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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]

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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]

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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]

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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]

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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]-64=79. What is the value [#permalink]

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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]

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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]

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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]

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

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New post 08 Jan 2014, 03:19
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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]

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New post 30 Jan 2016, 06:00
rajathpanta wrote:
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


Sum of the digits of (10^x)^y - 64 = 279.

We know that (X^a)^b = X^(ab)

Thus, the given form becomes = 10^(xy) - 64.

Now start with xy=2, 10^2 - 64 = 36, sum of the digits = 6+3=9 (realize that the sum of digits is 9*(xy-1))
xy=3 ---> 1000-64=936 = 18 (realize that the sum of digits is 9*(xy-1))
xy=4 ---> 10000-64=9936 = 27 (realize that the sum of digits is 9*(xy-1))
xy=5 ---> 100000-64 = 99936 (realize that the sum of digits is 9*(xy-1))... etc so this is your pattern. The sum of the digits = 9*(xy-1)

Now sum of the digits given = 279.

Thus, based on the pattern above, the sum of the digits must be ---> 9*(xy-1) = 279 ---> xy = 32.

E is thus the correct answer.

Hope this helps.

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

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New post 24 Feb 2016, 17:32
gosh..the structure of the question is hideous...
i solved it this way..
the sum is 279. the last 2 digits must be 3 and 6, or 9. so the rest will be a bunch of 9's. how many 9's? 30.
now, 30 of nines + 36 -> 32 digits +1 since we need to round up - 33 in total, we thus must have 10^32.

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

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Re: The sum of the digits of [(10^x)^y]-64=279. What is the   [#permalink] 03 Oct 2017, 17:36
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