GreenlightTestPrep wrote:
\(J = (2^{43})(3^{44})(5^{45})\)
\(K = (7^{46})(11^{47})(13^{48})\)
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 0K = (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 0So, we get:
Quantity A: 0
Quantity B: a positive integer that is NOT equal to 0
Answer:
Cheers,
Brent
_________________
Brent Hanneson – Creator of greenlighttestprep.com
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