Last visit was: 22 Apr 2026, 20:31 It is currently 22 Apr 2026, 20:31
Home
Messages
Tests
Schools
Events
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
User avatar
Smallwonder
User avatar
Joined: 31 Jan 2013
Last visit: 28 Oct 2013
Posts: 13
Own Kudos:
143
 [97]
Given Kudos: 18
Schools: ISB '15
WE:Consulting (Energy)
Schools: ISB '15
Posts: 13
Kudos: 143
 [97]
Find unit digit of N when N = 63^(1! + 2! + ... + 63!) + 18^(1! + 2! + 12 Sep 2013, 05:19
7
Kudos
Add Kudos
88
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

75% (hard)

Question Stats:

56% (02:18) correct 44% (02:21) wrong based on 578 sessions

History

Date
Time
Result
Not Attempted Yet
Find unit digit of N when \(N = 63^{(1! + 2! + ... + 63!)} + 18^{(1! + 2! + ... + 18!)} + 37^{(1! + 2! + ... + 37!)}\)

A) 2
B) 4
C) 6
D) 8
E) 0­­
ShowHide Answer
Official Answer
D
Most Helpful Reply
User avatar
Vetrik
User avatar
Joined: 04 Jul 2014
Last visit: 05 Jan 2022
Posts: 37
Own Kudos:
90
 [27]
Given Kudos: 40
Schools: Smeal" 20
Schools: Smeal" 20
Posts: 37
Kudos: 90
 [27]
Find unit digit of N when N = 63^(1! + 2! + ... + 63!) + 18^(1! + 2! + 09 Nov 2014, 10:22
17
Kudos
Add Kudos
10
Bookmarks
Bookmark this Post
Smallwonder
Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+...+18! + 37^1!+2!+...37!

A) 2
B) 437
D) 8
E) 0
Show more

The cyclicity of 3, 7 & 8 are 4 . Hence we need to divide the power by 4 and the reminder should be taken as the new power and from there unit digit of each of three terms needs to be found out. By adding the unit digit of the three terms we get the unit digit of the complete expression.

But how do find the reminder of the power divided by 4?

1 ! = 1 ; 2 ! = 2 ; 3 ! = 6 ...... 4 ! is 1*2*3*4 hence divisible by 4...similarly all the factorials above 4! will also be divisible by 4.

The power of 63 is 1+2+6+ 4 (1*2*3 + 1* 2*3*5 +.....) = 9 + 4(X) / 4 -> reminder is 1. Similarly the new powers of 18 & 37 are also 1.

Hence the unit digits are 63^ 1 + 18 ^ 1 + 37 ^ 1 -> 3+ 8+ 7 -> 18 ...so the last digit of is '8'

Hence Answer D
User avatar
mau5
User avatar
Verbal Forum Moderator
Joined: 10 Oct 2012
Last visit: 31 Dec 2024
Posts: 478
Own Kudos:
3,386
 [14]
Given Kudos: 141
Posts: 478
Kudos: 3,386
 [14]
Find unit digit of N when N = 63^(1! + 2! + ... + 63!) + 18^(1! + 2! + 12 Sep 2013, 05:30
10
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
Smallwonder
Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+...+18! + 37^1!+2!+...37!

A) 2
B) 4
C) 6
D) 8
E) 0
Show more

3,8 and 7 have a power cycle of 4, i.e. the units digit in each case will repeat after every 4th power.

For eg : 3^1 = 3, 3^2 = 9 , 3^3 = 27 , 3^4 = 81, 3^5 = 243

All the powers given(1!+2!+....), are multiples of 4. It is so because the last 2 digits of the total sum will be 00,for each one of them, which make them divisible by 4.

Thus, the given problem boils down to \(3^4+8^4+7^4 = 1+6+1 = 8\)

Thus, the units digit is 8.

