If the number 12341234B1234A, in which A and B represent digits, is di

Question

If the number 12341234B1234A, in which A and B represent digits, is divisible by 6, then what is the maximum value of |A−B|?

a) 9
b) 8
c) 7
d) 6
e) 5

Temurkhon · Accepted Answer

All number should be divisible by 3 and 2. We can see that sum of digits without A and B is already divisible by 3, 
so we should care that sum of A and B is divisible by 3

We should highest/lowest A - lowest/highest B. Maximum possibility is 9, so let's start testing from it

Picking: A=0 and B=9, it is divisible by 2 (because even) and divisible by 3 (sum is divisible by 3)

A

mvictor · Answer

tricky one, as I tried to make some assumptions which were way too unnecessary. clearly if A=0, then the number is divisible by 2. to be divisible by 3, B can be 0, 3, 6, 9. maximum value of B is 9, so |0-9|=9. note that A can't be 9, because in this case, the number would not be divisible by 2, and thus, not divisible by 6.
good question.

keats · Answer

Bunuel   VeritasPrepKarishma

Where am I going wrong?
a+b+30 should be divisible by 6. This means a+b should be divisible by 6. So, possibilities of a+b can be 6,12,18

Max difference between a and b can be (6-0) = 6

Please assist.

keats · Answer

chetan2u  Can you please help me with the above query ^^ Thank you in advance.

chetan2u · Answer

Hi
a+b+30 is div by 6...
Where you are going wrong is that a+b+30 should be div by 3...
So a+b can be 9, 12, 15 ,18 etc..

But we are looking at a-b....
Since a and b are single digit number, a-b can be max 9....
if we take a as 0, the number 1234...... will be EVEN and then b can take max value of 9 and still be div by 3..
A number div by 2 and 3 will be div by 6..

So |a-b|=|0-9|=9
A

stonecold · Answer

If a number is divisible by 6 it must be divisible by both 2 and 3
so A must be even.
A=> {0,2,4,6,8}
sum of digits must be a multiple of 3
30+A+B => let A=9 B=0 => |9-0|=9
Hence A

Roamer · Answer

Three times sum of 1+2+3+4 =10*3=30
Now, the expression becomes 30+A+B.What we want is a number divisible by 6( which as per divisibility rule should be divisible by both 2 and 3), ending with an even number or zero.
So, by hit and trial our number can be 30+0+9= 39
Hence, 9-0 = 9.
It can be maximum difference.Hence A.

Nightmare007 · Answer

If number divisible by 6- Sum of digits divisible by 3 and should be even.

1+2+3+... A + B should be divisible by 3

A+B should be divisible by 3

and B is even.

Max A-B ----

B is 0 
A is 9
Answer should be 9

A is Answer

KarishmaB · Answer

Responding to a pm:

A and B can take values from 0 to 9. We need them to be as far apart as possible so get the maximum value of |A - B| since absolute values show distance between the two. The maximum distance between them can be 9 (if one of them is 9 and other is 0) and the minimum distance between them would be 0 if both A and B must take the same values. Let's find out.

Divisibility by 6 - The number would be divisible by both 2 and 3.
To be divisible by 2, the number's units digit should be 0/2/4/6/8 - value of A

To be divisible by 3, the sum of all digits should be divisible by 3. 
1+2+3+4+1+2+3+4+1+2+3+4+A+B = 30 + A + B
(or we can make groups of multiples of 3 and keep ignoring them)

Since 30 is divisible by 3, we need A + B to be divisible by 3 too. If A = 0, B can be 9 which makes A + B divisible by 3.

Hence A should be 0 and B should be 9.

Answer (A)

johnymaung · Answer

A lot of people have taken the max value of A as 9, which is incorrect as the last digit needs to be even to be divisible by 2, thus A will be 0 and B will be 9, as posted by Karishma.

KrishnakumarKA1 · Answer

Solution:

Given:  The number is 12341234B1234A, in which A and B are digits.
 To find:  The maximum value of |A−B|?
 Approach:  Here A and B can take the value from 0 to 9 and also the number must be divisible by 6. If the number has to be divisible by 6 it has to be divisible by both ‘2’ and ‘3’ also. 
 Divisibility test for 2:  The unit digit has to be an even number i.e.; 0/2/4/6/8.
 Divisibility test for 3:  The sum of the digits of the number must be divisible by 3. Let’s add the digits of the given number:
 1+ 2+3+4+1+2+3+4+B+1+2+3+4+A=30+A+B 
Here we can see that 30 is divisible by “3” therefore “A+B” should be also divisible by “3”.
As we need the maximum difference of |A−B|; let's substitute A = 0 and B = 9; where we get the maximum difference and ‘A + B’ is also divisible by 3. 
Hence A should be ‘0’ and B should be ‘9’.

The correct answer option is B.

ScottTargetTestPrep · Answer

A number divisible by 6 must be divisible by the two prime factors of 6, which are 2 and 3. Since 12341234B1234A is divisible by 6, the number must be even (divisible by 2), and the sum of its digits must be a number divisible by 3.

Summing the digits, we have 30 + B + A.

We know that A must be an even number, so it can be any of 0, 2, 4, 6, 8. Additionally, we want the absolute value of the difference of A and B to be as large as possible. So let’s let A be as small as possible, so A = 0. This means that B must be as large as possible and still satisfy that (30 + A + B) is divisible by 3. So if A = 0, then the largest possible value for B would be 9. Thus, with A = 0 and B = 9, the maximum value of |A - B| is 9.

Answer: A

exc4libur · Answer

Any digits can take = {0,1,2,3,4,5,6,7,8,9};

A number divisible by 6 must be an EVEN divisible by 3; and,
A number is divisible by 3 if the sum of its digits is divisible by 3;

Thus A must be an even: {0,2,4,6,8}; and,
1234+1234+B+1234+A = 30+B+A must be divisible by 3; 
So A+B must be a multiple of 3;
The max of |A-B| is when we max(A,B) and min(A,B);
If A=0, then B must be the largest digit multiple of 3, B=9:
|A-B|=|0-9|=9.

Ans (A)

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