Question

The numbers which gives the remainder 1 on dividing by 3.a) Write the sequence of these numbers?b) Which is the smallest two digit term of this sequence?c) How many numebrs are there up to 25 in this sequence ?d) What are the possible remainders on dividing a number by 3?​

Answer 1

Given:

We have to divide the numbers with 3.

To find:

a) Write sequence of numbers which give remainder 1 when divided by 3.

b) Smallest two digit number in the sequence = ?

c) How many numbers are there up to 25?

d) Possible remainders when a number is divided by 3.

Solution:

a) Let us start with number 1.

When 1 is divided by 3, remainder is 1.

When 2 is divided by 3, remainder is 2.

When 3 is divided by 3, remainder is 0.

When 4 is divided by 3, remainder is 1.

When 5 is divided by 3, remainder is 2.

When 6 is divided by 3, remainder is 0.

When 7 is divided by 3, remainder is 1.

When 8 is divided by 3, remainder is 2.

When 9 is divided by 3, remainder is 0.

We can see that remainder 1 is left when 1, 4, 7 .... are divided by 3.

So, the required sequence is: {1, 4, 7, 10, 13, .....}

b) As seen from above sequence, the smallest 2 digit number is 10.

c) We can see that above sequence is an Arithmetic Progression (AP)

With first term, a = 1

Common difference, d = 3

nth term of an AP is:

$a_n=a+(n-1)d$

We are given the nth term is 25, we have to find n.

$25=1+(n-1)\times 3\\\Rightarrow 25=1+3n-3\\\Rightarrow 3n=27\\\Rightarrow n = 9$

So, up to 25, there are 8 terms and 25 is the 9th term in the sequence.

d) As explained the concept in part a), we can easily see that the possible remainders when we divide any number by 3 are: 0, 1 and 2.

After that, the remainder keep on repeating as 0, 1, 2 only.