### 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,...}

Labels: CAT number theory, CAT remainders, CAT Solutions, Remainders

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