Figure it Out (Page 108):
1. At what number is ‘idli-vada’ said for the 10th time?
Answer:
The first number for which the players said ‘idli-vada’ for the 1st time is 15, which is a multiple of 3, and also a multiple of 5.
We see that:
15 × 1 = 15,
15 × 2 = 30,
15 × 3 = 45,
And so on….
Continuing on in the same way, we see:
15 × 10 = 150
Therefore, ‘idli-vada’ is said for the 10th time at the number 150.
2. If the game is played for the numbers from 1 till 90, find out:
a. How many times would the children say ‘idli’ (including the times they say ‘idli-vada’)?
b. How many times would the children say ‘vada’ (including the times they say ‘idli-vada’)?
c. How many times would the children say ‘idli-vada’?
Answer:
a. The children say ‘idli’ when the number is a multiple of 3.
The multiples of 3 between 1 and 90 are: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90.
Total number of multiples of 3 from 1 to 90 = 30.
Therefore, the children said ‘idli’ 30 times.
b. The children say ‘vada’ when the number is a multiple of 5.
The multiples of 5 between 1 and 90 are: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90.
Total number of multiples of 5 from 1 to 90 = 18.
Therefore, the children said ‘vada’ 18 times.
c. The children say ‘idli-vada’ when the number is a multiple of both 3 and 5.
The multiples of 3 and 5 between 1 and 90 are: 15, 30, 45, 60, 75, 90.
Therefore, the children said ‘idli-vada’ 6 times.
3. What if the game was played till 900? How would your answers change?
Answer:
Total number of multiples of 3 from 1 to 90 = 30.
We observe that, 900 = 90 × 10.
Therefore, number of multiples of 3 from 1 to 900 = 30 × 10 = 300.
Therefore, the children said ‘idli’ 300 times.
Total number of multiples of 5 from 1 to 90 = 18.
Similarly, number of multiples of 5 from 1 to 900 = 18 × 10 = 180.
Therefore, the children said ‘vada’ 180 times.
Total number of multiples of 15 from 1 to 90 = 6.
Similarly, number of multiples of 15 from 1 to 900 = 6 × 10 = 60.
Therefore, the children said ‘idli-vada’ 60 times.
Figure it Out (Page 110)
2. Who am I?
a. I am a number less than 40. One of my factors is 7. The sum of my digits is 8.
b. I am a number less than 100. Two of my factors are 3 and 5. One of my digits is 1 more than the other.
Answer:
a. Let us take all numbers which are less than 40 with one factor as 7.
7 × 1 = 7
7 × 2 = 14
7 × 3 = 21
7 × 4 = 28
7 × 5 = 35
We see that the sum of the digits of 35 = 3 + 5 = 8.
Therefore, the required number is 35.
b. Two of the factors of the number are 3 and 5.
All such numbers less than 100 are as follows:
1 × (3 × 5) = 15
2 × (3 × 5) = 30
3 × (3 × 5) = 45
4 × (3 × 5) = 60
5 × (3 × 5) = 75
6 × (3 × 5) = 90
The only number in which one of the digits is 1 more than the other is 45.
Therefore, the required number is 45.
3. A number for which the sum of all its factors is equal to twice the number is called a perfect number. The number 28 is a perfect number. Its factors are 1, 2, 4, 7, 14 and 28. Their sum is 56 which is twice 28. Find a perfect number between 1 and 10.
Answer:
Factors of 6 = 1, 2, 3, 6.
Sum of the factors of 6 = 1 + 2 + 3 + 6 = 12.
12 is two times the number 6.
Therefore, a perfect number between 1 and 10 is 6.
4. Find the common factors of:
a. 20 and 28 b. 35 and 50
c. 4, 8 and 12 d. 5, 15 and 25
Answer:
a. 20 and 28
The factors of 20 = 1, 2, 4, 5, 10, 20.
The factors of 28 = 1, 2, 4, 7, 14, 28.
Therefore, the common factors of 20 and 28 = 1, 2, 4.
b. 35 and 50
The factors of 35 = 1, 5, 7.
The factors of 50 = 1, 2, 5, 10, 25, 50.
The common factors of 35 and 50 = 1, 5.
c. 4, 8 and 12
The factors of 4 = 1, 2, 4.
The factors of 8 = 1, 2, 4, 8.
The factors of 12 = 1, 2, 3, 4, 6, 12.
The common factors of 4, 8 and 12 = 1, 2, 4.
d. 5, 15 and 25
The factors of 5 = 1, 5.
The factors of 15 = 1, 3, 5, 15.
The factors of 25 = 1, 5, 25.
The common factors of 5, 15 and 25 = 1, 5.
