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What is the rightmost non-zero digit of 20! ?

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What is the rightmost non-zero digit of 20! ?  [#permalink]

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New post 16 Sep 2019, 08:54
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

45% (01:38) correct 55% (01:42) wrong based on 20 sessions

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Re: What is the rightmost non-zero digit of 20! ?  [#permalink]

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New post 16 Sep 2019, 11:45
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To find the non zero digit the formula is

2^A A! B!

where A = 20/5 = 4 (quotient) and B is 20/5 =0 (remainder)

Now putting A and B in the formula

2^4* 4! * 0!

From the cyclicity we know that the unit digit of 2^4 would be 6

And, 4! Would have unit digit 4 (4*3*2*1=24)

Now multiplying all the unit digits:

6*4*1 =24. Therefore 4 would be the rightmost non zero digit of 20!

C is the right answer.

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What is the rightmost non-zero digit of 20! ?  [#permalink]

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New post 16 Sep 2019, 21:29
first of all we need to express 20! in the form of prime factors--

power of 2 in 20! = \(\frac{20}{2}\) = \(\frac{10}{2}\) = \(\frac{5}{2}\) = \(\frac{2}{2}\) = 1---10+5+2+1 = 18

power of 3 in 20! = \(\frac{20}{3}\) = \(6/3\) = 2 --6+2 = 8

power of 5 in 20! = \(\frac{20}{5}\) = 4 --4

power of 7 in 20! = \(\frac{20}{7}\) = 2 ---2

power of 11 in 20! = \(\frac{20}{11}\) = 1 --1

power of 13 in 20! = \(\frac{20}{13}\) = 1 ---1

power of 17 in 20! = \(\frac{20}{17}\) = 1 --1

power of 19 in 20! = \(\frac{20}{19}\) = 1 --1

20! (can be written as ) = 2^18*\(3^8\)*\(5^4\)*\(7^2\)*11*13*17*19 =2^14*\(3^8\)*\(7^2\)*11*13*17*19*\(10^4\)

rightmost non-zero digit of 20! = unit digit of--2^14*\(3^8\)*\(7^2\)*11*13*17*19 = 4

hence C is the correct answer
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What is the rightmost non-zero digit of 20! ?   [#permalink] 16 Sep 2019, 21:29
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