Given:
A number when divided by 780 gives remainder 38.
To find:
The reminder that would be obtained by dividing same number by 26.
Solution:
According to Euclis' division algorithm,
...(i)
Where, q is quotient and 
is the remainder.
It is given that a number when divided by 780 gives remainder 38.
Substituting 
in (i), we get
So, given number is in the form of 
, where q is an integer.
On dividing by 26, we get
Since q is an integer, therefore (30q+1) is also an integer but 
is not an integer. Here 26 is divisor and 12 is remainder.
Therefore, the required remainder is 12.
Answer:
Step-by-step explanation:
You don't need to memorize all of the square roots. Looking at each square root, find a number close to it that you know -- The square root of five is close to the square root of four, which is 2. The square root of 28 is close to the square root of 25, which is 5. 28 is a slightly bigger than 25, so the square root will be slightly greater than 5.
n, n + 2, n + 4 - three consecutive even integers
the twice the sum of the second and third: 2[(n + 2) + (n + 4)]
twelve less than six times the first: 6n - 12
The equation:
2[(n + 2) + (n + 4)] = 6n - 12
2(n + 2 + n + 4) = 6n - 12
2(2n + 6) = 6n - 12 <em>use distributive property</em>
(2)(2n) + (2)(6) = 6n - 12
4n + 12 = 6n - 12 <em>subtract 12 from both sides</em>
4n = 6n - 24 <em>subtract 6n from both sides</em>
-2n = -24 <em>divide both sides by (-2)</em>
n = 12
n + 2 = 12 + 2 = 14
n + 4 = 12 + 4 = 16
<h3>Answer: 12, 14, 16</h3>
There are going to be two solutions because the "biggest" x is x^2 but you can factor to find what they actually are.
