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The sum of the digits of [(10^x)^y]-64=279. What is the

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The sum of the digits of [(10^x)^y]-64=279. What is the [#permalink] New post 28 Oct 2012, 09:18
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The sum of the digits of [(10^x)^y]-64=279. What is the value of xy ?

A. 28
B. 29
C. 30
D. 31
E. 32
[Reveal] Spoiler: OA

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Last edited by Bunuel on 29 Oct 2012, 00:59, edited 2 times in total.
Renamed the topic and edited the question.
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Re: The Sum of the digits of(10^x)^y [#permalink] New post 28 Oct 2012, 10:13
I spent some time on this question, got stuck and could not move towards a solution - Should not the sum of the digits of the number [(10^x)^y - 64] be a multiple of 9. Please clarify if the question formed the way it is now is the best way. I think I have misinterpreted something here.
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Re: The Sum of the digits of(10^x)^y [#permalink] New post 28 Oct 2012, 10:22
Pansi wrote:
I spent some time on this question, got stuck and could not move towards a solution - Should not the sum of the digits of the number [(10^x)^y - 64] be a multiple of 9. Please clarify if the question formed the way it is now is the best way. I think I have misinterpreted something here.


Well, It is a multiple of 9. How will you arrive at xy with that approach?

Try finding patterns. (thats the clue)
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Re: The Sum of the digits of(10^x)^y [#permalink] New post 28 Oct 2012, 18:44
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Pansi wrote:
I spent some time on this question, got stuck and could not move towards a solution - Should not the sum of the digits of the number [(10^x)^y - 64] be a multiple of 9. Please clarify if the question formed the way it is now is the best way. I think I have misinterpreted something here.


Well, simple reason is that the question is incorrect.

rajathpanta wrote:
Well, It is a multiple of 9. How will you arrive at xy with that approach?

Try finding patterns. (thats the clue)


Question is:
10^xy -64 = N,
where sum of digits of N=79

The pattern is like this:

100 -64 = 36
1000 -64 = 936
10000 -64 =9936

or,
1 followed by (n times 0) = (n-2)times 9 followed by 36

Therefore sumof digits on right side is always a multiple of 9 [9s and 6+3 =9]

However in question stem RHS is 79, which is not divisible by 9. And therefore you can not arrive at any of the answer choices given.

Rajathpanta- on a lighter note - if this too is from Aristotle, I'd suggest please change the source of questions. :D

Hope it helps!
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Re: The Sum of the digits of(10^x)^y [#permalink] New post 28 Oct 2012, 19:40
still confuse with question
any more explanation
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Re: The Sum of the digits of(10^x)^y [#permalink] New post 28 Oct 2012, 19:42
Vips0000 wrote:
Pansi wrote:
I spent some time on this question, got stuck and could not move towards a solution - Should not the sum of the digits of the number [(10^x)^y - 64] be a multiple of 9. Please clarify if the question formed the way it is now is the best way. I think I have misinterpreted something here.


Well, simple reason is that the question is incorrect.

rajathpanta wrote:
Well, It is a multiple of 9. How will you arrive at xy with that approach?

Try finding patterns. (thats the clue)


Question is:
10^xy -64 = N,
where sum of digits of N=79

The pattern is like this:

100 -64 = 36
1000 -64 = 936
10000 -64 =9936

or,
1 followed by (n times 0) = (n-2)times 9 followed by 36

Therefore sumof digits on right side is always a multiple of 9 [9s and 6+3 =9]

However in question stem RHS is 79, which is not divisible by 9. And therefore you can not arrive at any of the answer choices given.

Rajathpanta- on a lighter note - if this too is from Aristotle, I'd suggest please change the source of questions. :D

Hope it helps!



Hi Vips00,

This is from the veritas prep questions set!

Thanks.
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Re: The Sum of the digits of(10^x)^y [#permalink] New post 28 Oct 2012, 20:47
rajathpanta wrote:
Vips0000 wrote:
Pansi wrote:
I spent some time on this question, got stuck and could not move towards a solution - Should not the sum of the digits of the number [(10^x)^y - 64] be a multiple of 9. Please clarify if the question formed the way it is now is the best way. I think I have misinterpreted something here.


