A large no. N =1234567891011121314…99899100. What will be the remainder when 1st 100 digits of N is divided by 9?  Choose the correct choice from the following answers:	A) 2	B) 3	C) 5	D) 7	E) 8
Hence the first 100 digits of N goes like this 1234567891011121314.......5253545   We need to find the sum of the above number to check for divisibility by 9.   From 1 to 50, each of the digits 1 to 4 occurs 14 times, digit 5 occurs 6 times and each of 6 to 9 occurs 5 times   Sum of digits from 1 to 4 = 10   Sum of digits from 6 to 9 = 30   Therefore, sum of digits from 1 to 50 = 10 * 15 + 5 * 6 + 30 * 5= 
<gadget id="calculator">10 * 15 + 5 * 6 + 30 * 5</gadget>
<output>330</output>
330   Sum of digits from 51 to 54 plus a 5= 5 * 4 + ( 1 + 2 + 3 + 4) + 5 = 20 + 15= 
<gadget id="calculator">20 + 15</gadget>
<output>35</output>
35   Total sum= 330 + 35= 
<gadget id="calculator">330 + 35</gadget>
<output>365</output>
365   When 365 is divided by 9, we get 5 as the remainder!   5orrect Answer: 5   5
<result>C</result>
A large no. N =1234567891011121314…99899100. What will be the remainder when 1st 100 digits of N is divided by 9?