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:
$$n + (n + 1) + (n + 2) + (n + 3) = 34$$$$\Rightarrow 4n + 6 = 34$$
$$\Rightarrow 4n = 28$$
$$\Rightarrow n = 7$$
∴ The numbers are 7, 8, 9 and 10. Check: $7 + 8 + 9 + 10 = 34$.
2. Suppose p is the greatest of five consecutive numbers. Describe the other four numbers in terms of p. #
Solution
Each number is 1 less than the next one, so counting down from $p$:
∴ The other four numbers are $p - 1$, $p - 2$, $p - 3$ and $p - 4$.
3. For each statement below, determine whether it is always true, sometimes true, or never true. Explain your answer. Mention examples and non-examples as appropriate. Justify your claim using algebra. #
(i) The sum of two even numbers is a multiple of 3.
Sometimes true.
- Examples: $2 + 4 = 6$ and $4 + 8 = 12$ are multiples of 3.
- Non-examples: $2 + 6 = 8$ and $6 + 8 = 14$ are not.
Two even numbers are $2m$ and $2n$, and their sum is $2(m + n)$. This is a multiple of 3 only when $m + n$ is a multiple of 3.
(ii) If a number is not divisible by 18, then it is also not divisible by 9.
Sometimes true.
- Example: 30 is not divisible by 18, and not divisible by 9.
- Non-example: 27 is not divisible by 18, but it is divisible by 9.
The multiples of 9 are $9k$. When $k$ is odd ($9, 27, 45, \ldots$), the number is divisible by 9 but not by 18.
(iii) If two numbers are not divisible by 6, then their sum is not divisible by 6.
Sometimes true.
- Example: 9 and 11 are not divisible by 6, and $9 + 11 = 20$ is not either.
- Non-example: 8 and 10 are not divisible by 6, but $8 + 10 = 18$ is.
Write the numbers as $6m + r$ and $6n + s$, with remainders $r$ and $s$ between 1 and 5. Their sum is $6(m + n) + (r + s)$, which is divisible by 6 only when $r + s = 6$.
(iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3.
Always true. A multiple of 6 is $6x$ and a multiple of 9 is $9y$:
$6x + 9y = 3(2x + 3y)$
This is 3 times a whole number, so it is always a multiple of 3. For example, $12 + 18 = 30$.
(v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9.
Sometimes true.
- Example: $18 + 9 = 27$ is a multiple of 9.
- Non-example: $12 + 9 = 21$ is not.
The sum is $6x + 3y = 3(2x + y)$, which is a multiple of 9 only when $2x + y$ is a multiple of 3.
4. Find a few numbers that leave a remainder of 2 when divided by 3 and a remainder of 2 when divided by 4. Write an algebraic expression to describe all such numbers. #
Solution
If a number $x$ leaves remainder 2 on division by both 3 and 4, then $x - 2$ is a multiple of both 3 and 4. So $x - 2$ is a multiple of their LCM:
$LCM(3, 4) = 12$
So $x - 2 = 12n$, that is:
$$x = 12n + 2$$Taking $n = 0, 1, 2, 3, \ldots$ gives 2, 14, 26, 38, 50, …
Check: $14 = 3 \times 4 + 2$ and $14 = 4 \times 3 + 2$.
∴ All such numbers are given by $12n + 2$.
5. “I hold some pebbles, not too many, When I group them in 3’s, one stays with me. Try pairing them up — it simply won’t do, A stubborn odd pebble remains in my view. Group them by 5, yet one’s still around, But grouping by seven, perfection is found. More than one hundred would be far too bold, Can you tell me the number of pebbles I hold?” #
Solution
Let the number of pebbles be $p$. The riddle says:
- in 3s, 2s and 5s, 1 pebble is left over;
- in 7s, none is left over;
- $p$ is at most 100.
Since 1 is left over in each case, $p - 1$ is a multiple of 3, 2 and 5, so of their LCM:
$LCM(2, 3, 5) = 30$
So $p = 30k + 1$, which gives $p = 31, 61$ or $91$ (the next one, 121, is more than 100).
Now we check which of these is divisible by 7:
- $31 = 7 \times 4 + 3$ ✗
- $61 = 7 \times 8 + 5$ ✗
- $91 = 7 \times 13$ ✓
∴ The number of pebbles is 91.
6. Tathagat has written several numbers that leave a remainder of 2 when divided by 6. He claims, “If you add any three such numbers, the sum will always be a multiple of 6.” Is Tathagat’s claim true? #
Solution
Let the three numbers be $6a + 2$, $6b + 2$ and $6c + 2$. Their sum is:
$$(6a + 2) + (6b + 2) + (6c + 2) = 6a + 6b + 6c + 6$$$$= 6(a + b + c + 1)$$
This is 6 times a whole number. For example, $8 + 14 + 20 = 42 = 6 \times 7$.
∴ Yes, Tathagat’s claim is true: the three remainders of 2 add up to 6, which is itself a multiple of 6.
7. When divided by 7, the number 661 leaves a remainder of 3, and 4779 leaves a remainder of 5. Without calculating, can you say what remainders the following expressions will leave when divided by 7? Show the solution both algebraically and visually. (i) 4779 + 661 (ii) 4779 – 661 #
From the given remainders, we can write $4779 = 7p + 5$ and $661 = 7q + 3$ for some whole numbers $p$ and $q$.
(i) 4779 + 661
$$= 7p + 7q + 8$$
$$= 7p + 7q + 7 + 1$$
$$= 7(p + q + 1) + 1$$
∴ $4779 + 661$ leaves a remainder of 1 when divided by 7.
Visually: think of each number as full groups of 7 plus some loose counters. Putting them together, the full groups stay full, and the loose counters $5 + 3 = 8$ make one more group of 7 with 1 counter left over.
(ii) 4779 – 661
$$= 7p - 7q + 2$$
$$= 7(p - q) + 2$$
∴ $4779 - 661$ leaves a remainder of 2 when divided by 7.
Visually: taking away 661 removes its full groups of 7 and 3 of the 5 loose counters, leaving full groups of 7 and 2 loose counters.
8. Find a number that leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5. What is the smallest such number? Can you give a simple explanation of why it is the smallest? #
Solution
In each case the remainder is one less than the divisor. So if we add 1 to the number, it divides exactly by 3, 4 and 5. That means the number is one less than a common multiple of 3, 4 and 5.
$LCM(3, 4, 5) = 3 \times 4 \times 5 = 60$
Smallest number $= 60 - 1 = 59$
Check: $59 = 3 \times 19 + 2$, $59 = 4 \times 14 + 3$ and $59 = 5 \times 11 + 4$.
∴ The smallest such number is 59. Other such numbers are $119, 179, \ldots$, that is $60k - 1$.
Why it is the smallest: the number plus 1 must be a common multiple of 3, 4 and 5, and the smallest positive common multiple is 60. So no number smaller than $60 - 1 = 59$ works.