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Celestial09
IMO its D
although I did performing a lengthy calculation which took some time.. hence waiting for a crisp approach and also a short method to find 10ens digit for any three digit or even 7 digit number.
Thanks
Celestial

hi..
whereever u are asked for last digit / tens digit / hundreds digit, it means remainder when div by 10 / 100 / 1000....
basically what question asks us is to find the remainder when divided by 100.. and we should be concerned only with last two digits..

36^10=(36^2)^5= 96^5(as the remainder when 36^2 is divided by 100)
=(-4)^5(-4 is the remainder when 96 div by 100)
= -24 or 76..
so the tens digit is 7
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hi the ans is D.. and the solution can be..
basically what question asks us is to find the remainder when divided by 100..
36^10= 96^5(as the remainder when 36^2 is divided by 100)=(-4)^5= -24 or 76..
so the tens digit is 7
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IMO its D
although I did performing a lengthy calculation which took some time.. hence waiting for a crisp approach and also a short method to find 10ens digit for any three digit or even 7 digit number.
Thanks
Celestial
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Bunuel
What is the tens digit of 36^10?

A. 1
B. 3
C. 5
D. 7
E. 9

Kudos for a correct solution.

36^10 = 6^20
If you type powers of six they end in
6
36
16
96
76
56
the pattern is 3-->1-->9-->7-->5 so for 36 we start with 3-->9-->5-->1-->7 and repeat.
36^10 will come at 7. Answer D.
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Answer = D = 7

\(36^{10} = 6^{20}\)

\(6^1 = 06\)

\(6^2 = 36\)

\(6^3 = ..16\)

\(6^4 = ...96\)

\(6^5 = ...76\)

\(6^6 = ...56\)

\(6^7 = ...36\)

By following the pattern above, tens digit would be 7
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we go like this:
36^10=36^(2*5)
36^2=1296, since we need only the tens digit we take 96
96-100 leeds us to -4; the value of -4^5 is the same as 2^10, which leeds us to 1024. 100-24=76. we need the tenths digit, so we take 7.
D
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Bunuel
What is the tens digit of 36^10?

A. 1
B. 3
C. 5
D. 7
E. 9

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

Since this problem asks for a specific digit that isn't a units digit, you should see this as a pattern-recognition, "create your own number property" problem. And one instinct should tell you that you can simplify 36^10 by rephrasing it as (6^2)^10, or 6^20. This gives you smaller numbers to work with as you establish a pattern.

From there, find the pattern with a keen eye on the tens digit:

6^2=36
6^3=216
6^4=...96 (Just multiply the last two digits since we only care about the tens digit)
6^5=...76
6^6=...56
6^7=...36
and hence starts the cycle again:

3, 1, 9, 7, 5,

3, 1, 9, 7, 5,

and so on

Since the pattern starts with the 6^2 (6^1 has no tens digit) and the 20th power isn't that much to jot down, you may want to simply write out that pattern until you get to the 20th number:

0, 3, 1, 9, 7, 5, 3, 1, 9, 7, 5, 3, 1, 9, 7, 5, 3, 1, 9, 7

The tens digit, then, will be 7.
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chetan2u
Celestial09
IMO its D
although I did performing a lengthy calculation which took some time.. hence waiting for a crisp approach and also a short method to find 10ens digit for any three digit or even 7 digit number.
Thanks
Celestial

hi..
whereever u are asked for last digit / tens digit / hundreds digit, it means remainder when div by 10 / 100 / 1000....
basically what question asks us is to find the remainder when divided by 100.. and we should be concerned only with last two digits..

36^10=(36^2)^5= 96^5(as the remainder when 36^2 is divided by 100)
=(-4)^5(-4 is the remainder when 96 div by 100)
= -24 or 76..
so the tens digit is 7

Hello Chetan,

In the above steps, can you please clarify how did you reached out to -24 or 76?

36^10=(36^2)^5= 96^5 (96 is the remainder when 36^2 is divided by 100)
=(-4)^5(-4 is the remainder when 96 div by 100)
= -24 or 76..???
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chetan2u
Celestial09
IMO its D
although I did performing a lengthy calculation which took some time.. hence waiting for a crisp approach and also a short method to find 10ens digit for any three digit or even 7 digit number.
Thanks
Celestial

hi..
whereever u are asked for last digit / tens digit / hundreds digit, it means remainder when div by 10 / 100 / 1000....
basically what question asks us is to find the remainder when divided by 100.. and we should be concerned only with last two digits..

36^10=(36^2)^5= 96^5(as the remainder when 36^2 is divided by 100)
=(-4)^5(-4 is the remainder when 96 div by 100)
= -24 or 76..
so the tens digit is 7

Hello Chetan,

In the above steps, can you please clarify how did you reached out to -24 or 76?

36^10=(36^2)^5= 96^5 (96 is the remainder when 36^2 is divided by 100)
=(-4)^5(-4 is the remainder when 96 div by 100)
= -24 or 76..???