5. Find any three numbers that are multiples of 25 but not multiples of 50.
Answer:
The multiples of 25 are: 25, 50, 75, 100, 125, 150, 175, 200.
The multiples of 50 are: 50, 100, 150, 200.
Comparing, we get the three numbers that are multiples of 25 but not multiples of 50 are: 25, 75, 125, 175.
9. Find the smallest number that is a multiple of all the numbers from 1 to 10 except for 7.
Answer:
The smallest number that is a multiple of all the numbers from 1 to 10 except for 7 will be the LCM of 1, 2, 3, 4, 5, 6, 8, 9, 10. The LCM can be found as below:
LCM = 2 × 2 × 3 × 5 × 2 × 3 = 360.
Therefore, the smallest number that is a multiple of all the numbers from 1 to 10 except for 7 is 360.
Figure it Out (Page 114-115)
1. We see that 2 is a prime and also an even number. Is there any other even prime?
Answer:
No, there are no other even primes other than 2.
Reason: All even numbers other than 2 are divisible by numbers other than 1 and the number itself.
2. Look at the list of primes till 100. What is the smallest difference between two successive primes? What is the largest difference?
Answer:
Looking at the list of primes till 100, we can see that:
The smallest difference between two successive primes = 3 – 2 = 1.
The largest difference between two successive primes = 97 – 89 = 8.
5. Write three pairs of prime numbers less than 20 whose sum is a multiple of 5.
Answer:
The pairs of prime numbers less than 20 whose sum is a multiple of 5 are as follows:
(2, 3) = 2 + 3 = 5
(3, 7) = 3 + 7 = 10
(2, 13) = 2 + 13 = 15
(7, 13) = 7 + 13 = 20
(3, 17) = 3 + 17 = 20
6. The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 and 3. Find such pairs of prime numbers up to 100.
Answer:
Pairs of prime numbers up to 100 having the same digits are 17 and 71, 37 and 73, 79 and 97.
7. Find seven consecutive composite numbers between 1 and 100.
Answer:
Seven consecutive composite numbers between 1 and 100 are 90, 91, 92, 93, 94, 95, 96.
8. Twin primes are pairs of primes having a difference of 2. For example, 3 and 5 are twin primes. So are 17 and 19. Find the other twin primes between 1 and 100.
Answer:
Twin primes between 1 and 100 are: (5, 7), (11, 13), (29, 31), (41, 43), (59, 61), (71, 73).
10. Which of the following numbers is the product of exactly three distinct prime numbers: 45, 60, 91, 105, 330?
Answer:
The numbers 45, 60, 91, 105, 330 divided into prime factors are shown below:
45 = 3 × 3 × 5
60 = 2 × 2 × 3 × 5
91 = 7 × 13
105 = 3 × 5 × 7
330 = 2 × 3 × 5 × 7
Therefore, 105 is the product of exactly three distinct prime numbers 3, 5 and 7.
Figure it Out (Page 120)
1. Find the prime factorisations of the following numbers: 64, 104, 105, 243, 320, 141, 1728, 729, 1024, 1331, 1000.
Answer:
64 = 2 × 2 × 2 × 2 × 2 × 2
104 = 2 × 2 × 3 × 13
105 = 3 × 5 × 7
243 = 3 × 3 × 3 × 3 × 3
320 = 2 × 2 × 2 × 2 × 2 × 2 × 5
141 = 3 × 47
1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
729 = 3 × 3 × 3 × 3 × 3 × 3
1024 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
1331 = 11 × 11 × 11
1000 = 2 × 2 × 2 × 5 × 5 × 5
2. The prime factorisation of a number has one 2, two 3s, and one 11. What is the number?
Answer:
The number will be the product of its prime factors.
Therefore,
Number = 2 × 3 × 3 × 11 = 198
3. Find three prime numbers, all less than 30, whose product is 1955.
Answer:
Prime factorisation of 1955 = 5 × 17 × 23
Therefore, the three prime numbers are 5, 17 and 23.
4. Find the prime factorisation of these numbers without multiplying first
a. 56 × 25 b. 108 × 75 c. 1000 × 81
Answer:
a. 56 × 25
Prime factorisation is shown below:
56 = 2 × 2 × 2 × 7 and
25 = 5 × 5
Therefore,
56 × 25 = 2 × 2 × 2 × 5 × 5 × 7
b. 108 × 75
Prime factorisation is shown below:
108 = 2 × 2 × 3 × 3 × 3 and
75 = 3 × 5 × 5
Therefore,
108 × 75 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5
c. 1000 × 81
Prime factorisation is shown below:
1000 = 2 × 2 × 2 × 5 × 5 × 5
81 = 3 × 3 × 3 × 3
Therefore,
1000 × 81 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 5
Figure it Out (Page 122)
1. Are the following pairs of numbers co-prime? Guess first and then use prime factorisation to verify your answer.
a. 30 and 45 b. 57 and 85
c. 121 and 1331 d. 343 and 216
Answer:
a. 30 and 45
My guess is that the numbers are not co-prime.