D.
General Discussion
User avatar
shameekv
User avatar
Joined: 29 Aug 2013
Last visit: 11 Aug 2020
Posts: 50
Own Kudos:
186
 [8]
Given Kudos: 26
Location: United States
Concentration: Finance, International Business
GMAT 1: 590 Q41 V29
GMAT 2: 540 Q44 V20
GPA: 3.5
WE:Programming (Computer Software)
GMAT 2: 540 Q44 V20
Posts: 50
Kudos: 186
 [8]
Find unit digit of N when N = 63^(1! + 2! + ... + 63!) + 18^(1! + 2! + 12 Sep 2013, 05:56
3
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
Smallwonder
Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+...+18! + 37^1!+2!+...37!

A) 2
B) 4
C) 6
D) 8
E) 0
Show more

Considering 1! + 2! + 3! + 4! + 5! + ..... = 1 + 2 + 6 + (24 + 120 + 720 + XXX0 ... )

After 4! + any terms above will be divisible by 4 and hence the power term becomes
9 + (Term which is divisible by 4). Therefore the remainder will always be 1 in all the 3 cases.

Now considering the power cycle for 3,8 and 7 we get 3+8+7 = 18. Hence unit digit is 8.
User avatar
maggie27
User avatar
Joined: 25 Apr 2014
Last visit: 19 May 2015
Posts: 75
Own Kudos:
368
Given Kudos: 1,474
Posts: 75
Kudos: 368
Find unit digit of N when N = 63^(1! + 2! + ... + 63!) + 18^(1! + 2! + 04 Aug 2014, 11:17
Kudos
Add Kudos
Bookmarks
Bookmark this Post
mau5
Smallwonder
Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+...+18! + 37^1!+2!+...37!

A) 2
B) 4
C) 6
D) 8
E) 0

3,8 and 7 have a power cycle of 4, i.e. the units digit in each case will repeat after every 4th power.

For eg : 3^1 = 3, 3^2 = 9 , 3^3 = 27 , 3^4 = 81, 3^5 = 243

All the powers given(1!+2!+....), are multiples of 4. It is so because the last 2 digits of the total sum will be 00,for each one of them, which make them divisible by 4.

Thus, the given problem boils down to \(3^4+8^4+7^4 = 1+6+1 = 8\)

Thus, the units digit is 8.

D.
Show more

Hi ,
Can u please explain me how did we that last two digits of the total sum will be 00, for each of them?
User avatar
oss198
User avatar
Joined: 18 Jul 2013
Last visit: 01 Jan 2023
Posts: 68
Own Kudos:
431
 [2]
Given Kudos: 120
Location: Italy
Schools: LBS '21 Insead Sept19 Intake (A) IESE '21 (A$)
GMAT 1: 600 Q42 V31
GMAT 2: 700 Q48 V38
GPA: 3.75
Schools: LBS '21 Insead Sept19 Intake (A) IESE '21 (A$)
GMAT 2: 700 Q48 V38
Posts: 68
Kudos: 431
 [2]
Find unit digit of N when N = 63^(1! + 2! + ... + 63!) + 18^(1! + 2! + 24 Aug 2014, 02:19
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Does someone have an answer here? i still don't get how to solve it.
User avatar
mvictor
User avatar
Board of Directors
Joined: 17 Jul 2014
Last visit: 14 Jul 2021
Posts: 2,118
Own Kudos:
1,276
Given Kudos: 236
Location: United States (IL)
Concentration: Finance, Economics
Schools: Stanford '20 (WD)
GMAT 1: 650 Q49 V30
GPA: 3.92
WE:General Management (Transportation)
Products:
My Reviews
Schools: Stanford '20 (WD)
GMAT 1: 650 Q49 V30
Posts: 2,118
Kudos: 1,276
Find unit digit of N when N = 63^(1! + 2! + ... + 63!) + 18^(1! + 2! + 26 Feb 2016, 19:43
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I don't see how it can be 8..and this is how I tried to check...
power of 3 - units digit
1 - 3
2 - 9
3 - 7
4 - 1
so for every power divisible by 4, the units digit is 1.

now 8:
1 - 8
2 - 4
3 - 2
4 - 6
5 - 8
we can ID that for every power divisible by 4, the units digit is 6.

now, all the possible units digit for first one: 3,9,7,1
all possible units digit for second one: 8,4,2,6
3+8=11 - no
3+4=7 - no
3+2=5-no
3+6=9-no

9+8=17 - no
9+4=13 - no
9+2=11 - no
9+6=15 - no

7+8=15 - no
7+4=11 - no
7+2=9 - no
7+6 = 13 - no

1+8=9 - no
1+4=5 - no
1+2=3 - no
1+6 = 7 - no

all the possible answers - neither of the options is present. hm?

oh man..now I see where i made the mistake..I did not see 37^....
User avatar
ENGRTOMBA2018
User avatar
Joined: 20 Mar 2014
Last visit: 01 Dec 2021
Posts: 2,319
Own Kudos:
3,889
Given Kudos: 816
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE:Engineering (Aerospace and Defense)
Products:
My Reviews
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
Posts: 2,319
Kudos: 3,889
Find unit digit of N when N = 63^(1! + 2! + ... + 63!) + 18^(1! + 2! + 26 Feb 2016, 19:57
Kudos
Add Kudos
Bookmarks
Bookmark this Post
mvictor
I don't see how it can be 8..and this is how I tried to check...
power of 3 - units digit
1 - 3
2 - 9
3 - 7
4 - 1
so for every power divisible by 4, the units digit is 1.

now 8:
1 - 8
2 - 4
3 - 2
4 - 6
5 - 8
we can ID that for every power divisible by 4, the units digit is 6.

now, all the possible units digit for first one: 3,9,7,1
all possible units digit for second one: 8,4,2,6
3+8=11 - no
3+4=7 - no
3+2=5-no
3+6=9-no

9+8=17 - no
9+4=13 - no
9+2=11 - no
9+6=15 - no

7+8=15 - no
7+4=11 - no
7+2=9 - no
7+6 = 13 - no

1+8=9 - no
1+4=5 - no
1+2=3 - no
1+6 = 7 - no

all the possible answers - neither of the options is present. hm?

oh man..now I see where i made the mistake..I did not see 37^....
Show more

You are not taking 37^abc... into account while taking the sum. Your calculations are only taking 63^1!+2!+...+63! + 18^1!+2!....

Refer to find-unit-digit-of-n-when-n-159623.html#p1266283 for the solution.
User avatar
stonecold
User avatar
Joined: 12 Aug 2015
Last visit: 09 Apr 2024
Posts: 2,231
Own Kudos:
3,643
Given Kudos: 893
Schools: Boston U '20 (M)
GRE 1: Q169 V154
Schools: Boston U '20 (M)
GRE 1: Q169 V154
Posts: 2,231
Kudos: 3,643
Find unit digit of N when N = 63^(1! + 2! + ... + 63!) + 18^(1! + 2! + 17 Mar 2016, 07:00
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Superb Question...
Here the key is to write 1!+2+....as 9+4*p for some integer p
hence UD => UD of 3+7+8=> 8
hence D
User avatar
gauravk
User avatar
Joined: 29 Aug 2008
Last visit: 23 Jul 2020
Posts: 79
Own Kudos:
102
Given Kudos: 284
Schools: AGSM '18
Posts: 79
Kudos: 102
Find unit digit of N when N = 63^(1! + 2! + ... + 63!) + 18^(1! + 2! + 07 Nov 2016, 23:43
Kudos
Add Kudos
Bookmarks
Bookmark this Post
mau5
Smallwonder
Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+...+18! + 37^1!+2!+...37!

A) 2
B) 4
C) 6
D) 8
E) 0

3,8 and 7 have a power cycle of 4, i.e. the units digit in each case will repeat after every 4th power.

For eg : 3^1 = 3, 3^2 = 9 , 3^3 = 27 , 3^4 = 81, 3^5 = 243

All the powers given(1!+2!+....), are multiples of 4.It is so because the last 2 digits of the total sum will be 00,for each one of them, which make them divisible by 4.

