1234ab

Any multiple of 11 can be found by taking the sum of the 1st, 3rd, 5th, etc numbers in the number and subtracting the sum of the 2nd, 4th, 6th, etc numbers. The difference must be 0 or a multiple of 11.

The sum of all the numbers must be a multiple of 9. The first 4 add to 10 so a+b must be either 8 or 17 to meet this condition making the sum of all 6 numbers 18 or 27.

Considering the first conditions... (a+4) - (b+6) = 0 since the highest possible positive number made with this equation is 7 and the lowest is -11, where for (a+4) - (b+6) = -11 a = 0 and b = 9... This makes the sum of each number in 1234ab 19 which is not a multiple of 9.

(a - b) + (4 - 6) = 0 a - b - 2 = 0 a - b = 2

We know the difference between a and b to be 2. The sum of 9 and 7 is 16 so their sum must be 8 and not 17.

a + b = 8 a - b = 2

a = 8 - b

(8 - b) - b = 2 -2b = -6 b = 3

a - 3 = 2 a = 5

So ab is 53.

A number is a multiple of 9 and 11 means it is divisible by 9 and 11

So,a fast way to know a multiple 9 is sum all the digits if the total is divisible by 9 means the original number is also divisible by 9.

The number 1234ab has to total up to 18 or 27 to be divisible by 9

So we have narrow down our solutions to 1234ab where ab is 17, 26, 35, 44, 53, 62, 72, 89, 98

Now to look for a multiple of 11, add the digits in blocks of 2 from the right to left

For eg : 17 + 34 + 12 = 63 ( not multiple of 11 )

53 + 34 + 12 = 99 ( a multiple of 11 and 9 )

Therefore, our answer is 53...

We begin by giving a random value to a and b, say $a=0$ and $b=0$

Therefore if we divide 123400 by 11:

$\frac{123400}{11}=11218 + \frac{2}{11}$ $\longrightarrow$ $123400=11218\cdot 11 + 2$

$123398=11218\cdot 11$

we know 123398 is a multiple of 11, so every time we add 11 to it, it will be a multiple of 11 as well

$\longrightarrow$ $123398=11\cdot x$ $\longrightarrow$ $123398+11=11\cdot x\ +11$ $\longrightarrow$ $123398+11=11\cdot( x+1)$

so if we keep adding 11 to 123398 we will end up with a number which is a multiple of 9.

If it is a multiple of 9, it will be a multiple of 3, therefore the sum of their digits will be a multiple of three too

$-123398+11=123409 (1+2+3+4+0+9=19)$ NOT A MULTIPLE OF 3

$-123409+11=123420 (1+2+3+4+2+0=12)$ A MULTIPLE OF 3 BUT NOT 9

$-123420+11=123431(1+2+3+4+3+1=14)$ NOT A MULTIPLE OF 3

$-123431+11=123442(1+2+3+4+4+2=16)$ NOT A MULTIPLE OF 3

$-123442+11=123453(1+2+3+4+4+2=18)$ VUOLA !!)

$Answer =123453$

divisiblity test for 9:- sub of digits divisiblle by 9 and divisibility test for 11:- A number passes the test for 11 if the difference of the sums of alternating digits is divisible by 11

so 1+2+3+4+a+b should be divisile by 9 (1+3+a)-(2+4+b) should be divisible by 11

10+a+b by 9 -2+a-b by 11

putting a=5 b=3 satisfies answer