If m is an integer and m = 10^32 – 32, what is the sum of the digits

Question

If m is an integer and  m = 10^{32} – 32 , what is the sum of the digits of m ?

A. 257
B. 264
C. 275
D. 284
E. 292

PS24000.02

BrentGMATPrepNow · Accepted Answer

Let's look for a pattern...

m = 10^{2} – 32=100 - 32 = 68     
 m = 10^{3} – 32=1,000 - 32 = 968     
 m = 10^{4} – 32=10,000 - 32 = 9968     
 m = 10^{5} – 32=100,000 - 32 = 99968

Conclusion: The number of 9's before the 68 is TWO LESS THAN the exponent . 
We want  m = 10^{32} – 32

32 - 2 = 30
So, m will have 30 9's followed by 68

The sum of the digits = (30)(9) + 6 + 8 = 284

Answer: D

Cheers, 
Brent

Archit3110 · Answer

total zeroes ; 10^32 ; 32
and last two digits when subtracted would give 68 ; 6+8 ; 14
left with 30 places of 9 each ; 30*9 ; 270
so total ; 270+14 ; 284
OPTION D

GMATWhizTeam · Answer

Solution 
 • We know,  10^2 – 32 = 68 
 o Digits in the above expression are 6 and 8 
• And  10^3 – 32 = 968 
 o Digits in the above expression are 6, 8, and 9 
o Here, number of  9s = (3-2) =1  
• Also,  10^4 – 32 = 9968 
 o Digits in the above expression are 6, 8, 9 and 9
o Here number of  9s = (4-2) = 2   
And so on.
 • We can observe that digits in  10^n – 32  will be 6, 8 and  (n-2)  number of  9 s.
 o Where n is a positive integer. 
• So, the sum of the digits of  10^n – 32 = 6+ 8+ (n-2) *9 = 9n -4 ………Eq.(i) 
• Now, following the same sequence for  m = 10^ {32} – 32  and substituting  n = 32   in Eq.(i) we get,
• Sum of the digits of  m = 32*9 – 4 = 288 – 4 = 284  
Thus, the correct answer is  Option D.

GaurSaini · Answer

m = 10^32 - 32

10^3 - 32 = 968
10^4 - 32 = 9968
10^5 - 32 = 99968

Everytime we are getting (power - 2) numbers of 9 and 68.

Hence 32- 2 = 30

30*9 + 6 + 8 = 284

karnavora1 · Answer

10^2=100 
 10^3=1000

From this we understand that  10^3^2  will have 32 Zeros.
There are 2 inferences we can make from this
1. Once you subtract 32 from  10^3^2  you will be left with some number ending in 68. (6+8=14)
2. The rest of the digits will be 9.

10^32  will have 32 zeros and 1 which means there will 33 digits.
When we subtract 32 from it we'll lose 1 digit because of the subtraction and 2 digits will be 6 and 8, hence we'll be left with 30 "9s".

Total of the digits  =(9*30)+(6+8)=284 
Answer D.

hkalabdu · Answer

When 10^32 subtracted from 32 it will give us 31's 9 and one 10

10^32 - 32 = 30*9 +(9-3) + (10-2) = 270 + 6 +8 = 284

bumpbot · Answer

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If m is an integer and m = 10^32 – 32, what is the sum of the digits

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If m is an integer and m = 10^32 – 32, what is the sum of the digits

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