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# If x is a positive integer and 10^x – 74 in decimal notation

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Director
Joined: 22 Nov 2007
Posts: 807
If x is a positive integer and 10^x – 74 in decimal notation  [#permalink]

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09 Mar 2008, 01:09
4
26
00:00

Difficulty:

45% (medium)

Question Stats:

66% (02:12) correct 34% (02:13) wrong based on 349 sessions

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If x is a positive integer and 10^x – 74 in decimal notation has digits whose sum is 440, what is the value of x?

A. 40
B. 44
C. 45
D. 46
E. 50
Math Expert
Joined: 02 Sep 2009
Posts: 64174
Re: If x is a positive integer and 10^x – 74 in decimal notation  [#permalink]

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06 Apr 2014, 08:43
2
7
marcodonzelli wrote:
If x is a positive integer and 10^x – 74 in decimal notation has digits whose sum is 440, what is the value of x?

A. 40
B. 44
C. 45
D. 46
E. 50

10^x is a (x+1)-digit number: 1 followed by x zeros.

10^x - 74 is a x-digit number: x-2 9's and 26 in the end. Thus the sum of the digits is (x-2)*9+2+6=440 --> x = 50.

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Hope it helps.
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Manager
Joined: 17 Aug 2009
Posts: 125

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18 Dec 2009, 10:49
5
1
This might be a simpler method -

We know that since 10^x contains x number of zeros, therefore -
the last 0 will yield 10-4 = 6
the 2nd last zero will yield the number 9-7 = 2

Also, the rest will all be 9s. So all we need to do is calculate the number of 9s

let x be the number of 9s

9m + 6 + 2 = 440
m=48
Now we need to add 2 to the above number for the zeros for the last two digits that contributed to the subtraction
Therefore x= 48 + 2 = 50

Hope this helps
##### General Discussion
Director
Joined: 01 May 2007
Posts: 625
Re: exponents - decimal notation  [#permalink]

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09 Mar 2008, 06:59
2
I think its E 50....My reasoning is as such:

Assuming its (10^x) - 74

I plugged in for x a 2 and 3

I saw that no matter what you put in x you get 10000alot of zeros - 74...Which always left you with a answer that had a 6 in the units and a 2 in the tens slot, and a bunch of 9s in the other slots...So the equation really is...

What number will give you 440? So it becomes 9(z) + 2 + 6 = 440....So how many 9s do I need in my answer. I solved for z and got 48 and then added 2 to it for the 2 and 6. OA please?
SVP
Joined: 29 Aug 2007
Posts: 1734
Re: exponents - decimal notation  [#permalink]

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10 Mar 2008, 07:55
2
marcodonzelli wrote:
If x is a positive integer and 10^x – 74 in decimal notation has digits whose sum
is 440, what is the value of x?

40
44
45
46
50

Please, provide clear explanations. thanks a lot

10^x – 74 = k....................k26
sum of k+.....+k+2+6 = 440
sum of k+.....+k = 440 - 8
sum of k+.....+k = 432

all k must be of integer 9.
so no of k = 432/9 = 48

so x has to be 50 (=48+2) to have k........k26 = 9..........926 = 10^x - 74.
so it is E.
SVP
Joined: 29 Aug 2007
Posts: 1734

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15 Nov 2008, 22:28
fresinha12 wrote:
this one just stumped me..

if the sum of all digits of (10^x)-74 equals 440, what is the value of x?

40
44
45
46
50

10^x ends 0 in unit digit, tens digit, hundreds digit and son on. If so, then, deducting 74 from 10^x, results in 6 in unit digit, 2 in tens digit and 9 in rest of the digits. so:

440 = 2+6+(9x)
x = (440-8)/9
x = 432/9=48

lets see tha pattren:
10^2 - 74 = 100-74 = 26
10^3 - 74 = 1,000-74=926
10^4 - 74 = 10,000-74 = 9926
10^5 - 74 = 100,000-74 = 99926

so in 10^x, x has to be 50.
Manager
Joined: 29 Nov 2009
Posts: 95
Location: United States

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06 Dec 2009, 13:20
Right, once you realize that 9 x 48 is 430, you know that A-D can't work.
Intern
Joined: 18 Dec 2009
Posts: 8

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18 Dec 2009, 19:12
the last two digits will be 2&6. so
9(x-2)=400-(2+6)
=>x=50
Intern
Joined: 05 Mar 2014
Posts: 10
Schools: Ross '18

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06 Apr 2014, 07:07

We have (10^x)-74, where the sum of the digits is 440.

100-74 = 26, so we already have 2+6 = 8 from the 440.

Because we have a power of 10, we will have numbers like 100, 1000, 10000 and so on. This minus 74 rests 26 and a lot of 9s. E.g: 1000-74 = 9926.

So dividing the 432/9 = 48, that`s the number of 9s. and we have 2 other numbers (2 and 6) wich were 00 before the subtraction.

So we have 48 + 2 as an X = 50
Manager
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Posts: 124
Location: India
Concentration: Operations, Finance
GMAT Date: 05-10-2015
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Re: If x is a positive integer and 10^x – 74 in decimal notation  [#permalink]

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06 Apr 2014, 21:29
marcodonzelli wrote:
If x is a positive integer and 10^x – 74 in decimal notation has digits whose sum is 440, what is the value of x?

A. 40
B. 44
C. 45
D. 46
E. 50

Reason: Rightmost 2 digits of pattern 10^x - 74 will be 26 (100 - 74 = 26, 1000 - 74 = 926...so on). One observation, the number of digits of the result of the pattern is equal to x (26 when x = 2, 926 when x = 3...so on). Now apart from last 2 digits 2 and 6, other digits in the result will be 9. We need to find n, where n*9 + (2+6) = 440 (given), which gives n = 48. And final answer is 48 + 2 (number of right hand 2 digits 2 and 6) = 50.
Manager
Joined: 10 Jun 2015
Posts: 110
Re: If x is a positive integer and 10^x – 74 in decimal notation  [#permalink]

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14 Aug 2015, 21:47
marcodonzelli wrote:
If x is a positive integer and 10^x – 74 in decimal notation has digits whose sum is 440, what is the value of x?

A. 40
B. 44
C. 45
D. 46
E. 50

10^2 - 74=26
440-8(sum of 2 and 6)=432
432/9=48
so we need 10^48*10^2=10^50
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Re: If x is a positive integer and 10^x – 74 in decimal notation  [#permalink]

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05 Nov 2016, 19:43

We know that we have some multiple of 10 that we're subtracting 74 from, so outside of 2+6 (left over from 10^n) we'll be dealing with 9's. The question is how many, because that will determine the value of n.

2+6=8 --> subtract from 440 --> we're left with 432

432/9=48

Now add 2 because we have two slots occupied by 2 and 6.

Thus, E is correct.
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Re: If x is a positive integer and 10^x – 74 in decimal notation  [#permalink]

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12 Jul 2019, 15:29
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Re: If x is a positive integer and 10^x – 74 in decimal notation   [#permalink] 12 Jul 2019, 15:29