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If N = 1 + 11 + 111 + 1111 + …… + 111111111, then what is the sum of the digits of N?
- A45
- B18
- C36
- D5
Answer not set
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0
CCommunity
The last term contains 9 digits. The total number of times 1 has to be added to unit place is 9. So, there will be no carry over while adding these terms.
⇒ Sum of all the digits in N = Total numbers of times digit '1’ is used = 1 + 2 + 3 + ... + 9
? Sum of first n natural numbers = {n × (n + 1)}/2
⇒ Sum of all the digits in N = (9 × 10)/2 = 45
∴ The sum of the digits of N is 45.
⇒ Sum of all the digits in N = Total numbers of times digit '1’ is used = 1 + 2 + 3 + ... + 9
? Sum of first n natural numbers = {n × (n + 1)}/2
⇒ Sum of all the digits in N = (9 × 10)/2 = 45
∴ The sum of the digits of N is 45.
0
CCommunity
The last term contains 9 digits. The total number of times 1 has to be added to unit place is 9. So, there will be no carry over while adding these terms.
⇒ Sum of all the digits in N = Total numbers of times digit '1’ is used = 1 + 2 + 3 + ... + 9
? Sum of first n natural numbers = {n × (n + 1)}/2
⇒ Sum of all the digits in N = (9 × 10)/2 = 45
∴ The sum of the digits of N is 45.
⇒ Sum of all the digits in N = Total numbers of times digit '1’ is used = 1 + 2 + 3 + ... + 9
? Sum of first n natural numbers = {n × (n + 1)}/2
⇒ Sum of all the digits in N = (9 × 10)/2 = 45
∴ The sum of the digits of N is 45.