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If x is a positive integer, what is the units digit of (24)^(2x+1)

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If x is a positive integer, what is the units digit of (24)^(2x+1)  [#permalink]

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08 Sep 2019, 21:56
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55% (hard)

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64% (02:03) correct 36% (02:17) wrong based on 81 sessions

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If x is a positive integer, what is the units digit of $$24^{(2x+1)}33^{(x+1)}17^{(x+2)}9^{(2x)}$$

A. 4
B. 6
C. 7
D. 8
E. 9

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Re: If x is a positive integer, what is the units digit of (24)^(2x+1)  [#permalink]

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08 Sep 2019, 22:34
only way i guess to answer the q is to sub 1 into x, if this is a single choice question. ANSWER : 8
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Re: If x is a positive integer, what is the units digit of (24)^(2x+1)  [#permalink]

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08 Sep 2019, 22:55
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Re: If x is a positive integer, what is the units digit of (24)^(2x+1)  [#permalink]

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09 Sep 2019, 01:49
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$$24^{2x+1} * 33^{x+1} * 17^{x+2} * 9^{2x}$$

Instead of factorizing the individual numbers and taking values that have similar powers together first:

= $$24^{2x+1}*33^{x+1}*17^{x+1}*17^1*81^x$$

= $$24^{2x+1}*(33*17)^{x+1}*17^1*81^x$$ [Reducing the base to minimum possible value i.e. 1]

= $$4^{2x+1}*1^{x+1}*7^1*1^x$$ [Taking unit values of all the entities]

= $$4*1*7*1$$ [Unit value of 4 for odd powers (2x+1) = 4]

= 28

Unit value = 8

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Re: If x is a positive integer, what is the units digit of (24)^(2x+1)  [#permalink]

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09 Sep 2019, 03:58
let suppose X=1

24^3 * 33^2 * 17^3 * 9^2
unit digit of each will be ...4*....9*...3*...1= ......8
therefore, 8, D
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Re: If x is a positive integer, what is the units digit of (24)^(2x+1)  [#permalink]

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09 Sep 2019, 06:50
solve using cyclicity ;
and test with x=1 and 2
its seen that value of unit digit will always be 8
IMO D

If x is a positive integer, what is the units digit of 24(2x+1)33(x+1)17(x+2)9(2x)24(2x+1)33(x+1)17(x+2)9(2x)

A. 4
B. 6
C. 7
D. 8
E. 9
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Re: If x is a positive integer, what is the units digit of (24)^(2x+1)  [#permalink]

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09 Sep 2019, 06:57
IMO 8 is the right answer.
let x=1,
24^3 * 33^2 * 17^3 * 9^2
4*9*3*1= unit's digit is 8.
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If x is a positive integer, what is the units digit of (24)^(2x+1)  [#permalink]

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09 Sep 2019, 11:19
Quote:
If x is a positive integer, what is the units digit of $$24^{(2x+1)}33^{(x+1)}17^{(x+2)}9^{(2x)}$$

A. 4
B. 6
C. 7
D. 8
E. 9

$$9^{(2x)}…cycles=(9,1)…2x=even…units=1$$
$$4^{(2x+1)}…cycles=(4,6)…2x=even+1=even+odd=odd…units=4$$
$$3^{(x+1)}…cycles=(3,9,7,1)…x=even…x+1=odd…units=(3,7)…x=odd…units=(9,1)$$
$$7^{(x+2)}…cycles=(7,9,3,1)…x=even…x+2=even…units=(9,1)…x=odd…units=(3,7)$$
$$x=even…units=(1)(4)(3,7)(9,1)=(1)(4)(3)(9)=4•27=4•7=28=8$$
$$x=even…units=(1)(4)(3,7)(9,1)=(1)(4)(7)(1)=4•7=28=8$$
$$x=odd…units=(1)(4)(9,1)(3,7)=(1)(4)(9)(3)=4•27=4•7=28=8$$
$$x=odd…units=(1)(4)(9,1)(3,7)=(1)(4)(1)(7)=4•7=28=8$$

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Re: If x is a positive integer, what is the units digit of (24)^(2x+1)  [#permalink]

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20 Sep 2019, 05:42
lnm87 wrote:
$$24^{2x+1} * 33^{x+1} * 17^{x+2} * 9^{2x}$$

Instead of factorizing the individual numbers and taking values that have similar powers together first:

= $$24^{2x+1}*33^{x+1}*17^{x+1}*17^1*81^x$$

= $$24^{2x+1}*(33*17)^{x+1}*17^1*81^x$$ [Reducing the base to minimum possible value i.e. 1]

= $$4^{2x+1}*1^{x+1}*7^1*1^x$$ [Taking unit values of all the entities]

= $$4*1*7*1$$ [Unit value of 4 for odd powers (2x+1) = 4]

= 28

Unit value = 8

What is the whole operation good for?
Am I allowed to just plug in one from the beginning, without the long factoring process?

Because I would come up with 8 as well then.

Thx
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Re: If x is a positive integer, what is the units digit of (24)^(2x+1)  [#permalink]

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20 Sep 2019, 09:59
chrtpmdr wrote:

What is the whole operation good for?
Am I allowed to just plug in one from the beginning, without the long factoring process?

Because I would come up with 8 as well then.

Thx

Could not understand what you are asking.
But if it is that steps are to be reduced you can do what you feel good.
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If x is a positive integer, what is the units digit of (24)^(2x+1)  [#permalink]

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21 Sep 2019, 10:30
1
Tips to solve efficiently (not just by plugging x=1) are:
1) (..6)^x should end with ...6
2) (..1)^x should end with ...1

In light of the tips above, let us adjust the equation below:
=24^(2x+1) * 33^(x+1) * 17^(x+2) *9^(2x)
=(24^2)^x * 24 * {33*17}^(x+1) * 17 * (9^2)^x
=(..6)^x * 24 * {..1}^(x+1) * 17 * (81)^x
=..6 * (24*17) * ..1 * ..1
=..6 *.. 8 * ..1 * ..1
=..8

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If x is a positive integer, what is the units digit of (24)^(2x+1)   [#permalink] 21 Sep 2019, 10:30
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