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If 1 0^{50} - 74 is written as an integer in base 10 notation, what is the sum of the digits in that integer?

Explanation	Calculations	Help
$We subtract 74 from 100 by borrowing across zeros, so the last two digits come from 100 - 74 and each preceding zero becomes 9.$	$100 - 74 = 26$
$We have 48 nines followed by digits 2 and 6, so we sum these digits.$	$Sum of digits = 48 \times 9 + 2 + 6$
$We calculate the contribution of the nines.$	$48 \times 9 = 48 im es (10 - 1) = 480 - 48 = 432$
$We add the remaining digits to get the total.$	$432 + 2 + 6 = 440$

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440

If 1 0^{50} - 74 is written as an integer in base 10 notation, what is the sum of the digits in that integer?

Explanation	Calculations	Help
$We subtract 74 from 100 by borrowing across zeros, so the last two digits come from 100 - 74 and each preceding zero becomes 9.$	$100 - 74 = 26$
$We have 48 nines followed by digits 2 and 6, so we sum these digits.$	$Sum of digits = 48 \times 9 + 2 + 6$
$We calculate the contribution of the nines.$	$48 \times 9 = 48 im es (10 - 1) = 480 - 48 = 432$
$We add the remaining digits to get the total.$	$432 + 2 + 6 = 440$

Scroll horizontally to view all columns

440