GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
If 2^32 + 1 is exactly divisible by a certain number. Which one of the
[#permalink]
10 Apr 2021, 22:24
10 Apr 2021, 22:24
1
2
Bookmarks
00:00
Question Stats:
33% (02:02) correct
66% (01:09) wrong based on 3 sessions
Hide Show timer Statistics
If \(2^{32} + 1\) is exactly divisible by a certain number. Which one of the following is also divisible by the same number?
A. \(2^{16} + 1\)
B. \(2^{16} - 1\)
C. \(2^{96} + 1\)
D. \(7(2^{33})\)
E. None of the above
A. \(2^{16} + 1\)
B. \(2^{16} - 1\)
C. \(2^{96} + 1\)
D. \(7(2^{33})\)
E. None of the above
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
If 2^32 + 1 is exactly divisible by a certain number. Which one of the
[#permalink]
11 Apr 2021, 11:52
11 Apr 2021, 11:52
1
Explanation:
\(2\) has a repeating pattern of \(4\) i.e. the unit digit values repeat after every 4 powers
\(2^1 = 2\)
\(2^2 = 4\)
\(2^3 = 8\)
\(2^4 = 16\)
\(2^5 = 32\)
Now,\( 2^{32}\) will have a unit digit as \(6\)
How - Divide \(32\) by \(4\) and we get a remainder \(0\)
We are looking for a number which will have a unit digit as \(6\) again!
A. \(2^{16}\) has unit digit as \(6\) but the number is less than \(2^{32}\)
B. \(2^{16}\) same reason as above
C. \(2^{96}\) has unit digit as \(6\) and is greater than \(2^{32}\)
D. \(2^{33}\) has unit digit as \(2\)
E. None of the above
Hence, option C
Another Approach:
We can also use the algebraic identity \((a^3 + b^3) = (a + b)(a^2 + b^2 - ab)\)
\(2^{96} + 1 = (2^{32})^3 + 1^3 = (2^{32} + 1)(2^{64} + 1 - 2^{32})\)
Since, \(2^{96} + 1\) is divisible by \(2^{32} + 1\), it will also be divisible by that number
\(2\) has a repeating pattern of \(4\) i.e. the unit digit values repeat after every 4 powers
\(2^1 = 2\)
\(2^2 = 4\)
\(2^3 = 8\)
\(2^4 = 16\)
\(2^5 = 32\)
Now,\( 2^{32}\) will have a unit digit as \(6\)
How - Divide \(32\) by \(4\) and we get a remainder \(0\)
We are looking for a number which will have a unit digit as \(6\) again!
A. \(2^{16}\) has unit digit as \(6\) but the number is less than \(2^{32}\)
B. \(2^{16}\) same reason as above
C. \(2^{96}\) has unit digit as \(6\) and is greater than \(2^{32}\)
D. \(2^{33}\) has unit digit as \(2\)
E. None of the above
Hence, option C
Another Approach:
We can also use the algebraic identity \((a^3 + b^3) = (a + b)(a^2 + b^2 - ab)\)
\(2^{96} + 1 = (2^{32})^3 + 1^3 = (2^{32} + 1)(2^{64} + 1 - 2^{32})\)
Since, \(2^{96} + 1\) is divisible by \(2^{32} + 1\), it will also be divisible by that number
gmatclubot
If 2^32 + 1 is exactly divisible by a certain number. Which one of the [#permalink]
11 Apr 2021, 11:52
Moderators:
