Last visit was: 10 Jul 2025, 10:39 It is currently 10 Jul 2025, 10:39
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.
Close
Request Expert Reply
Confirm Cancel
User avatar
CareerGeek
Joined: 20 Jul 2017
Last visit: 09 Jul 2025
Posts: 1,295
Own Kudos:
4,073
 [29]
Given Kudos: 162
Location: India
Concentration: Entrepreneurship, Marketing
GMAT 1: 690 Q51 V30
WE:Education (Education)
GMAT 1: 690 Q51 V30
Posts: 1,295
Kudos: 4,073
 [29]
3
Kudos
Add Kudos
26
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
GMATinsight
User avatar
Major Poster
Joined: 08 Jul 2010
Last visit: 09 Jul 2025
Posts: 6,375
Own Kudos:
15,574
 [11]
Given Kudos: 128
Status:GMAT/GRE Tutor l Admission Consultant l On-Demand Course creator
Location: India
GMAT: QUANT+DI EXPERT
Schools: IIM (A) ISB '24
GMAT 1: 750 Q51 V41
WE:Education (Education)
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Schools: IIM (A) ISB '24
GMAT 1: 750 Q51 V41
Posts: 6,375
Kudos: 15,574
 [11]
1
Kudos
Add Kudos
9
Bookmarks
Bookmark this Post
General Discussion
User avatar
preetamsaha
Joined: 14 Oct 2019
Last visit: 18 Jan 2025
Posts: 338
Own Kudos:
485
 [2]
Given Kudos: 127
Status:Today a Reader; Tomorrow a Leader.
Location: India
GPA: 4
WE:Engineering (Energy)
Posts: 338
Kudos: 485
 [2]
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
User avatar
rishab0507
Joined: 12 Mar 2019
Last visit: 25 Feb 2021
Posts: 181
Own Kudos:
Given Kudos: 105
Posts: 181
Kudos: 106
Kudos
Add Kudos
Bookmarks
Bookmark this Post
preetamsaha
METHOD I : Euler no of 9 = 9(1-1/3) = 6
rem [(25^18) /9]
= rem [(25^0)/9] [ 18 is divisible by 6 ]
= rem [1/9]
= 1

METHOD II : rem [(25^18 )/9]
=rem [ (-2)^18 /9] [ rem (25/9) = -2 ]
=rem [ (+2)^18 /9]
=rem [ (2^6) * ( 2^6 )* (2^6 )/9]
=rem [ 64*64*64 /9]
=rem [1*1*1 /9]
=1

METHOD III : rem [(25^18 )/9]
= rem [ (27-2)^18/9] [ using binomial theorem ]
=rem [ (-2)^18 /9] [ considering only last term since rest all terms will be divisible by 9 ]
=rem [ (+2)^18 /9]
=rem [ (2^6) * ( 2^6 )* (2^6 )/9]
=rem [ 64*64*64 /9]
=rem [1*1*1 /9]
=1

correct answer is A

hi, can you tell about euler theorem, What i know is
E(z)=z∗[1–1/P]∗[1–1/Q], where P and Q are prime factors of denominator
do we need 1 factor or 2 prime factors. In case we don't have 2 prime factors, Is 1 prime factor ok
avatar
omavsp
Joined: 20 Aug 2017
Last visit: 28 Jan 2024
Posts: 35
Own Kudos:
13
 [1]
Given Kudos: 191
Products:
Posts: 35
Kudos: 13
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
25^18/9 =
(18+7)^18/ 9 =
7^18 > 49^9 =
(45 + 4)^9 =
4^9 = 64^3 =
(63 + 1)^3
1^3 = 1
1/9 = r1
User avatar
CrackverbalGMAT
User avatar
Major Poster
Joined: 03 Oct 2013
Last visit: 10 July 2025
Posts: 4,847
Own Kudos:
8,628
 [4]
Given Kudos: 225
Affiliations: CrackVerbal
Location: India
Expert
Expert reply
Posts: 4,847
Kudos: 8,628
 [4]
1
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
A method which can be used in such a case is using cyclicity of remainders. In this we see how many cycles are required to get a remainder of 1.

Interestingly, if you get a remainder of -1, then the double of the cycles gives a remainder of 1.

We get a remainder of -1, if the remainder is 1 less than the divisor (eg, 31 divided by 8 gives a remainder of 7 which is also -1)

We stop the cycle, when we get a remainder of 1 or 8. If we get a remainder of 8, then double that cycle will give a remainder of 1.

