NCERT Solutions Class 8 Maths Ganita Prakash Chapter 5 Number Play on Padhaiwise

NCERT Solutions for Class 8 Maths Ganita Prakash Chapter 5 Number Play Figure it Out Page 122

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 8 Maths Ganita Prakash Chapter 5 — Figure it Out (Pages 122–123)#

A number that leaves a remainder $r$ when divided by $n$ can be written as $nk + r$, where $k$ is a whole number. For example, numbers that are 3 more than a multiple of 5 are $5k + 3$: 3, 8, 13, 18, …

1. The sum of four consecutive numbers is 34. What are these numbers? #

Solution

Let the four consecutive numbers be $n$, $n + 1$, $n + 2$ and $n + 3$. Their sum is 34:

NCERT Solutions for Class 8 Maths Ganita Prakash Chapter 5 Number Play Figure it Out Page 126

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 8 Maths Ganita Prakash Chapter 5 — Figure it Out (Page 126)#

Every place value is one more than a multiple of 9: $10 = 9 + 1$, $100 = 99 + 1$, $1000 = 999 + 1$, … So a number leaves the same remainder on division by 9 as the sum of its digits. In particular, a number is divisible by 9 exactly when the sum of its digits is divisible by 9.

NCERT Solutions for Class 8 Maths Ganita Prakash Chapter 5 Number Play Figure it Out Page 131

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 8 Maths Ganita Prakash Chapter 5 — Figure it Out (Page 131)#

The digital root of a number is found by adding its digits again and again until one digit is left. For example, $7309 \to 7 + 3 + 0 + 9 = 19 \to 1 + 9 = 10 \to 1 + 0 = 1$.

The digital root equals the remainder when the number is divided by 9, except that multiples of 9 have digital root 9 (not 0).

NCERT Solutions for Class 8 Maths Ganita Prakash Chapter 5 Number Play Figure it Out Page 132

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 8 Maths Ganita Prakash Chapter 5 — Figure it Out (Pages 132–134)#

The divisibility tests used here:

2: the last digit is even. 4: the last two digits form a multiple of 4.
3 / 9: the sum of the digits is a multiple of 3 / 9.
5: the last digit is 0 or 5. 10: the last digit is 0.
11: (sum of digits in odd places) − (sum of digits in even places) is 0 or a multiple of 11.
A number is divisible by $b \times c$ when it is divisible by both $b$ and $c$, if $b$ and $c$ have no common factor (for example, $18 = 2 \times 9$ and $44 = 4 \times 11$).

1. If 31z5 is a multiple of 9, where z is a digit, what is the value of z? Explain why there are two answers to this problem. #

Solution

The sum of the digits must be a multiple of 9: