Let us assume the opposite, (1) \(\sqrt { 3 } \) is irrational.
Hence \(\sqrt { 3 } =\frac { p }{ q } \)
Where p and q (q ≠ 0) are co-prime (no common factor other than 1)
Hence, \(\sqrt { 3 } =\frac { p }{ q } \)
\(\sqrt { 3 } \)q = p
Squaring both side
\({ (\sqrt { 3 }q ) }^{ 2 }={ p }^{ 2 }\)
3q² = p²
\({ q }^{ 2 }=\frac { p }{ 3 } \)
Hence, 3 divides p² So 3 divides p also   .....(1)
Hence we can say
\(\frac{p}{3}\) = c where c is some integer
s, p =p²
Putting p = 3c
3q² = (3c)²
3q² = 9c²
q2 = \(\frac13\) x 9c²
q² = 3c²
\(\frac{9^2}{3}\) = c²
Hence 3 divides q²
So, 3 divides q also  ...(2)
By (1) and (2) 3 divides both p and q
By contradiction \(\sqrt { 3 } \) is irrational.

Let us start with taking a, where a is a +ve odd integer.
We apply the division algorithm with 'a' and 'b' = 4.
Since 0  ≤ r < 4, the possible remainders are 0,1,2,3.
That is, a can be 4q, or 4q + 1, or 4q + 2 or 4q + 3, where 1 is the quotient. However, since a is odd, a cannot be 4q or 4q + 2 (since they are both divisible by 2).
Any odd integer is of the form 4q + 1 or 4q + 3

98 \(\equiv \) (x + 4) (mod 5)
98 - (x + 4) = 5n , for some integer n.
94 - x  = 5n
94- x is a multiple of 5
Therefore , the least positive value of x must be 4
Since 94 - 4 = 90  is the nearest multiple of 5 less than 94.

1³ + 2³ + 3³ + ...+ 16³ = \(\left[ \frac { 16\times \left( 16+1 \right) }{ 2 } \right] ^{ 2 }\)= (136)² = 18496

First term a = 3; common difference d = 6 - 3 = 3 ; last term l  = 111
We know that, n  = \(\left( \frac { l-a }{ d } \right) +1\)
n = \(\left( \frac { 111-3 }{ 3 } \right) +1\) = 37
Thus the A.P. contain 37 terms 

First term = a = 20; common difference = d = 8
Arithmetic Progression is a, a + d , a + 2d , a + 3d,....
In this case , we get 20, 20 + 8, 20 + 2 (8), 20 + 3(8),....
So, the required A.P is 20, 28, 36, 44,.... 

3x \(\equiv \) 1 (mod 15) can be written as
3x - 1 = 15k for some integer k
3x = 15k + 1
\(x=\frac { 15k+1 }{ 3 } \)
\(x=5k+\frac { 1 }{ 3 } \)
Since 5k is an integer , 5k + \(\frac { 1 }{ 3 } \) cannot be an integer 
So there is no integer solution

We see that 6 boxes are required to pack 30 cakes with 4 cakes left over. This distribution of cakes can be understood as follows:

Let the amount received by the three children be in the form of A.P. is given by
a - d, a, a + d, Since, Sum of the amount is Rs. 207, we have
(a - d) + a + (a + d) = 207
3a = 207 gives a = 69
It is given that product of the two least amounts is 4623.
(a - d)a = 4623 
(69 - d)69 = 4623
d = 2
Therefore, amount given by the mother to her three children are
Rs. (69 - 2), Rs. 69, Rs. (69 + 2). That is Rs. 67, Rs. 69 and Rs. 71.   

Since 70000 is divisible by 7
70000 \(\equiv \) 0 (mod 7)
70000 + 4 \(\equiv \) 0 + 4 (mod 7)
70004 \(\equiv \)  4 (mod 7)
Therefore, the remainder when 70004 is divided 7 is 4 
Since 777 is divisible by 7
777 \(\equiv \) 0 (mod 7)
777 + 1 \(\equiv \) 0 + 1 (mod 7)
778 \(\equiv \) 1 (mod 7)
Therefore, the remainder when 778 is divided by 7 is 1.