\(R[\frac{(25)^{18}}{9}] = R[\frac{5^{36}}{9}]\)


Cycle 1: \(R[\frac{5}{9}] = 5\)

Cycle 2: We multiply the remainder of the previous cycle by 5 and get the next remainder. \(R[\frac{5 * 5}{9}] = R[\frac{25}{9}] = 7\)

Cycle 3: \(R[\frac{5 * 7}{9}] = R[\frac{35}{9}] = 8 \space or -1\)

Therefore, we shall get a remainder of 1 in the 6th cycle. This means that the remainder for every 6 powers of 5 divided by 9 = 1


\(R[\frac{5^6}{9}] = 1\)


Therefore \(R[\frac{5^{36}}{9}] = R[\frac{(5^6)^6}{9}] = R[\frac{1}{9}] = 1\)



Option A

Arun Kumar
User avatar
Fdambro294
Joined: 10 Jul 2019
Last visit: 06 Apr 2025
Posts: 1,353
Own Kudos:
705
 [1]
Given Kudos: 1,658
Posts: 1,353
Kudos: 705
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Concept 1: we can 'SPLIT' the Dividend and Multiply the Remainders, removing any Excess Remainders at the end by continually Dividing by the Divisor = 9


Concept 2: The Divisibility and Remainder Rule for 9:
if the Digits Sum to = 9k -------- # is Divisible by 9
if the Digits Sum to = 9k + 1 --------- Rem is +1 when Divided by 9
if the Digits Sum to = 9k + 2 ------- Rem is +2 when Divided by 9

***where k = NON Negative Integer


(1st) Re-write the Divisor/NUM

(25)^18 /9 -----> (25^2)^9 /9 -------> (625)^9 /9

(625)^9 /9 ----> Remainder of = ?


(2nd) Multiply the Remainders after "SPLITTING" the Exponential Term in the Dividend/NUM

[ (625)^9 /9 ] Rem of = (625/9)Rem of * (625/9)Rem of * (625/9)Rem of * (625/9)Rem of .......Multiplied 9 TIMES


Each "Part" Remainder of is ----> (625/9) Rem of ----> which yields the following Remainder:

6 + 2 + 5 = 13 = 9k + 4 ------> which means 625 when Divided by 9 yields a Remainder = +4


Excess Remainder = (4)^9 / 9


(3rd)Remove the Excess Remainder by finding a way to re-write the Dividend/NUM in terms of an Expression that is +1 or -1 away from a Multiple of 9

SPOT: (4)^3 = 64

64 = 63 + 1 ------- which is +1 MORE than the Multiple of 9 = 63


(4)^9 /9 ----> Rem of = ?

[(4)^3 * (4)^3 * (4)^3 ] / 9 ----> Rem of =

( 4^3 / 9)Rem of * ( 4^3 /9) Rem of * ( 4^3 /9) Rem of =

(64 / 9)Rem of * (64 / 9)Rem of * (64 / 9) Rem of =


EACH "Part" Remainder being Multiplied =

(63 + 1) / 9 ---> Remainder of = the same as -----> (+1) / 9 = Remainder of 1

1 * 1 * 1 = Remainder of 1

-A-
User avatar
Abhishek009
User avatar
Board of Directors
Joined: 11 Jun 2011
Last visit: 21 Apr 2025
Posts: 5,965
Own Kudos:
5,154
 [1]
Given Kudos: 463
Status:QA & VA Forum Moderator
Location: India
GPA: 3.5
WE:Business Development (Commercial Banking)
Posts: 5,965
Kudos: 5,154
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Dillesh4096
Find the remainder when \(25^{18}\) is divided by 9.

A. 1
B. 2
C. 3
D. 5
E. 6
625/9 = Rem 4

Thus we have 4^9/9 = 64^3/9 = Remainder 1, Answer is clean (A) 1
User avatar
believerinthy
Joined: 30 Nov 2019
Last visit: 16 Mar 2023
Posts: 29
Own Kudos:
6
 [1]
Given Kudos: 77
Concentration: Finance, Strategy
Products:
Posts: 29
Kudos: 6
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Can someone please help me with this question.

I am doing everything as told above but my cyclicity of 2^18/9 is of 4.
and when im diving 18/4 = 2, which leaves it's unit digit to be 4 and not 8 as told above.
avatar
DongminShin
Joined: 12 Nov 2020
Last visit: 27 Jun 2021
Posts: 9
Own Kudos:
8
 [1]
Given Kudos: 26
Posts: 9
Kudos: 8
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
omavsp
25^18/9 =
(18+7)^18/ 9 =
7^18 > 49^9 =
(45 + 4)^9 =
4^9 = 64^3 =
(63 + 1)^3
1^3 = 1
1/9 = r1

Thank you for this amazing approach!
I have never seen this kind of creative approach before👍

Posted from my mobile device
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 37,375
Own Kudos:
Posts: 37,375
Kudos: 1,010
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Moderators:
Math Expert
102619 posts
PS Forum Moderator
685 posts