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.
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Rich
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