Question

how many 3 digit numbers are there. which leaves remainder 1 on division by 5​

Answer 1

Answer:

smallest 3 digit no. which leaves remainder 1 on division by 5 is =100

Largest 3 digit no. which leaves remainder 1 on division by 5 is =997

Now, we can observe that it is forming an AP with

first element = 100

last element =997

common difference =3

let an=a+(n-1)d

so,

997 = 100 + (n-1)3

997 - 100 = 3n -3

897 + 3 = 3n

900 = 3n

900/3 = n

n = 300

Therefore, there are 300, 3-digit number which leaves remainder 1 on division by 5.

Answer 2

Answer:

largest 3-digit number which leave 1 as a remainder when divided by 5 is 996

smallest 3-digit number which leave 1 as a remainder when divided by 5 is 101

common difference is 5

tn=a+(n-1)d

996=101+(n-1)5

996=101+5n-5

996=96+5n

5n=996-96

5n=900

n=180

Total 180, 3-digit numbers leave 1 as a remainder when divided by 5