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SOLUTION:
\(10^{10}\) less any two digit numbers = \(99999999xx\)

We know that the sum of the digits of x equal 84.

Since the first 8 digits of x are 9 -> \(8*9= 72\)

84 (Sum required) - 72 = 12

The sum of the digits \(xx\) must be =12.

Numbers that have the sum of their digits = 12 are:
\(Z: [39;93;48;84;57;75;66]\)

Note that Z must be two-digits integers.

Thus if we want 93 as the last digits of x, we have to do \(100-7\) (where 7 is a one digit integer)
try with 84 (second biggest numbers) -> \(100-16\) (where 16 is a two digit integer).

Thus our new set is:
\(Z: [39;48;84;57;75;66]\)

6 Possibilities.


ANSWER D
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giannilorenz
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Shouldn't the question clarify that z is a positive two-digit integer? Otherwise there would be at least 6 more possibilities for z, right?
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aditee02
x = 10^10 - z, where z is a two-digit integer. If the sum of the digits of x equals 84, how many values for z are possible?

A. 3
B. 4
C. 5
D. 6
E. 7
I was confused for sometime on this one, but I hope this helps anyone who's confused.

its 99999999xx
so 9*8=72 remaining 12

so xx must sum up to 12
so the numbers for xx are 39,93,48,84, 57,75, 66

WE ARE GOING REVERSE

so 93 cant be a number as it will give us z=07 which isn't a two digit number
Rest all will give us Z as a two digit number!!

hence answer is 6
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