Number Theory...7^100

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Number Theory...7^100

by sakali » Wed Sep 08, 2010 1:44 pm
Define A = the sum of digits of the number 7^100 and B = sum of digits of A. What is the sum of digits of B?

A) 7
B) 16
C) 10
D) 11
E) Cannot be Determined

OA - A

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by TimeforGMAT » Wed Sep 08, 2010 2:44 pm
sakali wrote:Define A = the sum of digits of the number 7^100 and B = sum of digits of A. What is the sum of digits of B?

A) 7
B) 16
C) 10
D) 11
E) Cannot be Determined

OA - A
Good Q. I am not sure if there is a easier way to do this but this is how I came up with A
Number Sum of Digits Sum(Sum of Digits)
7^1=7 7 7
7^2=49 13 4
7^3=343 10 10
7^4 = 2401 7 7
7^5=16807 22 4
7^6 = 117649 28 10

Similarly proceeding, 7^100 will have 7 as Sum(sum of Digits)

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by sanju09 » Wed Sep 08, 2010 9:44 pm
sakali wrote:Define A = the sum of digits of the number 7^100 and B = sum of digits of A. What is the sum of digits of B?

A) 7
B) 16
C) 10
D) 11
E) Cannot be Determined

OA - A
Extremely ambiguous though brilliant question seen after a long time.

There is difference in the two phrases, sum of digits and ultimate sum of digits. Like, sum of digits in 189 is 18, but the ultimate sum of digits in 189 is 18 taken as 1 + 8 again so as to bring the sum down to a single digit, 9. With that the choices B through E become irrelevant. We need to look for a single digit, so B, C, D are out; good labor can make things possible some day, so E is out. If one of the choices is guaranteed as correct, then I would pick [spoiler]A[/spoiler] without minding anything anymore.

This could be a GMAT question if tailored well.
The mind is everything. What you think you become. -Lord Buddha



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by TimeforGMAT » Thu Sep 09, 2010 4:31 am
sanju09 wrote:
sakali wrote:Define A = the sum of digits of the number 7^100 and B = sum of digits of A. What is the sum of digits of B?

A) 7
B) 16
C) 10
D) 11
E) Cannot be Determined

OA - A
Extremely ambiguous though brilliant question seen after a long time.

There is difference in the two phrases, sum of digits and ultimate sum of digits. Like, sum of digits in 189 is 18, but the ultimate sum of digits in 189 is 18 taken as 1 + 8 again so as to bring the sum down to a single digit, 9. With that the choices B through E become irrelevant. We need to look for a single digit, so B, C, D are out; good labor can make things possible some day, so E is out. If one of the choices is guaranteed as correct, then I would pick [spoiler]A[/spoiler] without minding anything anymore.

This could be a GMAT question if tailored well.
Great explanation. But question didn't ask for ultimate sum of digits - only for first couple of iterations. So, lets say, 7^50's first sum of digits is 89, second sum would lead to 17 and ultimate would lead to 8. So, there is a probability in this case that answer is not a single digit. Right?

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by sanju09 » Thu Sep 09, 2010 5:09 am
TimeforGMAT wrote:
sanju09 wrote:
sakali wrote:Define A = the sum of digits of the number 7^100 and B = sum of digits of A. What is the sum of digits of B?

A) 7
B) 16
C) 10
D) 11
E) Cannot be Determined

OA - A
Extremely ambiguous though brilliant question seen after a long time.

There is difference in the two phrases, sum of digits and ultimate sum of digits. Like, sum of digits in 189 is 18, but the ultimate sum of digits in 189 is 18 taken as 1 + 8 again so as to bring the sum down to a single digit, 9. With that the choices B through E become irrelevant. We need to look for a single digit, so B, C, D are out; good labor can make things possible some day, so E is out. If one of the choices is guaranteed as correct, then I would pick [spoiler]A[/spoiler] without minding anything anymore.

