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If 10^15 – 49 is written as an integer in base 10 notation

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If 10^15 – 49 is written as an integer in base 10 notation  [#permalink]

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New post 10 Mar 2018, 23:21
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A
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C
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Question Stats:

62% (00:43) correct 38% (01:14) wrong based on 37 sessions

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If 10^15 – 49 is written as an integer in base 10 notation, what is the product of digits in that integer?

A) 9^13
B) 4* 9^13
C) 5 * 9^13
D) 51* 9^13
E) 5 * 9^14

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If 10^15 – 49 is written as an integer in base 10 notation  [#permalink]

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New post 10 Mar 2018, 23:34
saswata4s wrote:
If 10^15 – 49 is written as an integer in base 10 notation, what is the product of digits in that integer?

A) 9^13
B) 4* 9^13
C) 5 * 9^13
D) 51* 9^13
E) 5 * 9^14


The easiest way to solve this problem is by expanding the integer
\(10^{15} – 49 = 1000000000000000 - 49 = 999999999999951\)

The product of these digits is \(9^{13} * 5 * 1\) = 5 * \(9^{13}\)(Option C)
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Re: If 10^15 – 49 is written as an integer in base 10 notation  [#permalink]

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New post 10 Mar 2018, 23:52
saswata4s wrote:
If 10^15 – 49 is written as an integer in base 10 notation, what is the product of digits in that integer?

A) 9^13
B) 4* 9^13
C) 5 * 9^13
D) 51* 9^13
E) 5 * 9^14


\(10^15\) has 1 followed by 15 zeroes. When you subtract 49 from it... it will leave with a string of 9s followed by 51. Since the number is smaller than what we started with we now have a total of 15 digits. 13 of those 15 are 9s and 5 and 1.

Hence \(5*9^13\)

This way one could solve without explicitly writing the number down. This is helpful in scaling the same for bigger numbers - for ex. if we had 10^59 to begin with.

Please give kudos if you liked my explanation...

Best,
Gladi
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Re: If 10^15 – 49 is written as an integer in base 10 notation  [#permalink]

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New post 10 Mar 2018, 23:53
saswata4s wrote:
If 10^15 – 49 is written as an integer in base 10 notation, what is the product of digits in that integer?

A) 9^13
B) 4* 9^13
C) 5 * 9^13
D) 51* 9^13
E) 5 * 9^14


10^15 has 1 followed by 15 zeroes. When you subtract 49 from it... it will leave with a string of 9s followed by 51. Since the number is smaller than what we started with we now have a total of 15 digits. 13 of those 15 are 9s and 5 and 1.

Hence 5*9^13

This way one could solve without explicitly writing the number down. This is helpful in scaling the same for bigger numbers - for ex. if we had 10^59 to begin with.

Please give kudos if you liked my explanation...

Best,
Gladi
_________________
Regards,
Gladi



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Re: If 10^15 – 49 is written as an integer in base 10 notation  [#permalink]

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New post 11 Mar 2018, 00:47
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saswata4s wrote:
If 10^15 – 49 is written as an integer in base 10 notation, what is the product of digits in that integer?

A) 9^13
B) 4* 9^13
C) 5 * 9^13
D) 51* 9^13
E) 5 * 9^14



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Re: If 10^15 – 49 is written as an integer in base 10 notation  [#permalink]

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New post 12 Mar 2018, 03:45
saswata4s wrote:
If 10^15 – 49 is written as an integer in base 10 notation, what is the product of digits in that integer?

A) 9^13
B) 4* 9^13
C) 5 * 9^13
D) 51* 9^13
E) 5 * 9^14



Let's look for patter

100 - 49 = 51
1000- 49 = 951
10000 -49 =9951
100,000 - 49 = 99951

Notice that it is 9 is always less than number of Zero by 2.

We have 15 zero's, so we get 13 of 9's........9 ^13

Also 5 *1 =5

So finally we get: 5 *9^13

Answer: C

--== Message from the GMAT Club Team ==--

THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.


If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
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Re: If 10^15 – 49 is written as an integer in base 10 notation   [#permalink] 12 Mar 2018, 03:45
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