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

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

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New post 28 Jun 2017, 01:30
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C
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Question Stats:

48% (02:12) correct 52% (02:18) wrong based on 304 sessions

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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

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

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New post 28 Jun 2017, 01:49
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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


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

Hence Answer C
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If the number 5m15n, where m and n represent the thousands’ and unit  [#permalink]

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New post 02 Jul 2017, 01:10
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

Answer is C.

Shoot me if I'm incorrect. I'm not confident though. Generally, I don't get answers right.
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Re: If the number 5m15n, where m and n represent the thousands’ and unit  [#permalink]

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New post 02 Jul 2017, 05:55
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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New post 08 Jul 2017, 01:59
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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New post 08 Jul 2017, 04:55
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!
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If the number 5m15n, where m and n represent the thousands’ and unit  [#permalink]

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New post 08 Jul 2017, 05:26
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
\[
\begin{matrix}
n & m & divisibile? & |m -n| \\
0 & 7 &no & 7 \\
2 & 5 & yes & 3 \\
4 & 3 & no & \\
6 & 1 & yes & 5 \\
8 & -1 & NV & \\
\end{matrix}
\]


Therefore answer is C
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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 &nbs [#permalink] 19 Aug 2018, 00:23
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