IMO : D

10^50 = is a number which has 51 digits.

Thus, when you substract 74 you get : 9 followed by 49 "9's" and the number ends with xx99926.

It asks you to add all the digits of that number :

9 * 49 + 2 + 6 = 449

Am I right ?

Gdieterling wrote:IMO : D

10^50 = is a number which has 51 digits.

Thus, when you substract 74 you get : 9 followed by 49 "9's" and the number ends with xx99926.

It asks you to add all the digits of that number :

9 * 49 + 2 + 6 = 449

Am I right ?

yes OA is D...how do u quickly find out that there are 49 9's?? I had to do an analogy between 10^5 and then I could determine.. it was too time consuming

How do you get to 49 9's I am also curious?

As an example, 10^5 is 100,000 - 74 = 99,926 or 3 9's. The way I looked at it was take the number of zeros then subtract by 2, so 10^50 = 48 9's.
However answer is with 49 9's. What gives?

10^5 = 100 000 --> 10 followed with 4 zeros
When substracting one : 9 followed by 4 9's (99999)

10^10 = 10 000 000 000 --> 10 followed with 9 zeros
When susbtrasting one : 9 followed by 9 9's (9 999 999 999)

10^50 = 10 followed with 49 zeros
When substracting 1 : 9 followed by 49 9's

Then, by susbtracting 76 to 10^50 : 9 followed by 48 9's (which is 49 9's) :

49 * 9 = 441 + 2 + 6 = 449


Besides, I just thought of an interesting way to guess :
You know that the answer will be a multiple of 9 to which we add : 8.

We know that 9 has a dual cyclicality :

9^1 : 9
9^2 : 81
9^3 : 729
9^4 : XXX1

Thus, the solution we are looking for ends either with 7 (9 + 8) or with 9 (1 + 8), reducing the solutions to 449 or 467.

467 - 8 = 459 which is 51 * 9 and obviously impossible when 10^50 was containing 51 digits.

10^50 has fifty zeros.

Subtract 1 and we get fifty nines.
Subtract 74 and we get 48 nines, 2 and 6.

48 * 9 + 2 + 6 = 360 + 72 + 8 = 440

Answer is C

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