## If x3627y0 is divisible by 44, what is the maximum possible value of (x+y)?

For the number x3627y0 to be divisible by 44.

Factorise [math]44[/math such that the factors are prime to each other.

Only option is 44 = 4 \times 11

Divisibility Rule for 11:

The sum of the digits at even place is subtracted from the sum of the digits at odd places. If the result is either 0: or divisible by 11, then the original number is divisible by 11.

Now, considering the given problem

Sum of digits at odd places:

{S_{odd}} = x + 6 + 7 + 0 = 13 + x

Sum of digits at odd places:

{S_{even}} = 3 + 2 + y = 5 + y

{S_{odd}} – {S_{even}} = 13+x-5-y

{S_{odd}} – {S_{even}} = 8+x-y = 0 or 11 ——- (1)

Divisibility Rule for 4:

Last two digits of given number should be divisible by 4

That is y0/4

y can take values as

y = 0, 2,4,6,8 —————————-(2)

Now, from equation (1) and (2)

we have

or x-y =11-8

x-y =3

x=y+3

For,

y = 0;x = 3

y = 2;x = 5

y = 4;x = 7

y = 6;x = 9

y = 8;x = 11 ( cannot be answer, since it is two digit)

From these set of values, we have

max(x+y) = 15

x=8+y

Considering this only, we get values of (x,y) as

y = 0;x = 8

y = 2;x = 10 ( cannot be answer, since it is two digit)

y = 4;x = 12( cannot be answer, since it is two digit)

y = 6;x = 14( cannot be answer, since it is two digit)

y = 8;x = 16( cannot be answer, since it is two digit)

From these set of values, we have

max(x+y) = 8

From both sets, the maximum value of x+y =15

Therefore, the total number of ordered pairs are 4.