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What is the rightmost non-zero digit in 15!?

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What is the rightmost non-zero digit in 15!?  [#permalink]

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New post 25 Feb 2020, 01:04
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What is the rightmost non-zero digit in 15!?  [#permalink]

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New post 25 Feb 2020, 01:52
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What is the rightmost non-zero digit in 15!?

—>[ \(\frac{15}{5}\) ] ...= 3 (there are three zeroes at the end of 15!; \(5^3\))

—> \([\frac{15}{2}]+ [\frac{15}{4}]+ [\frac{15}{8}]+...= 7+3+1= 11\) (—> \(2^{11}\) )

—> \([\frac{15}{3}]+ [\frac{15}{9}]+...= 5+1 = 6\)( —> \(3^6\))

—> \([\frac{15}{7}]+...= 2\) (—> \(7^{2}\))

So, \(15! = 2^{11}* 3^{6} *5^{3} *7^{2}*11*13\)

And eliminate the zeroes:
—> \(\frac{15!}{1000}= 2^8*3^6*7^2*11*13 \)

Now, just find the units digits and multiply one another to get the last digit —> ...6*...9...*9...*1...*3 = ...18...* 81 = ...8

The rightmost nonzero digit of 15! is 8

The answer is D.

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Re: What is the rightmost non-zero digit in 15!?  [#permalink]

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New post 25 Feb 2020, 06:44
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Bunuel wrote:
What is the rightmost non-zero digit in 15!?

A. 2
B. 4
C. 6
D. 8
E. 9

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15/5 = 3 => A

Remainder is 0 =>B

So, Last Non zero digit will be \(2^A*A!*B!\)

= \(2^3*3!*0!\)

= 8* 6* 1

=>8 , Hence Answer must be (D) 8
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Re: What is the rightmost non-zero digit in 15!?  [#permalink]

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New post 25 Feb 2020, 12:00
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Bunuel wrote:
What is the rightmost non-zero digit in 15!?

A. 2
B. 4
C. 6
D. 8
E. 9



Solution


    • We know that 5! = 1*2*3*4*5 = 120
    • So rightmost non-zero digit in 5! = 2
    • If we multiply the rightmost non-zero digit of 5! to 6, the unit digit of the resulting number will give the rightmost non-zero digit of 6!
      o i.e. rightmost non-zero digit in 6! = unit’s digit of \((2*6) = 2\)
    • Following the same pattern as above we get,
      o Rightmost non-zero digit in 7! = unit' s of ((rightmost non-zero digit in 6! )*7)= unit’s digit of \((2*7) =4 \)
      o Rightmost non-zero digit in 8! = unit’s digit of \((4*8) = 2\)
      o Rightmost non-zero digit in 9! = unit’s digit of \((2*9) = \)unit’s digit of \(18 = 8\)
      o Rightmost non-zero digit in 10! = ten’s digit of \((2*9*10)\) =ten’s digit of \((18*10) = 8\)
         We have taken ten’s digit because we know that on multiplying a number by 10, the unit’s digit will always be 0 and we want right-most non-zero digit.
      o Rightmost non-zero digit in 11! = unit’s digit of \((8*1) = 8\)
      o Rightmost non-zero digit in 12! = unit’s digit of \((8*2) = 6\)
      o Rightmost non-zero digit in 13! = unit’s digit of \((6*3) = 8\)
      o Rightmost non-zero digit in 14! = unit’s digit of \((8*4) =\) unit’s digit of \(32= 2\)
      o Rightmost non-zero digit in 15! = ten’s digit of \((32*15) \)= ten’s digit of \((32*15) = 8\)
Thus, the correct answer is Option D.
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Re: What is the rightmost non-zero digit in 15!?  [#permalink]

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New post 25 Feb 2020, 23:09
Bunuel wrote:
What is the rightmost non-zero digit in 15!?