Prime factorisation:
30 = 2 × 3 × 5
45 = 3 × 3 × 5
Common factors = 3, 5.
Therefore, these numbers are not co-prime. My guess was correct.
b. 57 and 85
My guess is that the numbers are co-prime.
Prime factorisation:
57 = 3 × 19
85 = 5 × 17
There are no common factors.
Therefore, these numbers are co-prime. My guess was correct.
c. 121 and 1331
My guess is that the numbers are co-prime.
Prime factorisation:
121 = 11 × 11
1331 = 11 × 11 × 11
Common factor = 11
Therefore, these numbers are not co-prime. My guess was incorrect.
d. 343 and 216
My guess is that the numbers are co-prime.
Prime factorisation:
343 = 7 × 7 × 7
216 = 2 × 2 × 2 × 3 × 3 × 3
There are no common factors.
Therefore, these numbers are co-prime. My guess was correct.
2. Is the first number divisible by the second? Use prime factorisation.
a. 225 and 27 b. 96 and 24
c. 343 and 17 d. 999 and 99
Answer:
a. 225 and 27
We can say that if one number is divisible by another, then the prime factorisation of the second number is included in the prime factorisation of the first number.
Prime Factorisation:
225 = 3 × 3 × 5 × 5
27 = 3 × 3 × 3
225 has two 3’s and 27 has three 3’s as factors.
Therefore, the entire prime factorisation of 27 is not included in 225 and we conclude that 225 is not divisible by 27.
b. 96 and 24
We can say that if one number is divisible by another, then the prime factorisation of the second number is included in the prime factorisation of the first number.
Prime Factorisation:
96 = 2 × 2 × 2 × 2 × 2 × 3
24 = 2 × 2 × 2 × 3
The entire prime factorisation of 24 is included in 96 and hence we conclude that 96 is divisible by 24.
c. 343 and 17
We can say that if one number is divisible by another, then the prime factorisation of the second number is included in the prime factorisation of the first number.
Prime Factorisation:
343 = 7 × 7 × 7
17 = 1 × 17
Therefore, the entire prime factorisation of 17 is not included in 343 and we conclude that 343 is not divisible by 17.
d. 999 and 99
We can say that if one number is divisible by another, then the prime factorisation of the second number is included in the prime factorisation of the first number.
Prime Factorisation:
999 = 3 × 3 × 3 × 37
99 = 3 × 3 × 11
Therefore, the entire prime factorisation of 99 is not included in 999 and we conclude that 999 is not divisible by 99.
3. The first number has prime factorisation 2 × 3 × 7 and the second number has prime factorisation 3 × 7 × 11. Are they co-prime? Does one of them divide the other?
Answer:
Prime factorisation:
First Number = 2 × 3 × 7
Second Number = 3 × 7 × 11
The two numbers have two common factors: 3 and 7. Therefore, they are not co-prime.
Neither number divides the other because the prime factors of each are different.
4. Guna says, “Any two prime numbers are co-prime”. Is he right?
Answer:
The statement that any two prime numbers are co-prime is right. This is because the only common factor that prime numbers have is 1.
Figure it Out (Page 125, 126)
1. 2024 is a leap year (as February has 29 days). Leap years occur in the years that are multiples of 4, except for those years that are evenly divisible by 100 but not 400.
a. From the year you were born till now, which years were leap years?
b. From the year 2024 till 2099, how many leap years are there?
Answer:
a. I was born in the year 2011. Till now the leap years were 2012, 2016, 2020 and 2024.
b. From the year 2024 till 2099, the leap years were 2024, 2028, 2032, 2036, 2040, 2044, 2048, 2052, 2056, 2060, 2064, 2068, 2072, 2076, 2080, 2084, 2088, 2092, 2096.
Therefore, there were 19 leap years.
2. Find the largest and smallest 4-digit numbers that are divisible by 4 and are also palindromes.
Answer:
Largest 4-digit numbers that are divisible by 4 and are palindromes = 8888.
Smallest 4-digit numbers that are divisible by 4 and are palindromes = 2112.
3. Explore and find out if each statement is always true, sometimes true or never true. You can give examples to support your reasoning.
a. Sum of two even numbers gives a multiple of 4.
b. Sum of two odd numbers gives a multiple of 4.
Answer:
a. Sum of two even numbers gives a multiple of 4.
Sometimes true.
