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

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GMAT Forum Moderator
Joined: 28 May 2014
Posts: 475
GMAT 1: 730 Q49 V41
If 10^15 – 49 is written as an integer in base 10 notation [#permalink]

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10 Mar 2018, 23:21
00:00

Difficulty:

45% (medium)

Question Stats:

64% (00:43) correct 36% (01:20) wrong based on 36 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
[Reveal] Spoiler: OA

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Joined: 26 Feb 2016
Posts: 2247
Location: India
GPA: 3.12
If 10^15 – 49 is written as an integer in base 10 notation [#permalink]

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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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Joined: 16 Sep 2016
Posts: 67
WE: Analyst (Health Care)
Re: If 10^15 – 49 is written as an integer in base 10 notation [#permalink]

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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,
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Manager
Joined: 16 Sep 2016
Posts: 67
WE: Analyst (Health Care)
Re: If 10^15 – 49 is written as an integer in base 10 notation [#permalink]

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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,
_________________

Kudos will encourage many others, like me.

Give me a Kudos! if you found the post helpful...

Math Expert
Joined: 02 Sep 2009
Posts: 44282
Re: If 10^15 – 49 is written as an integer in base 10 notation [#permalink]

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11 Mar 2018, 00:47
1
KUDOS
Expert's post
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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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

Re: If 10^15 – 49 is written as an integer in base 10 notation   [#permalink] 12 Mar 2018, 03:45
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