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

Question

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

askhere · Accepted Answer

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

Bunuel · Answer

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Ravixxx · Answer

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:

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:

6 Possibilities.

ANSWER D

giannilorenz · Answer

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?

ADisHere · Answer

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

Kinshook · Answer

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?

Let z = 10a+b; where a is a single digit between 1 & 9 and b is a single digits between 1 & 9

x = 10000000000 - 10a - b
x = 99999999{9-a}{10-b}

Sum of digits of x = 9*8+9-a+10-b = 91-a-b = 84
a+b = 7
1. a=1; b=6; z=16
2. a=2; b=5; z=25
3. a=3; b=4; z=34
4. a=4; b=3; z=43
5. a=5; b=2; z=52
6. a=6; b=1; z=61

IMO D

Ulfat · Answer

10000000000-Z
=> {10000000000-99, 10000000000-98, ..., 10000000000-10}
=> {9999999901, or 9999999902, ..., or 9999999990}
=> {9x8+01, 9x8+02, ...9x8+90}
=> 72+01, 72+02 ..., 72+90

For sum of all digits to be 84, the sum of last 2 digits should be 84-72 = 12, as the sum of first 8 digits is 72.

12 can be achieved by adding the digits of 93 or 39 or 84 or 48 or 75 or 57 or 66.
Therefore Z could be 07, 61, 16, 52, 25, 43, 34.
But, since X is two-digit number, so Z can't be 07.
Thus there are only 6 possible values of z.

Kinshook · Answer

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?

Let z be ab; where a & b are tens and unit digits

z = 10a + b
x = 10^10 - (10a+b); 
x = 99999999(9-a)(10-b)

Sum of digits of x = 9*8+9-a+10-b = 91-a-b= 84
a+b = 91-84=7

Possible values of z = {70,61,52,43,34,25,16} : 6 values

IMO D

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

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

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