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

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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As

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

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

Try finding patterns. (thats the clue)

Well, simple reason is that the question is incorrect.



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 Bunuel,

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

Thanks


quote="Bunuel"]

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]

\(10^{xy}-64\) 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.
_________________

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
_________________

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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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.

Hi All,

This question has some awkward wording to it, but here's its intent: the number 10^(xy) - 64 has digits that add up to 279.

So, we need to figure out what THAT number is.

Here's what you need to "see" to solve this problem:

IF xy = 3, then 1,000 - 64 = 936 and the sum of digits = 18 = (2)(9)
If xy = 4, then 10,000 - 64 = 9,936 and the sum of digits = 27 = (3)(9)
If xy = 5, then 100,000 - 64 = 99,936 and the sum of digits = 36 = (4)(9)

Notice the pattern? The sum of digits is increasing by 9 every time.

So, how many times does 9 divide into 279? 31 times

As a reminder of the pattern:
xy = 3 --> (2)(9)
xy = 4 --> (3)(9)
xy = 5 --> (4)(9)

So, (31)(9) --> xy = 32

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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The Question can be solved as follows:



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