Correct Answer - Option 4 : 330
Concept used:
Sum of all numbers from 1 to n = n(n+1)/2
Calculation:
Sum of all numbers from 1 to 100 = 100(100 + 1)/2
⇒ Sum = 50 × 51
⇒ Sum = 5050
Number of 6’s comes in units place = 10
Number of 6’s comes in tens place = 10
Sum of these 6's = 10(6 × 10) + 10(6 × 1)
⇒ Required sum = 600 + 60
⇒ Required sum = 660
If all 6’s replaced by 9’s
Sum of those 9's = 10(9 × 10) + 10(9 × 1)
⇒ Sum of those 9’s = 900 + 90
⇒ Sum of those 9’s = 990
Algebraic sum = 5050 – 660 + 990
⇒ Sum = 5380
Required difference = 5380 – 5050
⇒ Required difference = 330
∴ The algebraic sum of all the numbers from 1 to 100 varies by 330.
