Question

If a 10 digit number 2094x843y2 is divisible by 88

Answer 1

Answer:

2094184312

Step-by-step explanation:

let z=2094x843y2

If z is divisible by 88 then it should be divisible by by its factors 11 and 8

for 2094x843y2 to be divisible by 8 the last 3 digits no 3y2 should be divible by 8

for this the values divisible by 8 possible are :

312,352,392

so y=1,5 or 9————————-(1)

………………

Now 2094x843y2 will be divisible by 11 if

sum of digits at odd positions-sum of digits at even positions=0 or multiple of 11

sum of digits at odd positions=2+3+8+4+0=17

sum of digits at even positions=y+4+x+9+2=x+y+15

So Difference D=x+y+15–17=x+y-2——————-(2)

x+y-2=11n

x+y=11n+2

x=11n+2-y

for x to be maximum y should be minimum

So from (1)

y=1

from(2)

D= x+1–2

D=x-1

x-1 should be equal to 0 or multiple of 11

As maximum possible value of x=9 then x-1 can not be multiple of 11

so x-1=0

x=1

Therefore Value

of 5x-7y=1*5–7*1

=5–7=-2

Ans : -2