A. 2
B. 4
C. 6
D. 8
E. 9


Solution


    • \(15! = 1*2*3*4*5*6*7*8*9*10*11*12*13*14*15\)
      o \(15! =1*2*3*(2^2)*5*(2*3)*7*(2^3)*(3^2)*(2*5)*11*(2^2*3)*13*(2*7)*(3*5) = 2^{11}*3^6*5^3*7^2*11*13\)
    • \(2^3*5^3\) will give three 0, and we want rightmost nonzero, so ignoring it we get
    • Rightmost non – zero digit of 15! = unit’s digit of \((2^8*3^6*7^2*11*13) ………. (i)\)
      o Unit’s digit of \((7^2*11*13) =\) Unit’s digit of \((9*1*3) = 7 ……. (ii)\)
      o Cyclicity of powers of 3 is 4, so units digit of \(3^6 = 9 ………. (iii)\)
      o Also, Cyclicity of powers of 2 is 4, so units digit of \(2^8 = 6 …. (iv)\)
    • Thus, from (i), (ii), (iii) and (iv) we can write,
      o Rightmost non – zero digit of \(15! =\) unit’s digit of \((7*9*6) = 8 \)
Thus, the correct answer is Option D.
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What is the rightmost non-zero digit in 15!?  [#permalink]

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New post 26 Feb 2020, 03:06
Bunuel wrote:
What is the rightmost non-zero digit in 15!?

A. 2
B. 4
C. 6
D. 8
E. 9

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\(15!= 1*2*3*4*5*6*7*8*9*10*11*12*13*14*15\)

\(15!= 2^{11}*3^6*5^3*7^2*11*13 = (2*5)^3*2^8*3^6*7^2*11*13 = 10^3*6*9*9*1*3 = 10^3*8\)

i.e. 8 will be the rightmost non-zero digit

Answer: Option D
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Re: What is the rightmost non-zero digit in 15!?  [#permalink]

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New post 28 Feb 2020, 07:41
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Bunuel wrote:
What is the rightmost non-zero digit in 15!?

A. 2
B. 4
C. 6
D. 8
E. 9

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15! = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 x 11 x 12 x 13 x 14 x 15

= 2 x 3 x 2 x 2 x 5 x 6 x 7 x 8 x 9 x 2 x 5 x 11 x 12 x 13 x 14 x 3 x 5

= (2 x 2 x 2 x 5 x 5 x 5) x 3 x 6 x 7 x 8 x 9 x 2 x 11 x 12 x 13 x 14 x 3

As we can see from the above, 15! has 3 factors of 5; therefore, we pair them with 3 factors of 2 so that the 3 factors of 2 and 3 factors 5 (i.e., the factors in the parentheses) become 1000. This tells us that 15! has 3 trailing zeros, and the rightmost non-zero digit (i.e., the first non-zero digit to the left of the 3 trailing zeros) will be determined by the units digit of the product of the factors outside of the parentheses. So let’s determine that units digit by recording the units digits as we multiply them, starting from 3 x 6 and proceeding to the right:

3 x 6 → 8

8 x 7 → 6

6 x 8 → 8

8 x 9 → 2

2 x 2 → 4

4 x 11 → 4

4 x 12 → 8

8 x 13 → 4

4 x 14 → 6

6 x 3 → 8

We see that the rightmost non-zero digit is 8.

Answer: D
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Re: What is the rightmost non-zero digit in 15!?  [#permalink]

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New post 28 Mar 2020, 03:40
Bunuel wrote:
What is the rightmost non-zero digit in 15!?

A. 2
B. 4
C. 6
D. 8
E. 9

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Asked: What is the rightmost non-zero digit in 15!?

15! = 15*14*13*12*11*10*9*8*7*6*5*4*3*2 = 6*14*13*12*11*9*8*7*6*3*1000

Rightmost non-zero digit = Unit digit of 6*4*3*2*9*8*7*6*3 = Unit digit of 2*2*2 = 8

IMO D
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Re: What is the rightmost non-zero digit in 15!?  [#permalink]

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New post 28 Mar 2020, 03:42
Abhishek009 wrote:
Bunuel wrote:
What is the rightmost non-zero digit in 15!?

A. 2
B. 4
C. 6
D. 8
E. 9

Project PS Butler


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15/5 = 3 => A

Remainder is 0 =>B

So, Last Non zero digit will be \(2^A*A!*B!\)

= \(2^3*3!*0!\)

= 8* 6* 1

=>8 , Hence Answer must be (D) 8



Hi Abhishek009
Please explain highlighted part.
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Re: What is the rightmost non-zero digit in 15!?   [#permalink] 28 Mar 2020, 03:42

What is the rightmost non-zero digit in 15!?

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