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21 Mar 2023, 01:30
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
57% (02:06) correct
43%
(02:28)
wrong
based on 254
sessions
History
Date
Time
Result
Not Attempted Yet
If k is a positive integer and \(\frac{269!}{2^k}\) is a positive integer, what is the greatest possible value of k ?
A. 262
B. 263
C. 264
D. 265
E. 266
A. 262
B. 263
C. 264
D. 265
E. 266
21 Mar 2023, 01:55
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We need to find the number of 2's in 269!.
An elegant method to solve such problem is -
269/2 = 134 (ignore remainder 1)
134/2 = 67
67/2 = 33 (ignore remainder 1)
33/2 = 16 (ignore remainder 1)
16/2 = 8
8/2 = 4
4/2 = 2
2/2 = 1
Now find 134 + 67 + 33 + 16 + 8 + 4 + 2 +1 = 265
Answer D
An elegant method to solve such problem is -
269/2 = 134 (ignore remainder 1)
134/2 = 67
67/2 = 33 (ignore remainder 1)
33/2 = 16 (ignore remainder 1)
16/2 = 8
8/2 = 4
4/2 = 2
2/2 = 1
Now find 134 + 67 + 33 + 16 + 8 + 4 + 2 +1 = 265
Answer D
General Discussion
02 Sep 2024, 05:26
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↧↧↧ Detailed Video Solution to the Problem ↧↧↧
Given that \(\frac{269!}{2^k}\) is a positive integer and we need to find the the greatest possible value of k
Lets solve the problem using two methods:
Method 1:
\(\frac{269!}{2^k}\)
\(\frac{134}{2}\) = 67
\(\frac{67}{2}\) = 33
\(\frac{33}{2}\) = 16
\(\frac{16}{2}\) = 8
\(\frac{8}{2}\) = 4
\(\frac{4}{2}\) = 2
\(\frac{2}{2}\) = 1
Total = 134 + 67 + 33 + 16 + 8 + 4 + 2 + 1 = 265
Method 2: (Similar to 1st Method)
\(\frac{269!}{2^k}\)
We need to divide 269 by all powers of 2 starting from 2^1 till the power which will give us a value lesser than 269 and add all the integer values after the division
256 < 269 < 512
=> We need to go from \(2^1\) till \(2^8\) (256)
\(\frac{269}{2^1} + \frac{269}{2^2} + \frac{269}{2^3} + ... + \frac{269}{2^8}\)
= 134 + 67 + 33 + 16 + 8 + 4 + 2 + 1 = 265
So, Answer will be D
Hope it helps!
Watch the following video to MASTER Exponents
Given that \(\frac{269!}{2^k}\) is a positive integer and we need to find the the greatest possible value of k
Lets solve the problem using two methods:
Method 1:
\(\frac{269!}{2^k}\)
- We need to divide 269 by 2 and take the integer value
- Divide it again by 2 and repeat the process till we get 1
- Add all the integer values we got above
\(\frac{134}{2}\) = 67
\(\frac{67}{2}\) = 33
\(\frac{33}{2}\) = 16
\(\frac{16}{2}\) = 8
\(\frac{8}{2}\) = 4
\(\frac{4}{2}\) = 2
\(\frac{2}{2}\) = 1
Total = 134 + 67 + 33 + 16 + 8 + 4 + 2 + 1 = 265
Method 2: (Similar to 1st Method)
\(\frac{269!}{2^k}\)
We need to divide 269 by all powers of 2 starting from 2^1 till the power which will give us a value lesser than 269 and add all the integer values after the division
256 < 269 < 512
=> We need to go from \(2^1\) till \(2^8\) (256)
\(\frac{269}{2^1} + \frac{269}{2^2} + \frac{269}{2^3} + ... + \frac{269}{2^8}\)
= 134 + 67 + 33 + 16 + 8 + 4 + 2 + 1 = 265
So, Answer will be D
Hope it helps!
Watch the following video to MASTER Exponents
