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
xy is positive integer
Ans-E
A. 28
B. 29
C. 30
D. 31
E. 32
xy is positive integer
Ans-E
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Sorry, I was confused for a second because I didn't see a definition of n.theCodeToGMAT wrote:[(10^x)^y]-64=279
To find: xy
10^xy - 64 = 279
Let xy = 2 ==> 36 ==> Sum = 9
Let xy = 3 ==> 936 ==> Sum = 18
Let xy = 4 ==> 9936 ==> SUm = 27
So, using formula for AP
9 + (n-1)9 = 279
9n = 279
n = 31
So, 31+1 32
[spoiler]{E}[/spoiler]
No, I just used the formula for AP term==> a + (n-1)dkackerarnav wrote:Shouldn't that term in bold above be n-2, by the pattern?theCodeToGMAT wrote:[(10^x)^y]-64=279
To find: xy
10^xy - 64 = 279
Let xy = 2 ==> 36 ==> Sum = 9
Let xy = 3 ==> 936 ==> Sum = 18
Let xy = 4 ==> 9936 ==> SUm = 27
So, using formula for AP
9 + (n-1)9 = 279
9n = 279
n = 31
So, 31+1 32
[spoiler]{E}[/spoiler]
Edit[email protected] wrote:Hi Mathsbuddy,
You made a conceptual math error in your work. Remember that we have to add up the digits in the calculation. When you change the calculation (by turning - 64 into "-100 +36"), you're changing the digits. Unless you do the necessary steps to "undo" those changes later, you're going to end up with the wrong power of 10.
GMAT assassins aren't born, they're made,
Rich
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