Ex.16.2 Q2 Playing with Numbers Solutions - NCERT Maths Class 8

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If \(31z5\) is a multiple of \(9,\) where \(z\) is a digit, what is the value of \(z\)?

You will find that there are two answers for the last problem. Why is this so?

 Video Solution
Playing With Numbers
Ex 16.2 | Question 2

Text Solution

What is known?

A puzzled number

What is unknown?

Value of the alphabet i.e. \(y.\)


If the sum of all digits of a number is equal to \(9,\) then the number is a multiple of \(9.\)


If a number is a multiple of \(9,\) then the sum of its digits will be divisible by \(9.\)

Sum of digits of

\[31z\,5 = 3 + 1 + z + 5 = 9 + z\]

Hence, \(9 + z\) should be a multiple of \(9.\)

This is possible, when \(9 + z\) is any one of these numbers \(0,\; 9,\; 18,\; 27,\) and so on ...

However, since \(z\)  is a single digit number, this sum can be either \(9\) or \(18.\)

\(9+z =9\\z=9-9\\z=0\)

Therefore, \(z\) should be either \(0\) or \(9.\)

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