Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 18 Jul 2019, 02:20 ### 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

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.  # If the number 5m15n, where m and n represent the thousands’ and unit

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 56261
If the number 5m15n, where m and n represent the thousands’ and unit  [#permalink]

### Show Tags

6
24 00:00

Difficulty:   95% (hard)

Question Stats: 47% (02:25) correct 53% (02:38) wrong based on 349 sessions

### HideShow timer Statistics If the number 5m15n, where m and n represent the thousands’ and unit digits, is divisible by 36, what is the maximum value of |m − n|?

(A) 1
(B) 3
(C) 5
(D) 6
(E) 8

_________________
Senior Manager  G
Joined: 28 May 2017
Posts: 282
Concentration: Finance, General Management
If the number 5m15n, where m and n represent the thousands’ and unit  [#permalink]

### Show Tags

13
2
Bunuel wrote:
If the number 5m15n, where m and n represent the thousands’ and unit digits, is divisible by 36, what is the maximum value of |m − n|?

(A) 1
(B) 3
(C) 5
(D) 6
(E) 8

The number 5m15n is divisible by 36, So the number has to be divisible by 9 as well.

So, 11+m+n has to be divisible by 9.
11+m+n = 18 or 27
(Any higher multiple is not possible because m & n are single digit positive integers)
m+n = 7 or 16

The number 5m15n is divisible by 36, So the number has to be divisible by 4 as well. For a number to be divisible by 4, last 2 digits should be divisible by 4.
Last 2 digits of 5m15n = 5n
So, n could be either 2 or 6

Let's have a look at probable solutions
n m n+m Possible (P)/ Not Possible (NP)
2 5 7 P
6 1 7 P
2 14 16 NP since m is a single digit positive integer
6 10 16 NP since m is a single digit positive integer

max |m-n| = |1-6| = 5

_________________
If you like the post, show appreciation by pressing Kudos button
##### General Discussion
Manager  B
Joined: 17 Jun 2015
Posts: 201
GMAT 1: 540 Q39 V26 GMAT 2: 680 Q46 V37 If the number 5m15n, where m and n represent the thousands’ and unit  [#permalink]

### Show Tags

For a number to be divisible by 36, it must have two 2s and two 3s i.e. to be divisible by 4 and 9. So an even number with the last two digits divisible by 4 and the sum of digits to be divisible by 9.

The sum of digits is: 5 + m + 1 + 5 + n = 11 + m + n (divisible by 9 and the last two digits (5,n) divisible by 4)

A number is divisible by 4, if the last two digits are divisible by 4. So, that means, the number has to end with 52 or 56.

5 + m + 1 + 5 + 2 = 13 + M
5 + m + 1 + 5 + 6 = 17 + N

a number is divisible by 9 if the sum of the digits is divisible by 9. However, the question asks for | m - n | to be maximum.

5 + m + 1 + 5 + 2 = 13 + M

The max value of | m - n | in this case is possible when m-9. But the number wouldn't be divisible by 9. The maximum possible, keeping into perspective the divisibility by 9, would be when m=5. In which case | m - n | would be 3.

5 + m + 1 + 5 + 6 = 17 + N
In the other case, similarly, the maximum value of | m - n | would be when m= 0. But then, the number wouldn't be divisible by 9. So we are looking for a number that is as far from 6 as possible. 1 suffices the condition.

