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The sum of the digits of 10^x−1 is equal to 3^8. What is the value of

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The sum of the digits of 10^x−1 is equal to 3^8. What is the value of  [#permalink]

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New post 10 Oct 2017, 21:32
1
17
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A
B
C
D
E

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  75% (hard)

Question Stats:

57% (02:15) correct 43% (02:12) wrong based on 219 sessions

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Re: The sum of the digits of 10^x−1 is equal to 3^8. What is the value of  [#permalink]

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New post 20 Oct 2017, 20:36
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1
Bunuel wrote:
The sum of the digits of 10^x−1 is equal to 3^8. What is the value of x?

A. 18
B. 243
C. 729
D. 2187
E. 6561



\(10^x-1\) will always lead to a number with digits as 9s...
so \(10^x-1 = 99999...x\) times
sum of digits =\(9+9+9+...x\) times = \(x*9\)

so \(x*9=3^8=3^6*9.......x=3^6=729\)

C
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Re: The sum of the digits of 10^x−1 is equal to 3^8. What is the value of  [#permalink]

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New post 10 Oct 2017, 22:20
1
lets generalize this sum,
consider x=1,
10^1 - 1= 9 ------9--- 3^2
x=2,
10^2 - 1=99 ----18---2*3^2
x=3,
10^3 - 1=999----27---3*3^2
x=4,
10^4 - 1=9999---36---4*3^2

similarly,
10^x-1= 999...x times= x*9---- x*3^2
It is given that,
x*3^2 = 3^8
hence, x=3^6
x=729

Answer is option C.

Kudos if it helps.
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Re: The sum of the digits of 10^x−1 is equal to 3^8. What is the value of  [#permalink]

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New post 20 Oct 2017, 12:18
Could you please clarify how to calculate this in a bit more comprehensive way?
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Re: The sum of the digits of 10^x−1 is equal to 3^8. What is the value of  [#permalink]

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New post 20 Oct 2017, 17:10
krikre wrote:
Could you please clarify how to calculate this in a bit more comprehensive way?



10^x - 1 = 3^8

It is worth noting when starting this problem that 10-1 = 9, 10^2 - 1 = 99, etc. So whatever X equals, is how many 9s that number will have.

Also:
3^8 = (3^2)^4 = 9^4

So to find out what X is we need to determine how many 9's are in 9^4

Well, (9) (9^1) has only one 9
(9)*9 (9^2) has nine, 9s
(9)*9*9 (9^3) has eighty-one 9s
and (9)*9*9*9 (9^4) has 729 9s.

Thus 10^729 - 1 will have the same amount of 9's as 9^4
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Re: The sum of the digits of 10^x−1 is equal to 3^8. What is the value of  [#permalink]

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New post 22 Jun 2018, 10:13
Here's my 2 cents !!

Let's test the condition here


10^2 - 1 = 99 so, sum = 9+9 = 2*9
10^3 - 1 = 999 sum = 3*9
10^4 -1 = 9999 sum = 4*9
.
From here we know that
10^x - 1 , sum = x*9 = 9x

we can equate as it is given the question

9x = 3^8
9x = 9^4
x= 9^3 = 729

Hence C is the answer.
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The sum of the digits of 10^x−1 is equal to 3^8. What is the value of  [#permalink]

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New post 22 Jun 2018, 11:33
First notice that 10^x will be a number that ends in 0. Once you minus the number by 1, you will get a number that has all 9's.

For example: 100-1 = 99 or 1000-1 =999

Next 3^8 also equals = 3^2 * 3^2 *3^2 *3^2 = 9*9*9*9 = 81*81 = 6561.

now you want to determine how many 9's are in 6561. This can be determined by dividing 6561/9.

6561/9 = 729

Answer C
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The sum of the digits of 10^x−1 is equal to 3^8. What is the value of &nbs [#permalink] 22 Jun 2018, 11:33
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