 # Number Theory Remainders – 2 more questions

Have given below two more questions on remainders.
Question 1:
A prime number p greater than 100 leaves a remainder q on division by 28. How many values can q take?.
Question 2:
How many positive integers are there from 0 to 1000 that leave a remainder of 3 on division by 7 and a remainder of 2 on division by 4?
Correct Answers:
Question 1: 12 different values
Question 2: 36 positive numbers
Explanatory Answers:
A prime number p greater than 100 leaves a remainder q on division by 28. How many values can q take?
q can be 1.
If q =2, number would be of the form 28n + 2 which is a multiple of 2
Similarly, when q =4, number would be of the form 28n + 4 which is again a multiple of 2. Any number of the form 28n + an even number will be a multiple of 2
When q =7, number would be of the form 28n + 7 which is a multiple of 7
So, the only remainders possible are remainders that share no factors with 28. Or numbers that are co-prime to 28.
There is a formula for this and a shorter way of finding the number of numbers co-prime to a given natural number. A more detailed discussion on this is provided here and here.
1,3,5,9,11,13,15,17,19,23,25 and 27. q can take 12 different values.
Qn2:
How many positive integers are there from 0 to 1000 that leave a remainder of 3 on division by 7 and a remainder of 2 on division by 4?
Number should be of the form 7n + 3 and 4m +2
The LCM of 7 and 4 is 28. So, let us see what are the possible remainders when we divide this number by 28
A number of the form 7n + 3 can be written as 28K + 3 or 28k + 10 or 28+ 17 or 28k +24.
A number of the form 4m + 2 can be written as 28l +2, 28l + 6, 28l + 10, 28l + 14, 28l + 18, 28l + 22, 28l + 26,
For a detailed discussion on how we get to this, look at this post.
Within these, the only common term is 28K + 10.
The numbers in this sequence are 10, 38, 66…..990.
We still need to figure out how many numbers are there in this sequence. We are going in steps of 28, so let us see if we can write these numbers in terms of 28p + r
10 = 28 * 0 + 10
38 = 28 * 1 + 10
66 = 28 * 2 + 10
94 = 28 * 3 + 10
…..
990 = 28 * 35 + 10
There are 36 numbers in this sequence.
Another way of looking at this question is to spot the number of common terms in two APs
{3,10,17,24 ,… } and {2,6,10,14,…}