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Its a multiplication of some random large numbers. But notice, one of these terms is 10. So no matter what other numbers are, your answer must be 0.
_________________

This is how I intrepretted. 101^102^103 102^103 would have 8 in the units digit. 101^**8 would have 1 in the units digit as 1 is the units digit for 101 and the power doesnt matter

102^103^104

103^104 would have 9 in the units digit . 102^**9 would be a 2 in units digit

108^109^110

109^110 would have 1 in units place. 108^**1 would have 2 in units digit

109^110^111

110^111 '0' in the units place. 109^**0 would have 1 in units digit..

As VIPS pointed out pls take care of ur post... Let it be legible.... Just edit this post...

Hi Shan, can you explain how did you come up with the answer 2 ? Hi 2013gmat,

Sorry ans was not 2 as i calculated... it should be 4 if my evaluation is correct....

Here is my explanation, i'm ready to correct myself in case of mistakes...

\(a^b^c =\) a^(bc) say 2^3^2 = \(2^6 = 64\)

I used this logic here...

101^102^103 = 101^10506 ====> unit digit ===> 1^6====> 1 102^103^104 = 102^10712 ====> unit digit ===> 2^2====> 4 108^109^110 = 108^11990 ====> unit digit ===> 8^0====> 1 109^110^111 = 109^12210 ====> unit digit ===> 9^0====> 1

1*4*1*1 = 4

correct me if im wrong

Shan, does the question say (a^m)^n OR a^(m)^n ? You have considered it as an former one but if it would have been the latter one.. then what wud be our approach ?
_________________

Concentration: Entrepreneurship, International Business

GMAT 1: 440 Q33 V13

GPA: 3

Re: How to find last digit of a number [#permalink]

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06 Dec 2012, 00:05

2013gmat wrote:

Shan, does the question say (a^m)^n OR a^(m)^n ? You have considered it as an former one but if it would have been the latter one.. then what wud be our approach ?

hmmm, i guess both should be same... Coz powers to powers will get multiplied....

\(a^b^c =\) a^(bc) say 2^3^2 = \(2^6 = 64\)

moreover, latter one, which you said will not lead to our desired result...

a^\((b)^c\) say 2^3^2 = \(2^9 = 512\)

If any discrepancy please let me know... happy to learn....
_________________

GMAT - Practice, Patience, Persistence Kudos if u like

Re: How to find last digit of a number [#permalink]

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06 Dec 2012, 02:06

shanmugamgsn wrote:

Hi 2013gmat,

Sorry ans was not 2 as i calculated... it should be 4 if my evaluation is correct....

Here is my explanation, i'm ready to correct myself in case of mistakes...

\(a^b^c =\) a^(bc) say 2^3^2 = \(2^6 = 64\)

I used this logic here...

101^102^103 = 101^10506 ====> unit digit ===> 1^6====> 1 102^103^104 = 102^10712 ====> unit digit ===> 2^2====> 4 108^109^110 = 108^11990 ====> unit digit ===> 8^0====> 1 109^110^111 = 109^12210 ====> unit digit ===> 9^0====> 1

1*4*1*1 = 4

correct me if im wrong

Yup. its wrong!

you can not simply take unit digit of powers to deduce unit digit, simple as that. for example 102^103^104 = 102^10712 ====> unit digit ===> 2^2====> 4 This is incorrect. if you can, use a calculator to see: 102^2 and 102^12 have different unit digits. instead, u need to find out cyclicity of 2. which is 4. so unit digit of 102^10712 will be same as 102^4 which will be same as unit digit of 2^4 => which is 6 Try it!
_________________

Concentration: Entrepreneurship, International Business

GMAT 1: 440 Q33 V13

GPA: 3

Re: How to find last digit of a number [#permalink]

Show Tags

06 Dec 2012, 03:57

Vips0000 wrote:

shanmugamgsn wrote:

Hi 2013gmat,

Sorry ans was not 2 as i calculated... it should be 4 if my evaluation is correct....

Here is my explanation, i'm ready to correct myself in case of mistakes...

\(a^b^c =\) a^(bc) say 2^3^2 = \(2^6 = 64\)

I used this logic here...

101^102^103 = 101^10506 ====> unit digit ===> 1^6====> 1 102^103^104 = 102^10712 ====> unit digit ===> 2^2====> 4 108^109^110 = 108^11990 ====> unit digit ===> 8^0====> 1 109^110^111 = 109^12210 ====> unit digit ===> 9^0====> 1

1*4*1*1 = 4

correct me if im wrong

Yup. its wrong!

you can not simply take unit digit of powers to deduce unit digit, simple as that. for example 102^103^104 = 102^10712 ====> unit digit ===> 2^2====> 4 This is incorrect. if you can, use a calculator to see: 102^2 and 102^12 have different unit digits. instead, u need to find out cyclicity of 2. which is 4. so unit digit of 102^10712 will be same as 102^4 which will be same as unit digit of 2^4 => which is 6 Try it!

Thanks VIPS... Ya it was mistake... \(a^n\) based on 'n' value (weight) 'a' is changed... (added to my notes ) Always its good to learn from others...
_________________

GMAT - Practice, Patience, Persistence Kudos if u like

The process has to be the one that is used by "shanmugamgsn"

shanmugamgsn wrote:

I used this logic here...

101^102^103 = 101^10506 ====> unit digit ===> 1^6====> 1 102^103^104 = 102^10712 ====> unit digit ===> 2^2====> 4 108^109^110 = 108^11990 ====> unit digit ===> 8^0====> 1 109^110^111 = 109^12210 ====> unit digit ===> 9^0====> 1

1*4*1*1 = 4

correct me if im wrong

That's not correct.

If exponentiation is indicated by stacked symbols, the rule is to work from the top down: \(a^m^n=a^{(m^n)}\) and not \((a^m)^n\). For more check Number Theory chapter of Math Book: math-number-theory-88376.html _________________

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