Find the value of .
Answer. 214
Solution. We have,
We know that
216 = 6.6.6 = 6^{3}
256 = 4.4.4.4 = 4^{4}
243 = 3.3.3.3.3 = 3^{5}
So,
= 4 × 6^{2} + 4^{3} + 2 × 3
= 4 × 36 + 64 + 6
= 144 + 70
= 214
Hence the answer is 214
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Simplify :
Answer. 2^{48}
Solution. Given,
We know that,
256 = 2.2.2.2.2.2.2.2 = 2^{8}
=
Hence the answer is 2^{48}
If and then find the value of x^{2} + y^{2}.
Answer. 98
Solution.We use the identity
So,
Using (a – b) (a + b) = a^{2}– b^{2}= 98
Hence the answer is 98.
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If then find the value of
Answer. 7
Solution.Given that :-
On rationalizing the denominator, we get
Using (a – b) (a + b) = a^{2} – b^{2}
Also,
Substituting the values of a and
We get,
= (3)^{2} = 9
Hence the correct answer is 7.
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If and then find the value of
Answer. 2.0632
Solution. Given that :
,
Using (a – b) (a + b) = a^{2} – b^{2
}
Putting the given values,
= 2.0632
Hence the answer is 2.0632.
Simplify :-
Answer. 1
Solution.
Rationalise the denominators:
Hence the answer is 1.
Express 0.6 + in the form where p and q are integers and .
Answer.
Solution. Let x = 0.6
Multiply by 10 on LHS and RHS
10x = 6
So, the from of 0.6 =
Let y =
Multiply by 10 on LHS and RHS
10y = 7.7777 …….
10y – y =
= 7.77777 ….. – 0.77777 ……
9y = 7
So the from of 0.7777 =
Let z = 0.47777…
Multiply by 10 on both side
10z = 4.7777 ….
10z – z =
9z = 4.3
So the from of 0.4777 …. =
Therefore, form of 0.6 + is,
Hence the answer is
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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 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.
For any positive integer n, prove that n^{3} – 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
.n = 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.
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.
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.
x (x + 1) (x + 2) is divisible by 2.
Since, x (x + 1) (x + 2) is divisible by 2 and 3.
x (x + 1) (x + 2) is divisible by 6.
Now the given number is:
P = n^{3} – n
P = n (n^{2} – 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
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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Integers
Find the value of .4 by (216) power -2 by 3 + 1 by (256) power -3 by 4 + 2 by (243) power -1 by 5
Simplify : (256) power -(4 power -3by 2)
If a=3+sqroot 5 by 2, then find the value of a power 2 + 1 by a power 2
Simplify :- 7sqroot 3 by sqroot10+sqroot3-2sqroot5 by sqroot6+sqroot5-3sqroot2by sqroot15+3sqroot2
Express 0.6 + 0.bar 7 +0.bar47 in the form p by q where p and q are integers and q not equal to 0.
For any positive integer n, prove that npower 3 – n is divisible by 6.
Prove that one of any three consecutive positive integers must be divisible by 3.