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

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Re: If the number 5m15n, where m and n represent the thousands’ and unit [#permalink]
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.
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Re: If the number 5m15n, where m and n represent the thousands’ and unit [#permalink]
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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.
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Re: If the number 5m15n, where m and n represent the thousands’ and unit [#permalink]
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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!
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If the number 5m15n, where m and n represent the thousands’ and unit [#permalink]
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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
$\begin{matrix} n & m & divisibile? & |m -n| \\ 0 & 7 &no & 7 \\ 2 & 5 & yes & 3 \\ 4 & 3 & no & \\ 6 & 1 & yes & 5 \\ 8 & -1 & NV & \\ \end{matrix}$

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Re: If the number 5m15n, where m and n represent the thousands’ and unit [#permalink]
GMATAspirer09 wrote:
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!

How we can say, m will be odd?
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Re: If the number 5m15n, where m and n represent the thousands’ and unit [#permalink]
GMATAspirer09 wrote:
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!

How we can say, m will be odd?

sampad, it just happens to be true because of the numbers that are given. IMO, this question is best solved by using the divisibility rule of 4 and 9 and won't take more than a minute if you know these rules.
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If the number 5m15n, where m and n represent the thousands’ and unit [#permalink]
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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

For the number to be divisible by 36, it must be divisible by 4 and 9.

An integer is divisible by 4 if its last two digits form a multiple of 4.
Here, the last two digits are 5n.
Between 50 and 59 there are two multiples of 4:
52 and 56
Thus, n=2 or n=6.

An integer is divisible by 9 if its digit sum is a multiple of 9.
Here, the digit sum = 5+m+1+5+n = 11+m+n

Case 1: n=2 --> digit sum = 11+m+2 = 13+m
The digit sum will be a multiple of 9 if m=5:
13+5 = 18
In this case, |m-n| = |5-2| = 3

Case 2: n=6 --> digit sum = 11+m+6 = 17+m
The digit sum will be a multiple of 9 if m=1:
17+1 = 18
In this case, |m-n| = |1-6| = 5

The greatest possible value for |m-n| is yielded by Case 2.

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If the number 5m15n, where m and n represent the thousands’ and unit [#permalink]
Did in a much simpler way (not sure if it's correct - Anyone could help here would be good)

36 = 2^2*3^2 -> max delta = 3^2-2^2 = 5

Is this a proper way of solving this, or was just a strike of luck?
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Re: If the number 5m15n, where m and n represent the thousands’ and unit [#permalink]
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

Solution:

If a number is divisible by 36, then it’s divisible by 4 and 9. Recall that to be divisible by 4, the last two digits of the number have to be divisible by 4. To be divisible by 9, the sum of the digits of the number has to be divisible by 9.

Since the tens digit is 5, then n has to be either 2 or 6 since 52 and 56 are each divisible by 4. Now let’s look at these two cases.

Case 1: n = 2

If n is 2, then the sum of the digits, including n but excluding m, is 5 + 1 + 5 + 2 = 13. Therefore, we see that m must be 5 in order to the number to be divisible by 9. In this case, |m - n| = |5 - 2| = 3.

Case 2: n = 6

If n is 6, then the sum of the digits, including n but excluding m, is 5 + 1 + 5 + 6 = 17. Therefore, we see that m must be 1 in order to the number to be divisible by 9. In this case, |m - n| = |1 - 6| = 5.

Therefore, the maximum value of |m - n| is 5.

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Re: If the number 5m15n, where m and n represent the thousands’ and unit [#permalink]
Tapesh03 wrote:
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.

hey, instead of 2 and 8 for 21, we can assume n,m to be 1 and 9 and thereby making the difference 8 right?
also, if the sum is 21, how would it be divisible by 2? I cant understand why the max difference is 6?
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Re: If the number 5m15n, where m and n represent the thousands and unit [#permalink]
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Re: If the number 5m15n, where m and n represent the thousands and unit [#permalink]
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