Well, simple reason is that the question is incorrect.

rajathpanta wrote:
Well, It is a multiple of 9. How will you arrive at xy with that approach?

Try finding patterns. (thats the clue)


Question is:
10^xy -64 = N,
where sum of digits of N=79

The pattern is like this:

100 -64 = 36
1000 -64 = 936
10000 -64 =9936

or,
1 followed by (n times 0) = (n-2)times 9 followed by 36

Therefore sumof digits on right side is always a multiple of 9 [9s and 6+3 =9]

However in question stem RHS is 79, which is not divisible by 9. And therefore you can not arrive at any of the answer choices given.

Rajathpanta- on a lighter note - if this too is from Aristotle, I'd suggest please change the source of questions. :D

Hope it helps!



Hi Vips00,

This is from the veritas prep questions set!

Thanks.


Hmmm, but you get that the question is incorrect and why?
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Re: The Sum of the digits of(10^x)^y [#permalink] New post 28 Oct 2012, 20:49
Aristocrat wrote:
still confuse with question
any more explanation

What is confusing? I already explained it in detail. If there is any particular thing you could not understand let me know, would try to explain further.
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Re: The sum of the digits of [(10^x)^y]-64=79. What is the value [#permalink] New post 29 Oct 2012, 00:59
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rajathpanta wrote:
The sum of the digits of [(10^x)^y]-64=79. What is the value of xy

A. 28
B. 29
C. 30
D. 31
E. 32


The question should read:
The sum of the digits of [(10^x)^y]-64=279. What is the value of xy

A. 28
B. 29
C. 30
D. 31
E. 32

Also, it should be mentioned that xy is a positive integers.

First of all (10^x)^y=10^{xy}.

10^{xy} has xy+1 digits: 1 and xy zeros. For example: 10^2=100 --> 3 digits: 1 and 2 zeros;

10^{xy}-64 will have xy digits: xy-2 9's and 36 in the and. For example: 10^4-49=10,000-49=9,951 --> 4 digits: 4-2=two 9's and 51 in the end;

We are told that the sum of all the digits of 10^{xy}-64 is 279 --> 9(xy-2)+3+6=279 --> 9(xy-2)=270 --> xy=32.

Answer: E.

Similar questions to practice:
the-sum-of-all-the-digits-of-the-positive-integer-q-is-equal-126388.html
10-25-560-is-divisible-by-all-of-the-following-except-126300.html
if-10-50-74-is-written-as-an-integer-in-base-10-notation-51062.html

Hope it's clear.
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Re: The sum of the digits of [(10^x)^y]-64=279. What is the [#permalink] New post 29 Oct 2012, 01:26
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Question is (10^x)^y - 64 . Let say (10^x)^y as Number1
Say Number1 - 64 = Number2 ==>
100 - 64 = 36 [ Number1: No. of zeroes = 2 , Number2: No. of 9's = zero ] and Sum of digits of Number 2 : 9*0 + (3+6) = 1*9 = 9
1000 - 64 = 936 [ Number1: No. of zeroes = 3 , Number2: No. of 9's = 1] and Sum of digits of Number 2 : 9*1 + (3+6) = 9 + 9 = 2*9 = 18
10000 - 64 = 9936 [ Number1: No. of zeroes = 4 , Number2: No. of 9's = 2] and Sum of digits of Number 2 : 9*2 + (3+6) = 18 + 9 = 3*9= 27
100000 - 64 = 99936 [ Number1: No. of zeroes = 5 , Number2: No. of 9's = 3] and Sum of digits of Number 2 : 9*3 + (3+6) = 27 + 9 =4*9= 36


so lets go from right to left for the sum of digits of number2 i.e given as 279
so 279 = 31*9 = 9*30 + (3+6) => Number2: Number of 9's = 30 ==> Number1: Number of zeros = 32

So the Number1 i.e. (10^x)^y = 10000.....(32 zeroes)

Now, as we now, 10^1 = 10 (1 zero)
10^2 = 100 (2 zeroes)
10^3 = 1000 (3 zeroes)

same way, 10000.....(32 zeroes) = 10^32

(10^x)^y = 10^(xy) = 10^32
==> xy = 32
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Re: The sum of the digits of [(10^x)^y]-64=79. What is the value [#permalink] New post 24 Nov 2013, 18:36
Hi Bunuel,

Could you please xplain the last bit oft he equations which takes us to a 279?

