Anurag,

I am a little confused, can you please recheck the question

I am sure any number ending in 2 raised to any number other than 0 will not give a units digit of 1.

Thank You

Thank You Anurag, that helps !!!

gmat7202011 wrote:Find the units digit of the expression (36472) raised to 123!


Thank You

Last number is 2 so lets multiply and we get: 2, 4, 8, 16, 32, 64, 128, . . . .do you see the repetition?

Remaineder of 123/4= 3. So we choose third which is 8.

Oh i thought ! Sign is put for exclamation. Lol.

pri187 wrote: Hi Anurag,
Im a little confused here.
123! is a multiple of 4, agreed.
So is 3 or any number less then or equal to 123 will be a multiple of 123!
if i consider 123! = 3n , then power of 2 is multiple of 3, so according to cyclicity of 2 the unit digit will be 8.
Please let know where am I going wrong

Hi, pri187.
I will take the liberty to answer your question, simply because I don´t have any clue whether Anurag is still active (at least in this forum). (*)

Let < N > denote the unit´s digit of N, for instance: <2> = 2 and <2^4> = 6 (because 2^4 = 16). With this notation in mind, please note that:
$$\left\langle {{2^{12}}} \right\rangle = \left\{ \matrix{
\left\langle {{{\left( {{2^3}} \right)}^4}} \right\rangle = \left\langle {{8^4}} \right\rangle = 6 \hfill \cr
\left\langle {{{\left( {{2^4}} \right)}^3}} \right\rangle = \left\langle {{6^3}} \right\rangle = 6 \hfill \cr} \right.$$
In other words, it´s not just a matter of divisibility (by 3 or by 4), but also the recognition that once we get the unit´s digit equal to 6, any "sequential" positive integer power will keep this last digit 6...

In our case, we may proceed as follows:
$$\left\langle {{2^{123!}}} \right\rangle = \left\{ \matrix{
\left\langle {{{\left( {{2^4}} \right)}^{\,{{123!} \over 4}}}} \right\rangle = \left\langle {{6^{{{123!} \over 4}}}} \right\rangle = 6 \hfill \cr
\left\langle {{{\left( {{2^3}} \right)}^{\,{{123!} \over 3}}}} \right\rangle = \left\langle {{8^{{{123!} \over 3}}}} \right\rangle = \left\langle {{{\left( {{8^4}} \right)}^{{{123!} \over {3 \cdot 4}}}}} \right\rangle = \left\langle {{6^{{{123!} \over {3 \cdot 4}}}}} \right\rangle = 6 \hfill \cr} \right.$$

I hope things are clearer now!

Regards,
Fabio.

(*) P.S.: in approximately 2012 I exchanged here many interesting posts with him, through which I feel honored to consider myself his friend since then.
(He is an outstanding expert and an even more impressive soul, I must say. His intellectual power never crossed his friendly nature. No ego at all.)
If someone here has any contact with him, please send to him my best wishes!

HI Fabio,
Thanks a lot for explaining it so well.
I have now understood the concept

fskilnik@GMATH wrote:

pri187 wrote: Hi Anurag,
Im a little confused here.
123! is a multiple of 4, agreed.
So is 3 or any number less then or equal to 123 will be a multiple of 123!
if i consider 123! = 3n , then power of 2 is multiple of 3, so according to cyclicity of 2 the unit digit will be 8.
Please let know where am I going wrong

Hi, pri187.
I will take the liberty to answer your question, simply because I don´t have any clue whether Anurag is still active (at least in this forum). (*)

Let < N > denote the unit´s digit of N, for instance: <2> = 2 and <2^4> = 6 (because 2^4 = 16). With this notation in mind, please note that:
$$\left\langle {{2^{12}}} \right\rangle = \left\{ \matrix{
\left\langle {{{\left( {{2^3}} \right)}^4}} \right\rangle = \left\langle {{8^4}} \right\rangle = 6 \hfill \cr
\left\langle {{{\left( {{2^4}} \right)}^3}} \right\rangle = \left\langle {{6^3}} \right\rangle = 6 \hfill \cr} \right.$$
In other words, it´s not just a matter of divisibility (by 3 or by 4), but also the recognition that once we get the unit´s digit equal to 6, any "sequential" positive integer power will keep this last digit 6...

In our case, we may proceed as follows:
$$\left\langle {{2^{123!}}} \right\rangle = \left\{ \matrix{
\left\langle {{{\left( {{2^4}} \right)}^{\,{{123!} \over 4}}}} \right\rangle = \left\langle {{6^{{{123!} \over 4}}}} \right\rangle = 6 \hfill \cr
\left\langle {{{\left( {{2^3}} \right)}^{\,{{123!} \over 3}}}} \right\rangle = \left\langle {{8^{{{123!} \over 3}}}} \right\rangle = \left\langle {{{\left( {{8^4}} \right)}^{{{123!} \over {3 \cdot 4}}}}} \right\rangle = \left\langle {{6^{{{123!} \over {3 \cdot 4}}}}} \right\rangle = 6 \hfill \cr} \right.$$

I hope things are clearer now!

Regards,
Fabio.

(*) P.S.: in approximately 2012 I exchanged here many interesting posts with him, through which I feel honored to consider myself his friend since then.
(He is an outstanding expert and an even more impressive soul, I must say. His intellectual power never crossed his friendly nature. No ego at all.)
If someone here has any contact with him, please send to him my best wishes!

pri187 wrote:HI Fabio,
Thanks a lot for explaining it so well.
I have now understood the concept

Hi, Pri187!

Thank you for the nice compliment!
I am glad I could be helpful.
(Try to answer in Anurag´s place is never easy.)

Regards,
Fabio.