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

Question

The sum of the digits of  -64=279. What is the value of xy ?

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

Bunuel · Accepted Answer

The question should read: The sum of the digits of -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.

goutamread · Answer

As

Question is  (10^x)^y - 64  . Let say  (10^x)^y  as Number1
Say Number1 - 64 = Number2 ==>
100 - 64 = 36   and Sum of digits of Number 2 : 9*0 + (3+6) = 1*9 = 9 
1000 - 64 = 936   and Sum of digits of Number 2 : 9*1 + (3+6) = 9 + 9 = 2*9 = 18
10000 - 64 = 9936   and Sum of digits of Number 2 : 9*2 + (3+6) = 18 + 9 = 3*9= 27
100000 - 64 = 99936   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

Pansi · Answer

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   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.

rajathpanta · Answer

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

Try finding patterns. (thats the clue)

Vips0000 · Answer

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

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!

Shibs · Answer

Hi Bunuel,

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

Thanks

quote="Bunuel"]

rajathpanta

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.

Bunuel · Answer

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.

Paris75 · Answer

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

ENGRTOMBA2018 · Answer

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.

mvictor · Answer

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.

EMPOWERgmatRichC · Answer

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:  E

GMAT assassins aren't born, they're made, 
Rich

GMATBusters · Answer

The Question can be solved as follows: https://gmatclub.com/forum/download/file.php?id=56020 photo_2020-07-16_13-50-28.jpg

saubhikbhaumik · Answer

but 10^32 - 64 is not 279. How come this is the answer then?

Bunuel · Answer

We are told that   the sum of all the digits   of  10^{xy}-64  is 279.

MoeKo · Answer

My dumb quick method

= 10^xy

1........00 - 64 = 9..... 36
last two digits: 3 + 6 = 9
279 - 9 = 270

So we need 9........ to be 270.
9*30 = 270. So number of 9 is 30. Don't forget 3 & 6.

Then 9*30 + 3 + 6 = 279

Answer is E.32

Kinshook · Answer

rajathpanta

The sum of the digits of  -64=279. What is the value of xy ?

Let xy = t

The sum of the digits of  -64=279
The sum of the digits of 10^xy - 64 = 10^t - 64 = 279

10^t has t zeros and 1 at t+1th digit
The sum of digits of 10^t - 64 = 999.....t-2 times 36 = 9(t-2) + 3 + 6 = 9(t-1) = 279
t - 1 = 279/9 = 31
t = xy = 32

IMO E

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=85636 --- Harsha GMAT-Club-Forum-t7jfrp4s.png

lehung97 · Answer

I paste it in image form as I cannot do the power in this chatbox. Hope the reasoning helps https://gmatclub.com/forum/download/file.php?id=85675 ' GMAT-Club-Forum-gqywn0rs.png

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

Prep Toolkit

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