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The numbers from 1 through 100 contain 10 instances of the digit 6 in the units position and 10 instances of the digit 6 in the tens position. Therefore, replacing any digit 6 in one of these numbers with a digit 9 increases the sum by:

10(9 - 6) + 100(9 - 6) = 110 X 3 = 330.

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14y ago
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Anonymous

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3y ago

Every one's value of 6 (6,16,26...96) which are replaced by 9 (and become 9,19,...99) is increased by 3 each= 10*3= 30

Similarly every ten's value of 6(60,61...69) which are replaced by 9(and become (90,91....99) is increased by 30 each = 10*30 =300

Therefore, 30+300= 330

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Q: If all the 6 are replaced by 9 then the algebraic sum of all the numbers from 1 to 100 both inclusive varies by?
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