Thus, the given problem boils down to \(3^4+8^4+7^4 = 1+6+1 = 8\)

Thus, the units digit is 8.

D.
Show more


I don't think the last digit of all powers would be 00, it would be 13 which will leave a remainder of 1 when divided by 4 and then the cyclicity rule can be applied.

We will get it at 3 + 7 + 8 = 18 and last digit would be 8.
avatar
aarushisingla
avatar
Joined: 23 May 2019
Last visit: 12 Sep 2022
Posts: 39
Own Kudos:
16
Given Kudos: 58
Posts: 39
Kudos: 16
Find unit digit of N when N = 63^(1! + 2! + ... + 63!) + 18^(1! + 2! + 26 Jun 2020, 15:12
Kudos
Add Kudos
Bookmarks
Bookmark this Post
mau5
Smallwonder
Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+...+18! + 37^1!+2!+...37!

A) 2
B) 4
C) 6
D) 8
E) 0

3,8 and 7 have a power cycle of 4, i.e. the units digit in each case will repeat after every 4th power.

For eg : 3^1 = 3, 3^2 = 9 , 3^3 = 27 , 3^4 = 81, 3^5 = 243

All the powers given(1!+2!+....), are multiples of 4. It is so because the last 2 digits of the total sum will be 00,for each one of them, which make them divisible by 4.

Thus, the given problem boils down to \(3^4+8^4+7^4 = 1+6+1 = 8\)

Thus, the units digit is 8.

D.
Show more

Hi,

What if we solve the question with another approach.?

What I have done is that I added 1! up to 4! which gives us 33 because after 4! the unit's digit will be 0 and so, if we add the numbers in each case the unit's digit will be 3.

Thus, it would look like 63^ 3 + 18^3 + 37^3 ( I have just written unit digit)
Hence, when we will solve the units digit will appear as 7+2+3 = 12 which gives us 2 as the answer. ( 3^3= 27, 8^3=512 , 7^3= 343)

Could you please tell me what is wrong with my approach?
User avatar
Oppenheimer1945
User avatar
Joined: 16 Jul 2019
Last visit: 21 Apr 2026
Posts: 786
Own Kudos:
663
Given Kudos: 236
Location: India
Schools: INSEAD '26 (D) ISB '27 (D) IIMB '26 (A) IIMA '26 (A) HEC '26 (D) LBS '26 IESE '27 (D) Fuqua '27 (WL) Tepper '27 (WL)
GMAT Focus 1: 645 Q90 V76 DI80
GPA: 7.81
Schools: INSEAD '26 (D) ISB '27 (D) IIMB '26 (A) IIMA '26 (A) HEC '26 (D) LBS '26 IESE '27 (D) Fuqua '27 (WL) Tepper '27 (WL)
GMAT Focus 1: 645 Q90 V76 DI80
Posts: 786
Kudos: 663
Find unit digit of N when N = 63^(1! + 2! + ... + 63!) + 18^(1! + 2! + 08 Feb 2024, 07:53
Kudos
Add Kudos
Bookmarks
Bookmark this Post
1!+2!+3!+4!...=4k+1

63^1!+2!+...+63! 
63^(4k+1): last digit (3 has a cycle of 3,9,7,1), =3


18^1!+2!+...+18! 
18^(4k+1): last digit (8 has a cycle of 8,4,2,6), =8

37^1!+2!+...37!
37^(4k+1): last digit (7 has a cycle of 7,9,3,1), =7

Unit's digit=3+8+7=18=8
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,965
Own Kudos:
1,117
Posts: 38,965
Kudos: 1,117
Find unit digit of N when N = 63^(1! + 2! + ... + 63!) + 18^(1! + 2! + 07 Apr 2026, 09:32
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderators:
Bunuel
Math Expert
109754 posts
Krunaal
Tuck School Moderator
853 posts