )^{mn}
= \frac{4}{6^{-2}}+\frac{1}{4^{-3}}+\frac{2}{3^{-1}}
= 4 × 62 + 43 + 2 × 3    
\because \frac{1}{\left ( a \right )^{-n}}= \left ( a \right )^{n}                                  
= 4 × 36 + 64 + 6
= 144 + 70
= 214

Hence the answer is 214

 

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Simplify : \left ( 256 \right )^{-\left ( 4^{\frac{-3}{2}} \right )}

Answer. 248
Solution.  Given, \left ( 256 \right )^{-\left ( 4^{\frac{-3}{2}} \right )}
We know that,
256 = 2.2.2.2.2.2.2.2 = 28

\left ( 256 \right )^{-\left ( 4^{\frac{-3}{2}} \right )}  = \left ( 2^{8} \right )^{\left ( -4 \right )\times \left ( -\frac{3}{2} \right )}                                   

\because \left ( \left ( a \right ) ^{m}\right )^{n}= \left ( a \right )^{mn}

=\left ( 2 \right )^{8\times \left ( -4 \right )\times \left ( -\frac{3}{2} \right )}
= 2^{8\times 4\times \frac{3}{2}}= 2^{8\times 2\times 3}= 2^{48}
Hence the answer is 248

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If  x= \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} and y= \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} then find the value of x2 + y2.

Answer.  98
Solution.We use the identity \left ( a+b \right )^{2}= a^{2}+b^{2}+2ab
So, \left ( \sqrt{a}+\sqrt{b} \right )^{2}= a+2\sqrt{ab}+b
x^{2}+y^{2}= \left ( \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \right )^{2}+\left ( \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \right )^{2}
= \frac{3+2\sqrt{6}+2}{3-2\sqrt{6}+2}+\frac{3-2\sqrt{6}+2}{3+2\sqrt{6}+2}
= \frac{5+2\sqrt{6}}{5-2\sqrt{6}}+\frac{5-2\sqrt{6}}{5+2\sqrt{6}}
= \frac{\left ( 5+2\sqrt{6} \right )^{2}+\left ( 5-2\sqrt{6} \right )^{2}}{\left ( 5-2\sqrt{6} \right ){\left ( 5+2\sqrt{6} \right )}}
 = \frac{\left ( 25+20\sqrt{6}+24 \right )+\left ( 25-20\sqrt{6} +24\right )}{25-24}

Using   (a – b) (a + b) = a2– b2= 98
Hence the answer is 98.

 

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If  a= \frac{3+\sqrt{5}}{2} then find the value of a^{2}= \frac{1}{a^{2}}

Answer. 7
Solution.Given that :- a= \frac{3+\sqrt{5}}{2}

\therefore \frac{1}{a}= \frac{2}{3+\sqrt{5}}
On rationalizing the denominator, we get

\frac{1}{a}= \frac{2\left ( 3-\sqrt{5} \right )}{\left ( 3+\sqrt{5} \right )\left ( 3-\sqrt{5} \right )}
Using   (a – b) (a + b) = a2 – b2

= \frac{6-2\sqrt{5}}{3^{2}-\sqrt{5}^{2}}                

= \frac{6-2\sqrt{5}}{9-5}

= \frac{6-2\sqrt{5}}{4}
= \frac{3-\sqrt{5}}{2}
Also, \left ( a+\frac{1}{a} \right )^{2}= a^{2}+\frac{1}{a^{2}}+2
Substituting the values of a and \frac{1}{a}
We get, \left ( \frac{3+\sqrt{5}}{2}+\frac{3-\sqrt{5}}{2} \right )^{2}= \left ( a^{2}+\frac{1}{a^{2}}+2 \right )
\therefore a^{2}+\frac{1}{a^{2}}+2= \left ( \frac{3+\sqrt{5}+3-\sqrt{5}}{2} \right )^{2}
= (3)2 = 9

\therefore a^{2}+\frac{1}{a^{2}}= 9-2= 7
Hence the correct answer is 7.

