I got C

10^50 has 50 zeros and 51 digits.

Subtract 74 and we have 50 digits.

50-2 = 48 digits are 9's

last two are 26

48(9) + 2 + 6 = 440

dtweah wrote:If 10^50 - 74 is an integer in base 10 notation, what is the sum of the digits in that integer?

A. 424
B. 433
C. 440
D. 449
E. 467

10^2-74 = 26 ----->sum of digits = 8
10^3 -74= 926 ----> sum of digits = 17
10^4- 74= 9926 ----> sum of digits = 26

so everytime you raise 10 by an additional power the sum of the digits in the answer is increased by 9 (ie 8 to 17 ...increased by 9 / 17 to 26 increased by 9)

let n = the power that 10 is raised to
the formula then is 8 + 9 (n -2) = sum of digits
for the first equation above: 8 + 9(2-2) = 8
for the second equation above: 8 + 9(3-2) = 17
for the third equation above: 8 + 9 (4-2)= 26

thus for the 50th power : 8 + 9 (50-2)= 440 (C)

dtweah wrote:If 10^50 - 74 is an integer in base 10 notation, what is the sum of the digits in that integer?

A. 424
B. 433
C. 440
D. 449
E. 467

10^1, 10^2, 10^3, 10^4... = 10, 100, 1000, 10000...


Any of those values minus 74 will result with 26 in the last 2 digits. Now we need to determine how many 9s there will be in front of the last 2 digits.

So after the subtraction there are n digits and the last two are 26 so...

10^3 - 74 = 926 1 nine
10^4 - 74 = 9926 2 nines
....^5.... = 99926 3 nines
....^6.... = 999926 4 nines

There are n-2 nines for each value.

n = 50 which gives us 48 nines followed by 26

48 * 9 + 2 + 6 = 440

C!