Thanks


quote="Bunuel"]
rajathpanta wrote:
The sum of the digits of [(10^x)^y]-64=79. What is the value of xy

A. 28
B. 29
C. 30
D. 31
E. 32


The question should read:
The sum of the digits of [(10^x)^y]-64=279. What is the value of xy

A. 28
B. 29
C. 30
D. 31
E. 32

Also, it should be mentioned that xy is a positive integers.

First of all (10^x)^y=10^{xy}.

10^{xy} has xy+1 digits: 1 and xy zeros. For example: 10^2=100 --> 3 digits: 1 and 2 zeros;

10^{xy}-64 will have xy digits: xy-2 9's and 36 in the and. For example: 10^4-49=10,000-49=9,951 --> 4 digits: 4-2=two 9's and 51 in the end;

We are told that the sum of all the digits of 10^{xy}-64 is 279 --> 9(xy-2)+3+6=279 --> 9(xy-2)=270 --> xy=32.

Answer: E.

Similar questions to practice:
the-sum-of-all-the-digits-of-the-positive-integer-q-is-equal-126388.html
10-25-560-is-divisible-by-all-of-the-following-except-126300.html
if-10-50-74-is-written-as-an-integer-in-base-10-notation-51062.html

Hope it's clear.[/quote]
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Re: The sum of the digits of [(10^x)^y]-64=79. What is the value [#permalink] New post 25 Nov 2013, 01:52
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Shibs wrote:
Hi Bunuel,

Could you please xplain the last bit oft he equations which takes us to a 279?

Thanks


quote="Bunuel"]
rajathpanta wrote:
The sum of the digits of [(10^x)^y]-64=79. What is the value of xy

A. 28
B. 29
C. 30
D. 31
E. 32


The question should read:
The sum of the digits of [(10^x)^y]-64=279. What is the value of xy

A. 28
B. 29
C. 30
D. 31
E. 32

Also, it should be mentioned that xy is a positive integers.

First of all (10^x)^y=10^{xy}.

10^{xy} has xy+1 digits: 1 and xy zeros. For example: 10^2=100 --> 3 digits: 1 and 2 zeros;

10^{xy}-64 will have xy digits: xy-2 9's and 36 in the and. For example: 10^4-49=10,000-49=9,951 --> 4 digits: 4-2=two 9's and 51 in the end;

We are told that the sum of all the digits of 10^{xy}-64 is 279 --> 9(xy-2)+3+6=279 --> 9(xy-2)=270 --> xy=32.

Answer: E.

Similar questions to practice:
the-sum-of-all-the-digits-of-the-positive-integer-q-is-equal-126388.html
10-25-560-is-divisible-by-all-of-the-following-except-126300.html
if-10-50-74-is-written-as-an-integer-in-base-10-notation-51062.html

Hope it's clear.


10^{xy}-64 will have xy digits: xy-2 9's and 36 in the and. Threfore the sum of the digits is 9(xy-2)+3+6=279.

Hope it's clear.
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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: The sum of the digits of [(10^x)^y]-64=279. What is the [#permalink] New post 08 Jan 2014, 03:19
Hi,

this is my process (edited to be the most efficient possible):

1000-64= 936. Whatever XY is you finish with 36 ==> 3+6=9

Therefore, 279-9=270 and 270/9=30

Now you add the last two digits (3 and 6)

Answer is 30+2=32

Hope it helps
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Re: The sum of the digits of [(10^x)^y]-64=279. What is the   [#permalink] 08 Jan 2014, 03:19
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