This could be a GMAT question if tailored well.
Great explanation. But question didn't ask for ultimate sum of digits - only for first couple of iterations. So, lets say, 7^50's first sum of digits is 89, second sum would lead to 17 and ultimate would lead to 8. So, there is a probability in this case that answer is not a single digit. Right?
Can you imagine all the digits involved in 7^100, it must turn out to be a huge list of digits when unfolded. Had it not been to the ultimate sum of digits, no option was correct in this post.
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

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Posts: 11
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by TimeforGMAT » Thu Sep 09, 2010 9:37 am
sanju09 wrote:
TimeforGMAT wrote:
sanju09 wrote:
sakali wrote:Define A = the sum of digits of the number 7^100 and B = sum of digits of A. What is the sum of digits of B?

A) 7
B) 16
C) 10
D) 11
E) Cannot be Determined

OA - A
Extremely ambiguous though brilliant question seen after a long time.

There is difference in the two phrases, sum of digits and ultimate sum of digits. Like, sum of digits in 189 is 18, but the ultimate sum of digits in 189 is 18 taken as 1 + 8 again so as to bring the sum down to a single digit, 9. With that the choices B through E become irrelevant. We need to look for a single digit, so B, C, D are out; good labor can make things possible some day, so E is out. If one of the choices is guaranteed as correct, then I would pick [spoiler]A[/spoiler] without minding anything anymore.

This could be a GMAT question if tailored well.
Great explanation. But question didn't ask for ultimate sum of digits - only for first couple of iterations. So, lets say, 7^50's first sum of digits is 89, second sum would lead to 17 and ultimate would lead to 8. So, there is a probability in this case that answer is not a single digit. Right?
Can you imagine all the digits involved in 7^100, it must turn out to be a huge list of digits when unfolded. Had it not been to the ultimate sum of digits, no option was correct in this post.
After carefully reading the question again, I have noticed that answer is looking for the third iteration of sum of digits and not the second. 7^100 will have less than 100 digits in it and assuming it has 99 digits and all of them are 9's (to get the highest number), sum of digits will be 99*9 (891). The number that is less than 891 and that yields highest sum of digits is 889 for which the sum is 25. So, second iteration can have a maximum count of 25. Third and final iteration, sum of numbers of any number less than or equal to 25 will be in between 1 -10. Option A or D.

Am I wrong in any of my analysis above.

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by sanju09 » Fri Sep 10, 2010 12:52 am
TimeforGMAT wrote:
sanju09 wrote:
TimeforGMAT wrote:
sanju09 wrote:
sakali wrote:Define A = the sum of digits of the number 7^100 and B = sum of digits of A. What is the sum of digits of B?

A) 7
B) 16
C) 10
D) 11
E) Cannot be Determined

OA - A
Extremely ambiguous though brilliant question seen after a long time.

There is difference in the two phrases, sum of digits and ultimate sum of digits. Like, sum of digits in 189 is 18, but the ultimate sum of digits in 189 is 18 taken as 1 + 8 again so as to bring the sum down to a single digit, 9. With that the choices B through E become irrelevant. We need to look for a single digit, so B, C, D are out; good labor can make things possible some day, so E is out. If one of the choices is guaranteed as correct, then I would pick [spoiler]A[/spoiler] without minding anything anymore.

This could be a GMAT question if tailored well.
Great explanation. But question didn't ask for ultimate sum of digits - only for first couple of iterations. So, lets say, 7^50's first sum of digits is 89, second sum would lead to 17 and ultimate would lead to 8. So, there is a probability in this case that answer is not a single digit. Right?
Can you imagine all the digits involved in 7^100, it must turn out to be a huge list of digits when unfolded. Had it not been to the ultimate sum of digits, no option was correct in this post.
After carefully reading the question again, I have noticed that answer is looking for the third iteration of sum of digits and not the second. 7^100 will have less than 100 digits in it and assuming it has 99 digits and all of them are 9's (to get the highest number), sum of digits will be 99*9 (891). The number that is less than 891 and that yields highest sum of digits is 889 for which the sum is 25. So, second iteration can have a maximum count of 25. Third and final iteration, sum of numbers of any number less than or equal to 25 will be in between 1 -10. Option A or D.

Am I wrong in any of my analysis above.
I think I should quit now
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com