Question

If x = 111 ….. 1 (20 digits), y = 333 …… 3 (10 digits) and z = 222 … 2 (10 digits), then what is x- y2z equal to

Answer 1

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Answer 2

Answer:

“The value of x-y^2/z is equal to 1⃣”.

Step-by-step explanation:

It is given that,

x = 1 1 1 ... 1 ( 2⃣0⃣ digits)

y = 3 3 3 ... 3 ( 1⃣0⃣ digits)

z = 2 2 2 ... 2 ( 1⃣0⃣ digits)

Now,

x = 1 + 10 + 10^2 + ... + 10^19

= 1 [10^(20)-1] / (10-1)

y = 3 ( 1 + 10 + 10^2 + ... + 10^9 )

= 3 * 1[ 10^( 10)-1] / (10-1)

z = 2 ( 1 + 10 + 10^2 + ... + 10^9 )

= 2 * 1[10^(10)-1] / (10-1)

Therefore,

x - y^2

= [10^(10)+1][10^(10)-1] / 9 - 9 * [10^(10)-1][10^(10)-1] / 81

= {[10^(10)+1][10^(10)-1] - [10^(10)-1][10^(10)-1]} / 9

= [10^(10)-1] {10^(10)+1 - 10^(10)+1} / 9

= 2 * [10^(10)-1] / 9

= z

Therefore,

x - y^2 = z

➡️ x - y^2 / z = 1

❤ HOPE IT HELPS! ❤

REGARDS.