Question

2. The numbers from 101 to 150 are written as, 101102103104105.146147148149150.

What will be the remainder when this total number is divided by 3?​

Answer 1

We know that a number is divisible by 3 if the sum of its digits is a multiple of 3.

if the sum is a multiple of 3 + 1 (like 3*n + 1) the remainder is 1

if the sum is a multiple of 3 + 2 (like 3*n + 2) the remainder is 2.

Then we need to find the sum of all the single digits in the number:

101102103104...149150

the sum of all the single digits is:

( 1 + 1) + (1 + 2) + (1 + 3) + ... + (1 + 4 + 9) + ( 1 + 5)

The method i used is:

(from 101 to 110)

first we sum the hundreds digit 10 times, and then we add the unit digits for each case, this is:

1*10 + (1 + 2+ 3 + 4+ 5+ 6+ 7+ 8 + 9 + 1) = 10 + 46

From (111 to 119)

first we add the sum of the hundreds and tens digits 10 times, and then the units:

2*9 + ( 1 + 2+ 3 + 4+ 5+ 6+ 7+ 8 + 9 )

now if we also add the number 120 (so we have the range 111 to 120)

the sum becomes:

2*9 + ( 1 + 2+ 3 + 4+ 5+ 6+ 7+ 8 + 9 ) + 1 + 2

2*10 + (1 + 2+ 3 + 4+ 5+ 6+ 7+ 8 + 9 + 1 ) = 20 + 46

You already can see the pattern here.

in the range from (121 to 130) we will have the sum equal to 30 + 46

in the range from (131 to 140) we will have the sum equal to 40 + 46

in the range from (141 to 150) we will have the sum equal to 50 + 46

Then the total sum is:

(10 + 20 + 30 + 40 + 50) + 5*46 = 380

We know that 380 is multiple of 3 (because 3 + 8+ 0 = 12)

Then the remainder is 0.