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

Question

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

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lacktutor · Accepted Answer

What is the rightmost non-zero digit in 15!?
 
—> \frac{15}{5}  ] ...= 3 (there are three zeroes at the end of 15!;  5^3 )

—>   +  +  +...= 7+3+1= 11  (—>  2^{11}  )

—>   +  +...= 5+1 = 6 ( —>  3^6 )

—>   +...= 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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GMATWhizTeam · Answer

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.

GMATWhizTeam · Answer

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.

GMATinsight · Answer

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

ScottTargetTestPrep · Answer

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

Kinshook · Answer

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

sanjitscorps18 · Answer

Let's calculate the zeroes in 15!

15/5 + 15/25 = 3 + 0 = 3

Hence we can take out 2^3 from 15! to make a 10 as 15! contains 5^3

Post this re-writing the factorial in expanded form

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

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

Removing the 10s and rewriting again

1 x 2 x 3 x 6 x 7 x 8 x 9 x 11 x 12 x 13 x 14 x 3

Multiplying the units digits we get the last number as 8

Option D

armaankumar · Answer

Prime Factorization of 15! = 2^11 * 3^6 * 5^3 * 7^2 * 11 * 13

Separating the pairs of 2s and 5s:
2^8 * 3^6 * 7^2 * 11 * 13 * (2*5)^3
Unit Digits: 
2^8: 6
3^6: 9
7^2: 9
11: 1
13: 3

Unit Digit of 6*9*9*1*3 = 8

bumpbot · Answer

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

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