Explanation: There are cases where sum of two even numbers give a multiple of 4:
4 + 8 = 12, which is a multiple of 4.
6 + 10 = 16, which is a multiple of 4.
There are cases where sum of two even numbers does not give a multiple of 4:
4 + 2 = 6, which is not a multiple of 4.
10 + 12 = 22, which is not a multiple of 4.
b. Sum of two odd numbers gives a multiple of 4.
Sometimes true.
Explanation: There are cases where sum of two odd numbers give a multiple of 4:
7 + 9 = 16, which is a multiple of 4.
3 + 17 = 20, which is a multiple of 4.
There are cases where sum of two odd numbers does not give a multiple of 4:
7 + 11 = 18, which is not a multiple of 4.
5 + 9 = 14, which is not a multiple of 4.
4. Find the remainders obtained when each of the following numbers are divided by i) 10, ii) 5, iii) 2.
78, 99, 173, 572, 980, 1111, 2345.
Answer:
78:
(i) 78 ÷ 10 leaves a remainder of 8.
(ii) 78 ÷ 5 leaves a remainder of 3.
(iii) 78 ÷ 2 leaves a remainder of 0.
99:
(i) 99 ÷ 10 leaves a remainder of 9.
(ii) 99 ÷ 5 leaves a remainder of 4.
(iii) 99 ÷ 2 leaves a remainder of 1.
173:
(i) 173 ÷ 10 leaves a remainder of 3.
(ii) 173 ÷ 5 leaves a remainder of 3.
(iii) 173 ÷ 2 leaves a remainder of 1.
572:
(i) 572 ÷ 10 leaves a remainder of 2.
(ii) 572 ÷ 5 leaves a remainder of 2.
(iii) 572 ÷ 2 leaves a remainder of 0.
980:
(i) 980 ÷ 10 leaves a remainder of 0.
(ii) 980 ÷ 5 leaves a remainder of 0.
(iii) 980 ÷ 2 leaves a remainder of 0.
1111:
(i) 1111 ÷ 10 leaves a remainder of 1.
(ii) 1111 ÷ 5 leaves a remainder of 1.
(iii) 1111 ÷ 2 leaves a remainder of 1.
2345:
(i) 2345 ÷ 10 leaves a remainder of 5.
(ii) 2345 ÷ 5 leaves a remainder of 0.
(iii) 2345 ÷ 2 leaves a remainder of 1.
5. The teacher asked if 14560 is divisible by all of 2, 4, 5, 8 and 10. Guna checked for divisibility of 14560 by only two of these numbers and then declared that it was also divisible by all of them. What could those two numbers be?
Answer:
The required numbers are 8 and 10.
Explanation:
If 14560 is divisible by 8, then it will be divisible 2, 4 because 2,4 are factors of 8.
If 14560 is divisible by 10, then it will be divisible by 2, 5 because 2, 5 are factors of 10.
6. Which of the following numbers are divisible by all of 2, 4, 5, 8 and 10: 572, 2352, 5600, 6000, 77622160.
Answer:
| Divisibility by 2 | Divisibility by 4 | Divisibility by 5 | Divisibility by 8 | Divisibility by 10 | Divisibility by All | |
| 572 | Number ends with 2, hence it is divisible by 2. | The number formed by the last two digits, i.e. 72 is divisible by 4, hence number is divisible by 4. | Number does not end with 0 or 5, hence number is not divisible by 5. | Last 3-digits of number 572 is not divisible by 8, so number is not divisible by 8. | Last digit of the number ends with 2 and not with 0, hence number is not divisible by 10. | No |
| 5600 | Number ends with 0, hence it is divisible by 2. | The number formed by the last two digits, i.e. 00 is divisible by 4, hence number is divisible by 4. | Number ends with 0, hence number is divisible by 5. | Last 3-digits of number i.e. 600 is divisible by 8, so number is divisible by 8. | Last digit of the number ends with 0, hence number is divisible by 10. | Yes |
| 77622160 | Number ends with 0, hence it is divisible by 2. | The number formed by the last two digits, i.e. 60 is divisible by 4, hence number is divisible by 4. | Number ends with 0, hence number is divisible by 5. | Last 3-digits of number i.e. 160 is divisible by 8, so number is divisible by 8. | Last digit of the number ends with 0, hence number is divisible by 10. | Yes |
Therefore, the numbers are divisible by all of 2, 4, 5, 8 and 10 are 5600, 6000, 77622160.
7. Write two numbers whose product is 10000. The two numbers should not have 0 as the units digit.
Answer:
Prime factorisation = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5 = (2 × 2 × 2 × 2) × (5 × 5 × 5 × 5) = 16 × 625.
Therefore, two numbers whose product is 10000 and which have 0 as the units digit are 16 and 25.
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