Hi,
When we multiply -4*-4=16..
And 16*16=...56
Nowfifth -4 is left..
When 56 is multiplied by -4, we get -24 as last two digits..
But remainder cannot be NEGATIVE, so add to 100..
100+(-24)=76
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Check out the blog posts on my website given below.
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VeritasPrepKarishma

Hello Karishma,

I was able to get through powers of 11 note in the link provided. However, I have some queries related to
Last two digits of 6^58
Clearly the cyclicity of ten's digit is 5

3 1 9 7 5 3 1 9 7 5

"The new cycle with tens digit of 3 begins at the powers of 2, 7, 12, 17, 22, 27 etc. So the new cycle will also begin at power of 57 and 6^58 will have 1 as the tens digit."

Can you elaborate this more? How the new cycle for 6^58 will start with at a power of 57?
( Sorry for asking this but somehow I am unable to grasp this concept )


regards,
MoM
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VeritasPrepKarishma

Hello Karishma,

I was able to get through powers of 11 note in the link provided. However, I have some queries related to
Last two digits of 6^58
Clearly the cyclicity of ten's digit is 5

3 1 9 7 5 3 1 9 7 5

"The new cycle with tens digit of 3 begins at the powers of 2, 7, 12, 17, 22, 27 etc. So the new cycle will also begin at power of 57 and 6^58 will have 1 as the tens digit."

Can you elaborate this more? How the new cycle for 6^58 will start with at a power of 57?
( Sorry for asking this but somehow I am unable to grasp this concept )


regards,
MoM

Note that 6^1 has no tens digit.
The cyclicity is 3, 1, 9, 7, 5 but it starts from 6^2.
6^2 has tens digit of 3.
6^7 has tens digit of 3.
6^12 has tens digit of 3.
...

So the new cycle starts at 2, 7, 12, 17, 22, 27, 32, 37.... Note the symmetry (after every 5 powers).
The new cycle will start at 52 and then at 57 too.
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\(36^{10}\) = \(6^{20}\)=\((2*3)^{20}\)
Now we have 2 separate cases
\(2^{20}\)=\((2^{10})^2\)=\(1024^2\) 24 raise to even power will always end in 76.
\(3^{20}\)=\((3^4)^5\)=\(81^5\) Units digit 1 raised to any power will give 1. Tens digit is equal to the units digit of the product of 8 and 5.
So we have 76*01=76
Tens digit is 7.
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Bunuel
What is the tens digit of 36^10?

A. 1
B. 3
C. 5
D. 7
E. 9

Kudos for a correct solution.


36^10 = 4^10*9^10 = = 2^20*3^20 = 76*01 = 76



Answer Option D

Posted from my mobile device
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Bunuel
Bunuel
What is the tens digit of 36^10?

A. 1
B. 3
C. 5
D. 7
E. 9

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

Since this problem asks for a specific digit that isn't a units digit, you should see this as a pattern-recognition, "create your own number property" problem. And one instinct should tell you that you can simplify 36^10 by rephrasing it as (6^2)^10, or 6^20. This gives you smaller numbers to work with as you establish a pattern.

From there, find the pattern with a keen eye on the tens digit:

6^2=36
6^3=216
6^4=...96 (Just multiply the last two digits since we only care about the tens digit)
6^5=...76
6^6=...56
6^7=...36
and hence starts the cycle again:

3, 1, 9, 7, 5,

3, 1, 9, 7, 5,

and so on

Since the pattern starts with the 6^2 (6^1 has no tens digit) and the 20th power isn't that much to jot down, you may want to simply write out that pattern until you get to the 20th number:

0, 3, 1, 9, 7, 5, 3, 1, 9, 7, 5, 3, 1, 9, 7, 5, 3, 1, 9, 7

The tens digit, then, will be 7.

Hi the cyclicity is hence 5
6^20
if we divide 20 by 5 remainder is 0
how do we know that tens digit would be 7?
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Bunuel
Bunuel
What is the tens digit of 36^10?

A. 1
B. 3
C. 5
D. 7
E. 9

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

Since this problem asks for a specific digit that isn't a units digit, you should see this as a pattern-recognition, "create your own number property" problem. And one instinct should tell you that you can simplify 36^10 by rephrasing it as (6^2)^10, or 6^20. This gives you smaller numbers to work with as you establish a pattern.

From there, find the pattern with a keen eye on the tens digit:

6^2=36
6^3=216
6^4=...96 (Just multiply the last two digits since we only care about the tens digit)
6^5=...76
6^6=...56
6^7=...36
and hence starts the cycle again:

3, 1, 9, 7, 5,

3, 1, 9, 7, 5,

and so on

Since the pattern starts with the 6^2 (6^1 has no tens digit) and the 20th power isn't that much to jot down, you may want to simply write out that pattern until you get to the 20th number:

0, 3, 1, 9, 7, 5, 3, 1, 9, 7, 5, 3, 1, 9, 7, 5, 3, 1, 9, 7

The tens digit, then, will be 7.

Hi the cyclicity is hence 5
6^20
if we divide 20 by 5 remainder is 0
how do we know that tens digit would be 7?

6^1 has no tens digit, so we should divide 19 by 5, not 20. 19 gives remainder of 4 upon division by 5. The fourth number in the pattern is 7.
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Tip: Do not try to multiply. I did and it was a waste of time since I got "6" as tens digit, which isn´t even in the answer choices :D

As explained below, the best method is to convert 36^10 into 6^20 and then see a pattern:

6^2 = 36
6^3 = 216
6^4 = 1296
6^5 = ...76
6^6 = ...56
6^7 = ...36

--> so 6^20 will be ...76
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