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

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

I would cut out of 20! all the 10s and then calculate the unit digit of the resulting number. It appears to be 4.
Hence c




Parvezahmed2210 Can you please elaborate more the formula used?
Is the 5 a constant that can be always applied?

https://learningroots.in/cat-and-omet/qu ... ero-digit/

Posted from my mobile device

Given: 20! = (20)(19)(18)(17)(16)(15)(14)(13)(12)(11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)

Since one 2 and one 5 multiply to get 10 (causing the product to end in 0), let's the first factor out all of the 5's to get:
20! = (5)(4)(19)(18)(17)(16)(5)(3)(14)(13)(12)(11)(5)(2)(9)(8)(7)(6)(5)(4)(3)(2)(1)

There are four 5's in this product.
So, let's highlight four 2's that can pair with the four 5's to get all of our zeros.
20! = (5)(4)(19)(18)(17)(16)(5)(3)(14)(13)(12)(11)(5)(2)(9)(8)(7)(6)(5)(4)(3)(2)(1)

(5)(4)(5)(5)(2)(5)(2) = 10,000
So, those colored values are the ones that contribute to the product ending in four zeros.
Since our goal now is to identify the rightmost non-zero digit, we can ignore all of those colored values.
This results in the following product: (19)(18)(17)(16)(3)(14)(13)(12)(11)(9)(8)(7)(6)(4)(3)(1)
So the rightmost non-zero digit will be the unit's digit of the above product.

From here it's just a matter of keeping track of the unit's digit as we multiply each successive value in the product.
For example, 19 x 18 has units digit 2.
Now multiply 2 by the next number to get 2 x 17, which has units digit 4
Now multiply 4 by the next number to get 2 x 16, which has units digit 2
And so on.

When we do all of this, we end up with units digit 4.

Answer: C

Cheers,
Brent

Logically, the Formula has the Effect of Discounting/Removing all of the 5 Prime Factors that create the Trailing Zeroes in the N!

Step 1: Take the Number Value (here 20) that is in the Factorial! and Divide it by 5 ---- Set up the Remainder Equation

20/5 = 4 + 0 /5

20 = 5 (4) + 0

Step 2: for the Remainder Equation ---- N = 5 (a) + b

To find the Right-most NON-Zero Digit of any N! ---- Find the Units Digit of: (2^a * a! * b!) ----

Step 3: Calculate the Right-most NON-Zero Digit of 20!

in this example: a = 4 and b = 0 (remember 0! = 1)

Units Digit of (2^4th * 4! * 0!) = 16 * 24 = Units Digit of 4

This is the Right Most NON-Zero Digit of 20! ----- 4

it works for Lower Valued Factorials. If the Factorial becomes a higher value (such as 79!) you write the Remainder Equation by dividing 25 instead of 5 and use a base of 4 instead of 2.

hope that makes sense??? it's not something that is going to come up on the GMAT, it seems like more of an Indian CAT type Rule

Brent's Way seems more like the "GMAT way" to logic through this problem, rather than learning an obscure formula.

Hi,
Your method is as usual very easy to understand. However, in this case, I think it is too long so there is a huge margin for error. Do you have any shorter method ? Thanks in advance.

Response:
I don't know of a shorter method which will get us to the correct answer; however, this method looks long only because I wrote everything in detail and provided explanations along the way. During the test, it is possible to skip a lot of steps and arrive at the answer faster.

For instance, we don't need to write every factor of 20! to find the prime factorization of 20!. We could start from the greatest prime factor of 20! (which is 19) and determine the exponents of the prime factors all the way down to 2. The exponents of 19, 17, 13, and 11 will be 1 since 20! contains only one factor of each. There are two factors of 7 (one in 7 and one in 14), so the exponent of 7 will be 2. There are four factors of 5 (one in each of 5, 10, 15, and 20). It just remains to determine the exponents of 3 and 2, for which we can use the shortcut where we divide 20 by 3 and keep dividing the quotients by 3 until we obtain a zero quotient. Same goes for 2. Using this process, we obtain the prime factorization of 20! easily.

Once we write 20! = 2^18 x 3^8 x 5^4 x 7^2 x 11 x 13 x 17 x 19 and we remove the factors of 10, it remains to find the units digit of 2^14 x 3^8 x 7^2 x 11 x 13 x 17 x 19. Here, we observe that the units digit of 11 is 1, so we can just remove it from the product without changing the value of the units digit. Further, the units digit of the product 13 x 17 is also 1, so we can remove those factors as well. Finally, the units digit of 7^2 is 9 and thus, the units digit of 7^2 x 9 is 1. Removing all these factors, it just remains to calculate the units digit of 2^18 x 3^8. At this point, we can just use units digit patterns of 2 and 3 to find the answer.

While this solution is still a bit longer compared to the solutions of other questions, I think it is a reasonable enough solution.

Thanks BrentGMATPrepNow for the detailed solution.

I am wondering whether problems such as this can be solved within 2 minutes in exam. Can you please recommend how to approach this or similar problem in exam situation w.r.t timing ?

Keep in mind that explaining the solution takes a lot longer than actually solving the question.
Although this is a time-consuming question, I think it can be answered in under 2 minutes if we start our calculations quickly.

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