VJesus12 wrote: ↑Sat Jul 11, 2020 11:43 pm
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
[spoiler]OA=D[/spoiler]
Source: Official Guide
Let's look for a pattern...
m = 10^2 – 32 = 100 - 32 = 68
[when the exponent is 2, there are 0 9's before the 68]
m = 10^3 – 32 = 1,000 - 32 = 968
[when the exponent is 3, there is 1 9's before the 68]
m = 10^4 – 32 = 10,000 - 32 = 9968
[when the exponent is 4, there are 2 9's before the 68]
m = 10^5 – 32 = 100,000 - 32 = 99968
[when the exponent is 5, there are 3 9's before the 68]
Conclusion: The number of 9's before the 68 is TWO LESS THAN the exponent.
We want m = 10^32 – 32[/m]
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
