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?
Integer notation
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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 ?
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 ?
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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 consumingGdieterling 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 ?
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?
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?
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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.
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.
- PussInBoots
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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
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