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Re: Find the last digit in the expression (36472)^123!*34767^76 ? [#permalink]
Jai1903 wrote:
\(((36472)^{123!})∗((34767)^{76!})\)

Since we have to find last digit, we will find remainder when divided by 10.

For \(((34767)^{76!})\):
34767 fetches 7. Since we have 76! in power thus \(7^2\) gives 49 which gives -1 as remainder. Since 76! is multiple of 4 we get 1 as remainder.

For \(((36472)^{123!})\):
36472 fetches 2. Now we get \({2^123!}\)
We notice every 4th power of 2 fetches 6. Thus, \({2^123!}\) fetches 6.

Hence answer is 6


hello could you explain this in more steps, I did not understand how did we deduce that 76! by 4, by 7's cyclicality right ? but 76! would be too tough to calculate and why did we take \(7^2\) ?
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Find the last digit in the expression (36472)^123!*34767^76 ? [#permalink]
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SOLUTION:

Its a question testing basics from the concepts of Unit Digit

In ((36472)^123!) ,the last two digits of 123! would be 00 as it is a factorial and hence we can say that it is divisible by 4.The unit digit depends on the unit digit of 2^(123!)

Cyclicity of 2 is 4 and hence the unit digit for ((36472)^123!) would be 6 (2^1=2; 2^2 = 4; 2^3 = 8 and 2^4=..6 )

In ((34767)^76!),the last two digits of 76! would be 00 and hence divisible by 4.The unit digit depends on the unit digit of 7^(76!)

Cyclicity of 7 is 4 and hence the unit digit for ((34767)^76!) would be 1 (7!=7 ; 7^2=..9 ; 7^3=..3 ;7^4=..1)

Hence the Unit digit in the expression ((36472)^123!)∗((34767)^76!)

= 6 x 1 = 6 (OPTION E)

Hope this helps :thumbsup:
Devmitra Sen(Math)
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Find the last digit in the expression (36472)^123!*34767^76 ? [#permalink]
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