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# The tens digit of 9^20 is 0, what is the tens digit og 9^19

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The tens digit of 9^20 is 0, what is the tens digit og 9^19 [#permalink]  29 Oct 2005, 08:40
The tens digit of 9^20 is 0, what is the tens digit og 9^19?
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First I did by brute force by multiply the last 2 digits with 9. After the brute force, I think the following can be a short-cut.

The last digit of power of '9' alternates between 1 and 9. Hence even powers always wnd with 1 (i.e., 9 power 2, 4, 6 ... so on)

Hence last digit of 2^20 is 1 and 2^19 is 9.

Let x be the tens digit of 2^19 and y be the tens digit of 2^20.

y = Tens digit of (9 * 9) + ones digit of x * 9.

y =0 & Tens digit of (9 * 9) = 8

0 = 8 + ones digit of x * 9

The only possible value of x is 8, so that 0 = 8 + 2 (72 - Ones digit is 2)

Pls lmk if I have gone wrong.
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Confirm that the answer is 8 , you got it first, bubby
Here is my approach:
9^n has two cases of last digit : 9 or 1
with n is even, last digit is 9
n is odd, last digit is 1
----> 9^20 has two last digits of 01 ( 0 is provided by the stmt)
write 9^20 this way: .....01
9^19: ...C9( C is the tens digit)

...C9*9=.....01----> 9C+8=....B0 ( B is a digit) ----> C can only be 8.
Test: ....89*9 = ....01
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OK lets see...

9^even= unit digit is 9
9^odd=unit digit is 1

so 9^20=9^19*9

9*9=1 unit digit carry 8 towards the ten digit. now if ten digit is 0, then then ten digit of 9^19 must be 2, cause 8+2=10*9=90 <---ten digit here is 0...

So 2 it is..
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fresinha12 wrote:

OK lets see...

9^even= unit digit is 9
9^odd=unit digit is 1

so 9^20=9^19*9

9*9=1 unit digit carry 8 towards the ten digit. now if ten digit is 0, then then ten digit of 9^19 must be 2, cause 8+2=10*9=90 <---ten digit here is 0...

So 2 it is..

It should be 8 , buddy . Your approach is correct and explanation is much clearer than mine ^_^
Try testing it : .....21*9 = .......89
the answer must be 8 .......89*9= .......01
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ma bad...too early inthe morning for me

8*9=72 +8=80<--ten digit is zero...

laxieqv wrote:
fresinha12 wrote:

OK lets see...

9^even= unit digit is 9
9^odd=unit digit is 1

so 9^20=9^19*9

9*9=1 unit digit carry 8 towards the ten digit. now if ten digit is 0, then then ten digit of 9^19 must be 2, cause 8+2=10*9=90 <---ten digit here is 0...

So 2 it is..

It should be 8 , buddy . Your approach is correct and explanation is much clearer than mine ^_^
Try testing it : .....21*9 = .......89
the answer must be 8 .......89*9= .......01
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Just to facilitate the answer choices request:

A - 5
B - 6
C - 7
D - 8
E - 9
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Great explanations guys!

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fresinha12 wrote:

OK lets see...

9^even= unit digit is 9
9^odd=unit digit is 1

Should it not be opposite?

9^even= unit digit is 1
9^odd=unit digit is 9
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