Question

it's my holiday homework
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Answer 1

1y^3 y^6 is divisible by 11

3,6 can't be exponents as if y^3 is digit then y^3 can't be digit except 0,1 which is not divisible by 11

1 y 3 y 6

if its divisible by 11 then

(1+3+6) - (y+y) is divisible by 11

10 - 2y is divisible by 11

y positive 10-2y max value is 10

so it would be single digit

now only digit 0 is divisible by 11

So 10-2y = 0

y= 10/2 = 5

So y = 5

⭐Dr.Dhruv⭐

Answer 2

Answer:

★━━━━━━━━━━━━━━━━━★

1y^3 y^6 is divisible by 11

3,6 can't be exponents as if y^3 is digit then y^3 can't be digit except 0,1 which is not divisible by 11

1 y 3 y 6

if its divisible by 11 then

(1+3+6) - (y+y) is divisible by 11

10 - 2y is divisible by 11

y positive 10-2y max value is 10

so it would be single digit

now only digit 0 is divisible by 11

So 10-2y = 0

y= 10/2 = 5

So y = 5

★━━━━━━━━━━━━━━━━━★