How to find last digit of a number : GMAT Quantitative Section
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# How to find last digit of a number

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

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03 Dec 2012, 23:28
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How to find last digit of below number.

(101^102^103) * (102^103^104) * (108^109^110) * (109^110^111)

Official answer is not provided of this question.

Last edited by mangomangodolly on 05 Dec 2012, 03:52, edited 1 time in total.
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Re: How to find last digit of a number [#permalink]

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03 Dec 2012, 23:47
mangomangodolly wrote:
How to find last digit of below number.

$$101^102^103 * 102^103^104 * 108^109^110 * 109^110^111$$

Official answer is not provided of this question.

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

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04 Dec 2012, 00:03
101$$^102$$$$^103$$ * 102$$^103$$$$^104$$ * 108$$^109$$$$^110$$ * 109$$^110$$$$^111$$

This is like

101 is having power of 102 and again 102 is having power of 103

102 is having power of 103 and again 102 is having power of 104

108 is having power of 109 and again 102 is having power of 110

109 is having power of 110 and again 102 is having power of 111
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Re: How to find last digit of a number [#permalink]

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04 Dec 2012, 00:09
mangomangodolly wrote:
101$$^102$$$$^103$$ * 102$$^103$$$$^104$$ * 108$$^109$$$$^110$$ * 109$$^110$$$$^111$$

This is like

101 is having power of 102 and again 102 is having power of 103

102 is having power of 103 and again 102 is having power of 104

108 is having power of 109 and again 102 is having power of 110

109 is having power of 110 and again 102 is having power of 111

use braces to formulate question again in proper form... read the post on formatting math expressions.
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Re: How to find last digit of a number [#permalink]

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04 Dec 2012, 03:02
mangomangodolly wrote:
How to find last digit of below number.

$$101^102^103 * 102^103^104 * 108^109^110 * 109^110^111$$

Official answer is not provided of this question.

IMO unit digit must be 2..

As VIPS pointed out pls take care of ur post... Let it be legible....
Just edit this post...
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Re: How to find last digit of a number [#permalink]

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04 Dec 2012, 03:42
shanmugamgsn wrote:
mangomangodolly wrote:
How to find last digit of below number.

$$101^102^103 * 102^103^104 * 108^109^110 * 109^110^111$$

Official answer is not provided of this question.

IMO unit digit must be 2..

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

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04 Dec 2012, 04:21
mangomangodolly wrote:
How to find last digit of below number.

$$101^102^103 * 102^103^104 * 108^109^110 * 109^110^111$$

Official answer is not provided of this question.

IMO 4

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..

So I guess its 4 (Not sure)
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Re: How to find last digit of a number [#permalink]

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04 Dec 2012, 07:26
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2013gmat wrote:
shanmugamgsn wrote:
mangomangodolly wrote:
How to find last digit of below number.

$$101^102^103 * 102^103^104 * 108^109^110 * 109^110^111$$

Official answer is not provided of this question.

IMO unit digit must be 2..

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

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05 Dec 2012, 20:49
shanmugamgsn wrote:
shanmugamgsn wrote:
mangomangodolly wrote:
How to find last digit of below number.

$$101^102^103 * 102^103^104 * 108^109^110 * 109^110^111$$

Official answer is not provided of this question.

IMO unit digit must be 2..

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

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05 Dec 2012, 23:05
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....
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Re: How to find last digit of a number [#permalink]

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06 Dec 2012, 01:06
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!
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Re: How to find last digit of a number [#permalink]

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06 Dec 2012, 02:57
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...
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Re: How to find last digit of a number [#permalink]

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07 Dec 2012, 16:36
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
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Re: How to find last digit of a number [#permalink]

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08 Dec 2012, 04:10
Fistail wrote:
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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Re: How to find last digit of a number   [#permalink] 08 Dec 2012, 04:10
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