Question

34. Jf 3125 is a multiple of 9 where z is a digit ,what is the value of z? Write all
the possible values.​

Answer 1

Answer:

Numbers can be written in general form. Hence, a two digit number ab will be written as ab = 10a + b.

The general form of numbers is helpful in solving puzzles or number games.

Rules to be followed while solving the puzzles:. 

Each letter in the puzzle must stand for just one digit. Each digit must be represented by just one letter.  The first digit of a number cannot be zero.

 

The reasons for the divisibility of numbers by 10, 5, 2, 9 or 3 can be given when numbers are written in general form

 

Tests of Divisibility

Divisibility by 10

A number is divisible by 10 when its one’s digit is 0.

Divisibility by 5

If the ones digit of a number is 0 or 5, then it is divisible by 5.

 

Divisibility by 2

If the one’s digit of a number is 0, 2, 4, 6 or 8 then the number is divisible by 2

 

Divisibility by 9 and 3

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

A number  is divisible by 3 if the sum of its digits is divisible by 3.

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Solution:

Since 31z5 is a multiple of 3.

 According to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.

Here z is a digit.

3+1+z+5=9+z

9 + z = 9

z= 0

 

3+1+z+5=9+z

9 + z = 12

z = 3

 

3+1+z+5=9+z

9 + z = 15

z = 6

3+1+z+5=9+z

9 + z = 18

z = 9

 

Since z is a single digit number, the sum of the digits can be 9 or 12 or 15 or 18 and thus, the value of z comes to 0 or 3 or 6 or 9 respectively.

Therefore, z can have any of the four different values.

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