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If \sqrt{2}= 1\cdot 414 and \sqrt{3}= 1\cdot 732 then find the value of \frac{4}{3\sqrt{3}-2\sqrt{2}}+\frac{3}{3\sqrt{3}+2\sqrt{2}}

Answer.     2.0632
Solution.   Given that :
\sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732
\frac{4}{3\sqrt{3}-2\sqrt{2}}+\frac{3}{3\sqrt{3}+2\sqrt{2}}
= \frac{4\left ( 3\sqrt{3}+2\sqrt{2} \right )}{\left ( 3\sqrt{3}-2\sqrt{2} \right )\left ( 3\sqrt{3}+2\sqrt{2} \right )}+\frac{3\left ( 3\sqrt{3}-2\sqrt{2} \right )}{\left ( 3\sqrt{3}+2\sqrt{2} \right )\left (3\sqrt{3}-2\sqrt{2} \right )}
= \frac{4\left ( 3\sqrt{3}+2\sqrt{2} \right )+3\left ( 3\sqrt{3}-2\sqrt{2} \right )}{\left ( 3\sqrt{3}-2\sqrt{2} \right )\left ( 3\sqrt{3}+2\sqrt{2} \right )}
Using   (a – b) (a + b) = a2 – b2
= \frac{12\sqrt{3}+8\sqrt{2}+9\sqrt{3}-6\sqrt{2}}{\left ( 3\sqrt{3} \right )^{2}-\left ( 2\sqrt{2} \right )^{2}}
= \frac{21\sqrt{3}+2\sqrt{2}}{27-8}
= \frac{21\sqrt{3}+2\sqrt{2}}{19}
Putting the given values,= \frac{21\left ( 1\cdot 732 \right )+2\left ( 1\cdot 414 \right )}{19}
=\frac{39\cdot 2014}{19}
= 2.0632
Hence the answer is 2.0632.

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Simplify :- \frac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}

Answer. 1
Solution. \frac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}
Rationalise the denominators:
\Rightarrow \left ( \frac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}}\times \frac{\sqrt{10}-\sqrt{3}}{\sqrt{10}-\sqrt{3}} \right )-\left ( \frac{2\sqrt{5}}{\sqrt{6}+\sqrt{3}} \times \frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}}\right )-\left ( \frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2} } \times\frac{\sqrt{15}-3\sqrt{2}}{\sqrt{15}-3\sqrt{2}}\right )
\Rightarrow \frac{7\sqrt{3}\left ( \sqrt{10} -\sqrt{3}\right )}{10-3}-\frac{2\sqrt{5}\left ( \sqrt{6}-\sqrt{5} \right )}{6-5}-\frac{3\sqrt{2}\left ( \sqrt{15}-3\sqrt{2} \right )}{15-8}
\left [ \because a^{2}-b^{2} = \left ( a+b \right )\left ( a-b \right )\right ]
\Rightarrow \frac{7\sqrt{3}\left ( \sqrt{10} -\sqrt{3}\right )}{7}-\frac{2\sqrt{5}\left ( \sqrt{6}-\sqrt{5} \right )}{1}-\frac{3\sqrt{2}\left ( \sqrt{15}-3\sqrt{2} \right )}{3}
\Rightarrow \frac{7\sqrt{30}-21}{7}-\frac{2\sqrt{30}-10}{1}+\frac{3\sqrt{30}-18}{3}
\Rightarrow \frac{21\sqrt{30}-63-42\sqrt{30}+210+21\sqrt{30}-126}{21}
\Rightarrow \frac{21}{21}= 1
Hence the answer is 1.

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Express 0.6 + 0\cdot \bar{7}+0\cdot \overline{47} in the form \frac{p}{q} where p and q are integers and q\neq 0.

Answer. \frac{167}{90}
Solution. Let x = 0.6
Multiply by 10 on LHS and RHS
10x = 6
x= \frac{6}{10}

x= \frac{3}{5}
So, the \frac{p}{q} from of 0.6 = \frac{3}{5}
Let y = 0\cdot \bar{7}
Multiply by 10 on LHS and RHS

10y = 7.7777 …….
10y – y = 7\cdot \bar{7}-0\cdot \bar{7}

          = 7.77777 ….. – 0.77777 ……
9y = 7

y= \frac{7}{9}
So the \frac{p}{q} from of 0.7777 = \frac{7}{9}
Let z = 0.47777…
Multiply by 10 on both side
10z = 4.7777 ….
10z – z = 4\cdot \bar{7}-0\cdot4\bar{7}

9z = 4.3

z\approx \frac{4\cdot 3}{9}
z= \frac{43}{90}
So the \frac{p}{q}from of 0.4777 …. = \frac{43}{90}
Therefore, \frac{p}{q} form of 0.6 + 0\cdot \bar{7}+0\cdot4\bar{7} is,

x+y+z= \frac{3}{5}+\frac{7}{9}+\frac{43}{90}
= \frac{\left ( 54+70+43 \right )}{90}
= \frac{167}{90}
Hence the answer is \frac{167}{90}

 

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Show that one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5, where n is any positive integer.

