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Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
The units digit of J vs The units digit of K
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04 May 2017, 06:55
04 May 2017, 06:55
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00:00
Question Stats:
66% (02:00) correct
33% (02:09) wrong based on 218 sessions
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\(J = (2^{43})(3^{44})(5^{45})\)
\(K = (7^{46})(11^{47})(13^{48})\)
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