Hence, the maximum possible value for |m-n| = 5

Shoot me if I'm incorrect. I'm not confident though. Generally, I don't get answers right.
_________________
Fais de ta vie un rêve et d'un rêve une réalité
Intern  B
Joined: 21 Jun 2013
Posts: 25
Re: If the number 5m15n, where m and n represent the thousands’ and unit  [#permalink]

### Show Tags

I think answer is D and this is how I solved it!
5m51n - Since the no. is divisible by 36, it should be divisible by 2 & 3
Thus, N should be even and sum of integers should be divisible by 3 i.e m+n+5+5+1 should be multiple of 3

5+5+1 already makes 11 & thus we can look for the following multiples of 3 & keep looking for max. value of m-n
12 -> n = 0; m = 1 ; m-n = 1 (taking n=0 because we have to find maximum value of m-n)
15 -> n = 0; m = 4 ; m-n = 4
18 -> n = 0; m = 7; m-n = 7
21 -> n = 2; m = 8; m-n = 6 (n can't be zero as m has to be single digit & can't take 10 as value)

since we got max difference as 6, D is correct. please weight in my answer and suggest if went wrong anywhere.
Manager  B
Joined: 20 Jun 2014
Posts: 50
GMAT 1: 630 Q49 V27 GMAT 2: 660 Q49 V32 Re: If the number 5m15n, where m and n represent the thousands’ and unit  [#permalink]

### Show Tags

Since the number has to be divisible by 36 => 2^2 and 3^2
so lets check how it can be made divisible by 9 first
5+m+1+n+5 = 11+m+n
For it to be divisble by 9 => m+n = 7(ie. 18-11) or 16(ie. 27-11)
m-n has to be maximised so if we consider 16 then m & n (in any order) can be {8,8}, {9,7} the latter case is not possible since the number has to be divisible by 4 also and in units's place we cannot have odd number.
Also 0 is not maximum value (in case m,n are 8,8) so lets ignore it.
Lets see how we can get 7 from m ,n => {3,4},{6,1} {2,5} in any order
since 6-1 =5 is maximum value , we can check it for divisibility
lets check 51156 it is clearly divisible by 4 from last two digits. Hence |m-n| max = 5
Option C.

Experts please let me know if the above solution has any flaws or assumptions.
Intern  B
Joined: 16 Feb 2017
Posts: 19
Re: If the number 5m15n, where m and n represent the thousands’ and unit  [#permalink]

### Show Tags

2
Bunuel wrote:
If the number 5m15n, where m and n represent the thousands’ and unit digits, is divisible by 36, what is the maximum value of |m − n|?

(A) 1
(B) 3
(C) 5
(D) 6
(E) 8

Finding the answer with logic and no calculation as below

For a number to be divisible by 36, it should be divisible by 6 and 6.
And for a number to be divisible by 6, it should be divisible by both 2 and 3.

Now all multiple of 6 are EVEN.
This means that 'n' has to be even.
With this we can say that n can be either of 0,2,4,6,8
and m can be either of 1,3,5,7,9

Next, we have been asked for |m-n|
Going to Odd/Even addition/subtraction rules $$O-E=O$$

This means that the greatest odd value from 1 to 9 shall be our answer.

As 9 and 7 are not present, are C has value=5, it becomes our answer.

Hope this helps!
Current Student S
Joined: 23 Jul 2015
Posts: 152
If the number 5m15n, where m and n represent the thousands’ and unit  [#permalink]

### Show Tags

When nothing clicks, use brute force

Divisibility rule for 36 =
Divisibility rule for 9 - sum of digits must be divisible by 9
&&
Divisibility rule for 4 - last 2 digits divisible by 4 & unit digit cannot be odd

--> it follows, m+n = 7 & n is even
$\begin{matrix} n & m & divisibile? & |m -n| \\ 0 & 7 &no & 7 \\ 2 & 5 & yes & 3 \\ 4 & 3 & no & \\ 6 & 1 & yes & 5 \\ 8 & -1 & NV & \\ \end{matrix}$

Non-Human User Joined: 09 Sep 2013
Posts: 11689
Re: If the number 5m15n, where m and n represent the thousands’ and unit  [#permalink]

### Show Tags

1
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.
_________________ Re: If the number 5m15n, where m and n represent the thousands’ and unit   [#permalink] 19 Aug 2018, 00:23
Display posts from previous: Sort by

# If the number 5m15n, where m and n represent the thousands’ and unit  