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Note that the last digit of (36472) raised to any power is same as that of 2 raised to the same power.gmat7202011 wrote:Find the units digit of the expression (36472) raised to 123!
Last number is 2 so lets multiply and we get: 2, 4, 8, 16, 32, 64, 128, . . . .do you see the repetition?gmat7202011 wrote:Find the units digit of the expression (36472) raised to 123!
Thank You
Hi Anurag,Anurag@Gurome wrote:Note that the last digit of (36472) raised to any power is same as that of 2 raised to the same power.gmat7202011 wrote:Find the units digit of the expression (36472) raised to 123!
Now units digit of powers of 2 has a cycle of 2, 4, 8, 6, 2, 4, 8...
Thus if the power of 2 is multiple of 4, units digit is 6.
Now 123! factorial is obviously a multiple of 4, hence units digit of 2 raised to the power 123! is 6. And hence the units digit of (36472) raised to the power 123! is also 6.
Hi, pri187.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
fskilnik@GMATH wrote:Hi, pri187.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
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, Pri187!pri187 wrote:HI Fabio,
Thanks a lot for explaining it so well.
I have now understood the concept