By Euclid’s division –
Any positive integer can be written as:
n = bm + r
Here b = 5
r is remainder when we divide n by 5, therefore:
0 \leq r < 5, r = 0, 1, 2, 3.4
n = 5m + r, therefore n can have values:
n = 5m, 5m + 1, 5m + 2, 5m + 3, 5m + 4.
Here m is natural number
Case 1 : Let n is divisible by 5, it means n can be written as :
n = 5m,
Now, n +4 = 5m + 4;it gives remainder 4, when divided by 5
Now, n + 8 = 5m + 8= 5(m +1) +3; it gives remainder 3, when divided by 5
Now, n + 12 = 5m + 12= 5(m +2) +2; it gives remainder 2, when divided by 5
Now, n + 16 = 5m + 16= 5(m +3) +1; it gives remainder 1, when divided by 5

Case 2: Let n + 4 is divisible by 5, it means n + 4 can be written as :
n + 4 = 5m,
Now, n = 5m – 4 = 5(m – 1) + 1 ; it gives remainder 1, when divided by 5
Now, n + 8 = 5m + 4; it gives remainder 4, when divided by 5
Now, n + 12 = 5m + 8 = 5(m +1) +3; it gives remainder 3, when divided by 5
Now, n + 16 = 5m + 12 = 5(m +2) +2; it gives remainder 2, when divided by 5
Similarly, we can show for other cases.

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For any positive integer n, prove that n3 – n is divisible by 6.

Let three consecutive positive integers be, x, x + 1 and x + 2.
Divisibility by 3: Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2
.\thereforen = 3p or 3p + 1 or 3p + 2, where p is some integer.
So, we can say that one of the numbers among x, x + 1 and x + 2 is always divisible by 3.
\Rightarrow x (x + 1) (x + 2) is divisible by 3.

Divisibility by 2: Whenever a number is divided 2, the remainder obtained is 0 or 1.
\therefore  n = 2q or 2q + 1, where q is some integer.
So, we can say that one of the numbers among x, x + 1 and x + 2 is always divisible by 2.\Rightarrow
x (x + 1) (x + 2) is divisible by 2.
Since, x (x + 1) (x + 2) is divisible by 2 and 3.
\thereforex (x + 1) (x + 2) is divisible by 6.
Now the given number is:
P = n3 – n
P = n (n2 – 1)
P = n (n – 1) (n + 1)
P = (n – 1) n (n + 1)
Therefore, P is the product of three consecutive numbers.
Now P can be written as:
P = x (x + 1) (x + 2) which is is divisible by 6.
Hence P is divisible by 6.
Hence proved

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Prove that one of any three consecutive positive integers must be divisible by 3.

Let three consecutive integers are n, n + 1, n + 2 when n is a natural number
  i.e., n = 1, 2, 3, 4, ….
for n = 1,         chosen numbers :(1, 2, 3) and 3 is divisible by 3   
for n = 2,         chosen numbers :(2, 3,4) and 3 is divisible by 3    
for n = 3,         chosen numbers :( 3, 4, 5) and 3 is divisible by 3  
for n = 4,         chosen numbers :(4, 5, 6) and 6 is divisible by 3
for n = 5,         (chosen numbers :(5, 6, 7) and 6 is divisible by 3   
Proof:

Case 1 : Let n is divisible by 3, it means n can be written as :
n = 3m,
Now, n +1 = 3m + 1 ; it gives remainder 1, when divided by 3
Now, n +2 = 3m + 2; it gives remainder 2, when divided by 3

Case 2: Let n + 1 is divisible by 3, it means n +1 can be written as :
n +1 = 3m,
Now, n = 3m – 1= 3(m – 1) + 2 ; it gives remainder 2, when divided by 3
Now, n +2 = 3m + 1; it gives remainder 1, when divided by 3

Case 3: Let n + 2 is divisible by 3, it means n +2 can be written as :
n +2 = 3m,
Now, n = 3m – 2= 3(m – 1) + 1 ; it gives remainder 1, when divided by 3
Now, n +1 = 3m – 1=3(m – 1) + 2  ; it gives remainder 2, when divided by 3

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