Integer notation

This topic has expert replies
Master | Next Rank: 500 Posts
Posts: 128
Joined: Thu Jul 30, 2009 1:46 pm
Thanked: 1 times

Integer notation

by fruti_yum » Thu Sep 03, 2009 3:25 pm
If 10^50 - 74 is written as an integer in base 10 notation, what is the sum of the digits in that integer?

btw... the two blocks that appear.. it's a negative sign
A. 424
B. 433
C. 440
D. 449
E. 467


How do i even approach this question?

Junior | Next Rank: 30 Posts
Posts: 19
Joined: Thu Aug 06, 2009 8:20 am
Thanked: 2 times

by Gdieterling » Thu Sep 03, 2009 3:46 pm
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 ?

Master | Next Rank: 500 Posts
Posts: 128
Joined: Thu Jul 30, 2009 1:46 pm
Thanked: 1 times

by fruti_yum » Thu Sep 03, 2009 4:25 pm
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

Master | Next Rank: 500 Posts
Posts: 175
Joined: Mon Feb 09, 2009 3:57 pm
Thanked: 4 times

by tom4lax » Thu Sep 03, 2009 6:50 pm
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?

Junior | Next Rank: 30 Posts
Posts: 19
Joined: Thu Aug 06, 2009 8:20 am
Thanked: 2 times

by Gdieterling » Fri Sep 04, 2009 1:28 am
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.

User avatar
Master | Next Rank: 500 Posts
Posts: 157
Joined: Tue Oct 07, 2008 5:47 am
Thanked: 3 times

by PussInBoots » Mon Sep 14, 2009 11:11 pm
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

https://www.beatthegmat.com/base-10-nota ... 20304.html