If R = 1! + 2! + 3! …. 199!, what is the units digit of R?

Question

A. 0
B. 1
C. 2
D. 3
E. 4

chetan2u · Accepted Answer

Few concepts..
R=1!+2!+3!+4!+5!+6!+7!+8!+9!+10!+.....
1) units digit...
All terms after 4! contain one 5 and one 2 atleast, so units digit of all terms after 4! Will be 0..
So units digit will depend on 1!+2!+3!+4! Only
2) last two digits..
All terms after 9! contain two 5 and two 2 atleast, so last 2- digit of all terms after 9! Will be 00.
So last two digits will depend on 1!+2!+...+9! Only

Units digit here =1+2+3*2+4*3*2=1+2+6+24=33
Hence 3..

Some more questions on same concept..
 https://gmatclub.com/forum/if-n-is-a-positive-integer-what-is-the-last-digit-of-94088.html ..
 https://gmatclub.com/forum/what-is-63908.html

pushpitkc · Answer

1! = 1 | 2! = 2 | 3! = 6 | 4! = 24 | 5! = 120 | 6! = 720

From 5! to 199! the units digit will end with 0.
R = 1 + 2 + 6 + 24 + 120 + 720 ....... = 33 + 840 +... (will always end with 3)

Therefore, the units digit of R is  3(Option D)

GMATWhizTeam · Answer

Solution

•  1! = 1  hence unit’s digit of  1! = 1 
• So, units’ digit of 2! = unit’s digit of  1! *2 = 1*2 =2 
• Units’ digit of 3! = units’ digit of  (2! *3)  = unit’s digit of  (2*3) = 6 
• Unit’s digit of 4! = units’ digit of  (3! *4) =  unit’s digit of  (6*4) = 4  
• Unit’s digits of 5! = unit’s digit of  (4! *5) =  unit’s digit of  (4*5) = 0 
• So, Unit’s digit of 6! = unit’s digit of  (5! *6) =  unit’s digit of  (0*6) = 0  
• Thus, unit’s digit of all the numbers from 5! To 199! will be 0.
• Hence, unit’s digit of R = (unit’s digit of 1! ) + (unit's digit of 2!) + (Unit's digit of 3!) + (Unit’s digit of 4!)
 o Or, unit’s digit of R = unit’s digit of  (1 +2 +6 +4) = 3   
Thus, the correct answer is  Option D.

Alternate solution,

• We know that 5! = 1*2*3*4*5
 o Since, it contains 2*5 unit’s digit must be 0.
o Hence for all n! where  n ≥ 5  and n is positive integer, unit’s digit will be 0. 
• Thus, unit’s digit of R = Unit’s digit of  (1! +2! +3! +4!)  = Unit’s digit of  (1+2+6+24 ) = 3  
Thus, the correct answer is  Option D.

KalyanChamarthi · Answer

1! = 1
2! = 2
3! = 6
4! = 24 
Total = 33
Now
from 5! all the numbers until 199! - unit digits is 0

So when you add 33 with any number with unit digits 0 will always be "3"

Ans : D ( 3)

Abhishek009 · Answer

1! = 1
2! = 2
3! = 6
4! = 24

Rest all will result in trailing 0's in the end

5! = 120
6! = 720

So, Add 1! + 2! + 3! + 4!= 1 + 2 + 6 + 24 => 33, Thus Answer must be (D) 3

Elon_Musk · Answer

Step 1: Understand when factorials end in 0 
For any n≥5 n! is divisible by 10 (since n! contains both 2 and 5 → product ends in 0. So, units digit of n! for n ≥ 5 is 0

That means: 
5! = 120 ⇒units digit 0
6! = 720 ⇒ units digit 0⋮
199! ⇒units digit 0

So all terms from 5! to 199! contribute nothing to the units digit of R.

Step 2: Add only the first few factorials (from 1! to 4!) 
1! = 1
2! = 2
3! = 6
4! = 24
Now add them:

1+2+6+24=33
So, the units digit of R = units digit of 33 = 33

Final Answer: D. 3

If R = 1! + 2! + 3! …. 199!, what is the units digit of R?

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If R = 1! + 2! + 3! …. 199!, what is the units digit of R?

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