x = 10^10 - z, where z is a two-digit integer. If the sum of the digit

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This is a very good question.
x = 10^10 - z

Now what is 10^2 = 100 ; 1 followed by 2 zeros
10^3 = 1000 ; 1 followed by 3 zeros
Similarly 10^10 is 1 followed by 10 zeros

Now you are subtracting a 2 digit number from this. Values that z can take is any number between 10 to 99.
Now let us try to understand step by step the range of values x can take.

when you subtract 100-10 you get 90 & when you subtract 100-99 you get 1
when you subtract 1000-10 you get 990 & when you subtract 1000-99 you get 901
when you subtract 10000-10 you get 9990 & when you subtract 10000-99 you get 9901 i.e numbers are between 9901 and 9990

So you can see a pattern here:
When you subtract 10 from a number which is '1 followed by 10 zeros' you will get 9999999990 i.e 8 '9's followed by 90
& When you subtract 99 from a number which is '1 followed by 10 zeros' you will get 9999999901 i.e. 8 '9's followed by 01

So now we know range of values x can take i.e 9999999901 to 9999999990
The first 8 digit gives you a sum of 8*9 = 72. You need 12 more to make it 84. So focus on last 2 digits of 99999999|01 to 99999999|90 i.e 01 to 90.
How many numbers are there between 01 and 90 that gives you a sum of 12?

Start writing it down.
39.
48
57
66
75
84

93 is out of the range. So don't consider that.

So one value of x = 9999999939. How did you get this? By subtracting 61 from 10^10. So z = 61.
Likewise there are 6 possible values of x : 9999999939, 9999999948, 9999999957, 9999999966, 9999999975, 9999999984
As there are 6 possible values of x which gives you a sum of 84, there are 6 possible values of z as well.

Hope you understood.
Consider giving Kudos if you understood.

Cheers
Ambarish
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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

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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