The natural numbers between 300 and 600 which are divisible by 7 are 301, 308, 315, …, 595.
The sum of all natural numbers between 300 and 600 is 301 + 308 + 315 +...+ 595
The terms of the above series are in A.P.
First term a = 301; common difference d = 7; Last term l = 595.
\(n=\left( \frac { l-a }{ d } \right) +1=\left( \frac { 595-301 }{ 7 } \right) +1=43\)
Since, \({ S }_{ n }=\frac { n }{ 2 } \left[ a+l \right] \), we have \({ s }_{43 }=\frac { 43 }{ 2 } \left[ 301+595 \right] \) = 19264 

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.

Since the remainders are 4, 5 respectively the required number is the HCF of the number 445 - 4 = 441, 572 - 5 = 567.
567  = 441  x 1 + 126
441 = 126 x 3 + 63 
126 = 63 x 2 + 0
Therefore HCF of 441, 567 = 63 and so the required number is 63

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

Value of the first box from bottom = 5 x 2 = 10
Value of n = 5 x 10 = 50
Value of the second box from bottom = 3 x 50 = 150
Value of m = 2 x 150 = 300
Thus, the required numbers are
m = 300, n = 50 

Let the positive integer be 'n'
So n = 88 (p) + 61, where p be an integer
n = 88 (p) + (5 x 11 + 6)
n = 8 x 11 x p + 5 x 11 + 6
n = 11 (8p + 5) + 6
Dividing both the sides by 11, we get
\(\frac{n}{11}=(8 p+5)+\frac{6}{11}\)
When the same number n is divided by 11 the remainder will be 6.