The sum of the digits of integer z is 186 and z = 10^(n – 4). What is

Hi SajjadAhmad,

Based on the answers choices (and the correct answer), the "intent" of this question is that that Z = (10^N) - 4. We're also told that the sum of the DIGITS of Z is 186. We're asked for the value of N.

Consider the following examples...

IF....
N = 1, then Z = 10 - 4 = 6 and the sum of the digits is 6
N = 2, then Z = 100 - 4 = 96 and the sum of the digits is 15
N = 3, then Z = 1000 - 4 = 996 and the sum of the digits is 24

Notice that when N increases by 1, the sum of the digits of Z increases by 9. To answer this question, we just have to determine how many additional "9s" we have to add to get the sum of the digits up to 186.

Since N = 1 led to a sum of 6, we need twenty 9s to get the sum up to 186. Thus, N would have to be 1 + 20 = 21

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Hello
try finding the pattern...
10^2-4=96
10^3-4=996
10^4-4=9996
conclude that sum of digits=9(n-1)+6 --> 9(n-1)+6=186 n=21 answer:c

For any n>2, \(10^n-4\) will have 96 in the end. Sum of 96 = 15.

186 - 15 = 171.

All the numbers in \(10^n-4\), apart from 96, should to be 9. So numbers of 9's will be

\(\frac{171}{9}= 19\)

19(9's) + 2(96) = 21.

Answer C.

Kudos if helped..

[quote="SajjadAhmad"]The sum of the digits of integer z is 186 and \(z = 10^{n}\) -4. What is the value of positive integer n?

(A) 19
(B) 20
(C) 21
(D) 22
(E) 23

I have tried to write 10 raise to power n minus 4 in question according to (Writing Mathematical Formulas on the Forum) but failed so z= 10 raise to power n minus 4 which is highlited[/quote

sum of digits =186 =180+6
repeatition of 9 = 180/9=20 times
10-4=6
whereas 10^2-4=96
power of 10 must be one greater than the power of 9.
power of 10 = n= 20+1=21
ans C

It needs to be clear at the outset that the question states ((10^n) – 4)
(10^1) – 4 = 6 (i.e. 0 times 9 & 6)
(10^2) – 4 = 96 (i.e. 1 times 9 & 6)
(10^3) – 4 = 996 (i.e. 2 times 9 & 6)
(10^4) – 4 = 9996 (i.e. 3 times 9 & 6)
(10^5) – 4 = 99996 (i.e. 4 times 9 & 6)

Hence we can conclude that
(10^n) – 4 = (n-1) times 9 & 6

Total sum of digits = 186 & 6 is always present hence sum of 9’s should be = 186 -6 = 180
i.e. 9(n-1)=180 that implies n = 21

Hence C