Question

If the 9 digit number 807x6y9z8 is divisible by 99, the value of
x+y+z=?

Answer 1

Step-by-step explanation:

$3 \sqrt{3}$

Answer 2

SOLUTION

TO DETERMINE

If the 9 digit number 807x6y9z8 is divisible by 99, the value of x+y+z

CONCEPT TO BE IMPLEMENTED

Divisibility by 9 :

A number is divisible by 9 if the sum of digits is divisible by 9.

Divisibility by 11 :

A number is divisible by 11 if the difference of the sum of the digits at even places and the sum of the digits at odd places is divisible by 11.

EVALUATION

Here the given number is 807x6y9z8

Since the number is divisible by 99

So the number is divisible by both 9 and 11

As 807x6y9z8 is divisible by 9

So sum of digits is divisible by 9

Sum of the digits = 38 + x + y + z

⇒ 38 + x + y + z is divisible by 9 - - - (1)

Again 807x6y9z8 is divisible by 11

Then the difference of the sum of the digits at even places and the sum of the digits at odd places is divisible by 11.

Thus

( 0 + x + y + z ) - ( 8 + 7 + 6 + 9 + 8 ) is divisible by 11

⇒ x + y + z - 38 is divisible by 11 - - - - (2)

(1) and (2) holds when x + y + z = 16

FINAL ANSWER

x + y + z = 16

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