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: 10 Apr 2015
Posts: 6218
Given Kudos: 136
The units digit of J vs The units digit of K
[#permalink]
04 May 2017, 06:55
04 May 2017, 06:55
2
7
Bookmarks
00:00
Question Stats:
70% (01:59) correct
29% (02:16) wrong based on 206 sessions
Hide Show timer Statistics
\(J = (2^{43})(3^{44})(5^{45})\)
\(K = (7^{46})(11^{47})(13^{48})\)
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
\(K = (7^{46})(11^{47})(13^{48})\)
Quantity A |
Quantity B |
The units digit of J |
The units digit of K |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: The units digit of J vs The units digit of K
[#permalink]
06 May 2017, 05:39
06 May 2017, 05:39
3
GreenlightTestPrep wrote:
\(J = (2^{43})(3^{44})(5^{45})\)
\(K = (7^{46})(11^{47})(13^{48})\)
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
\(K = (7^{46})(11^{47})(13^{48})\)
Quantity A |
Quantity B |
The units digit of J |
The units digit of K |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
J = (2)(2)(2)...(2)(2)(3)(3)(3)...(3)(3)(5)(5)(5)...(5)(5)
= (2)(2)(2)...(2)(2)(3)(3)(3)...(3)(3)(5)(5)(5)...(5)(5)
= 10(2)(2)...(2)(2)(3)(3)(3)...(3)(3)(5)(5)...(5)(5)
Since J = some multiple of 10, the units digit of J is 0
K = (7)(7)(7)(7)...(7)(7)(11)(11)(11)...(11)(11)(13)(13)(13)....(13)(13)
Since there are no 2's or 5's hiding in the prime factorization of K, we can conclude that K is not a multiple of 10, which means the units digit of K is NOT 0
So, we get:
Quantity A: 0
Quantity B: a positive integer that is NOT equal to 0
Answer:
Show: ::
B
Cheers,
Brent
Re: The units digit of J vs The units digit of K
[#permalink]
13 Dec 2017, 21:13
13 Dec 2017, 21:13
2
2
Bookmarks
Can we not simply multiply the pattern of all numbers?
QA
We know that 2 has the 4 following patterns: 2,4,8,6 so (2^43) must have an ending of 8
3 has the following pattern 3,9,7,1 so 3^44 must be 1
5 has always the same pattern 5.
QB
7: 7,9,3,1 so 7^46 is 9
11 is like 5. The same pattern so always 1.
13: 3,9,7,1 so 7^48 so the ending is 1.
QA : (....8)(....1)(....5) = end with 0
QB: (....9)(....1)(....1) = end with 9
So B is greater
QA
We know that 2 has the 4 following patterns: 2,4,8,6 so (2^43) must have an ending of 8
3 has the following pattern 3,9,7,1 so 3^44 must be 1
5 has always the same pattern 5.
QB
7: 7,9,3,1 so 7^46 is 9
11 is like 5. The same pattern so always 1.
13: 3,9,7,1 so 7^48 so the ending is 1.
QA : (....8)(....1)(....5) = end with 0
QB: (....9)(....1)(....1) = end with 9
So B is greater
