Question

(When two numbers seprately divisible by 33 leaves
21 and 28 as remainder. Find the remainder if the
sum of those two number is divided by 33.)​

Answer 1

Answer:

Step-by-step explanation:

First number= 33a + 21  (as remainder is 21)

Second number= 33b+28 (as remainder is 28)

Adding the first and second numbers,

33a+21+33b+28

33a + 33b + 49

upon taking the commons,

33(a+b) + 49

Let's now neglect the value of a and b.....

We know that 33 is obviously divisible by 33 and so it's product with any integer is also divisible. So, 33(a+b) is also divisible by 33

To find remainder,

49/33

Quotient= 1 and remainder= 49-33=16

Therefore, remainder is 16.

I hope this helps you