Can anyone help me with this problem please? I know it should be easy but I haven't done this in a while.
Find all numbers b, 1950<b<2000, which are congruent to a modulo m where a=1776 and m=40.
This means that when you divide your number by 40, it will have the same remainder as a.
So 1776 divided by 40 = 44 with a remainder of 16
If you think about it for a little bit, you will realize that if you add 40 to 1776, then divide that answer by 50, it will be 45 with a remainder of 16. And if you add 40 again and divide, you will have 46 with a remainder of 16.
So you should be able to see that if you have a number, b, where $\displaystyle b\equiv a~ (mod ~m)$ then $\displaystyle b = a+nm$ for some integer n. (ie ...-2, -1, 0, 1, 2...)
So we could take a few different approaches to finding values between 1950 and 2000 which are congruent to a. Possible ideas are to find a number that 40 will divide (this is any number which will have a remainder of zero when divided by 40), then adding 16 to that in order to get the proper remainder. Then adding or subtracting 40 to find other numbers within the range you have been given.
Or, you could keep adding 40 to 1776 until you find you are in the appropriate range.
If I were doing it, I'd probably say that we need to add about 200, to get in our range, then 40*5=200, Then using the equation $\displaystyle b=a+nm$ where n=5, we add 200 to 1776 and get 1976. This is in our range, and when we divide it, it should have a remainder of 16. We check to see if it is true, and it is, because it equals 49 with a remainder of 16.
Then we check to see if there are any other values, add 40, and we get 2016. This is out of the range, so subtract 40 instead, and get 1936, this is out of the range as well, so 1976 is the only number b, which is between 1950 and 2000, that is congruent to 1976 mod 40.