Question 1

What is the smallest number whose prime factorisation has:
a. three different prime numbers?
b. four different prime numbers?

Accepted Answer

a. The smallest number with three different prime numbers is 2 × 3 × 5 = 30.
b. The smallest number with four different prime numbers is 2 × 3 × 5 × 7 = 210. — To find the smallest number with a given number of different prime factors, multiply the smallest primes together. For three primes: 2, 3, 5; for four primes: 2, 3, 5, 7.

Question 2

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

Accepted Answer

a. No, because 30 = 2 × 3 × 5 and 45 = 3 × 3 × 5 share common prime factors 3 and 5.
b. Yes, 57 = 19 × 3 and 85 = 17 × 5 have no common prime factors.
c. No, 121 = 11 × 11 and 1331 = 11 × 11 × 11 share prime factor 11.
d. Yes, 343 = 7 × 7 × 7 and 216 = 2 × 2 × 2 × 3 × 3 × 3 have no common prime factors. — Two numbers are co-prime if their greatest common divisor is 1, i.e., they share no common prime factors. Prime factorise each and check common factors.

Question 3

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

Accepted Answer

a. No, because 225 = 3 × 3 × 5 × 5 and 27 = 3 × 3 × 3. 225 lacks the third 3 needed.
b. Yes, 96 = 2 × 2 × 2 × 2 × 2 × 3 and 24 = 2 × 2 × 2 × 3; 96 contains all prime factors of 24.
c. No, 343 = 7 × 7 × 7 and 17 is prime and not a factor of 343.
d. No, 999 = 3 × 3 × 3 × 37 and 99 = 3 × 3 × 11; 999 lacks factor 11. — A number is divisible by another if its prime factors include all prime factors of the divisor with at least the same powers.

Question 4

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?

Accepted Answer

No, they are not co-prime because they share prime factors 3 and 7. No, neither divides the other because neither's prime factors are a subset of the other's. — Co-prime numbers share no common prime factors. Here, common factors exist. Divisibility requires all prime factors of divisor to be in dividend.

Question 5

Guna says, “Any two prime numbers are co-prime”. Is he right?

Accepted Answer

Yes, any two prime numbers are co-prime because they have no common factors other than 1. For example, 2 and 3; 3 and 11. — Prime numbers have only 1 and themselves as factors, so two different primes cannot share any common factor except 1.

Question 6

Is 8536 divisible by 4?

Accepted Answer

Yes, 8536 is divisible by 4 because the last two digits, 36, are divisible by 4. — A number is divisible by 4 if the number formed by its last two digits is divisible by 4.

Question 7

Consider these statements:
1. Only the last two digits matter when deciding if a given number is divisible by 4.
2. If the number formed by the last two digits is divisible by 4, then the original number is divisible by 4.
3. If the original number is divisible by 4, then the number formed by the last two digits is divisible by 4.
Do you agree? Why or why not?

Accepted Answer

Yes, all three statements are true. For example, numbers like 124, 364, 4028 are divisible by 4 because their last two digits form numbers divisible by 4. — Divisibility by 4 depends on the last two digits because 100 is divisible by 4, so the rest of the number contributes multiples of 4.

Question 8

Find numbers between 120 and 140 that are divisible by 8. Also find numbers between 1120 and 1140, and 3120 and 3140, that are divisible by 8. What do you observe?

Accepted Answer

Between 120 and 140 divisible by 8: 128, 136.
Between 1120 and 1140 divisible by 8: 1128, 1136.
Between 3120 and 3140 divisible by 8: 3128, 3136.
Observation: The numbers divisible by 8 in these ranges have the same last three digits pattern. — Divisibility by 8 depends on the last three digits. Numbers with last three digits divisible by 8 are divisible by 8.

Question 9

Change the last two digits of 8560 so that the resulting number is a multiple of 8.

Accepted Answer

Change last two digits to 52, so 8552 is a multiple of 8. — Since divisibility by 8 depends on last three digits, changing last two digits to 52 makes 8552 divisible by 8.

Question 10

Consider this statement:
1. Only the last three digits matter when deciding if a given number is divisible by 8.
2. If the number formed by the last three digits is divisible by 8, then the original number is divisible by 8.
3. If the original number is divisible by 8, then the number formed by the last three digits is divisible by 8.
Do you agree? Why or why not?

Accepted Answer

Yes, all three statements are true. Examples include 8576, 7648, 5024 which are divisible by 8. — Divisibility by 8 depends on last three digits because 1000 is divisible by 8, so the rest of the number contributes multiples of 8.

Question 11

is a leap year (as February has 29 days). Leap years occurs 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?

Accepted Answer

b. From the year 2024 till 2099, the number of leap years is 19.
a. (Answer depends on student's birth year.) — Leap years are years divisible by 4 except those divisible by 100 but not by 400. Counting such years between 2024 and 2099 gives 19 leap years.

Question 12

Find the largest and smallest 4-digit numbers that are divisible by 4 and are also palindromes.

Accepted Answer

Largest 4-digit palindrome divisible by 4: 9999.
Smallest 4-digit palindrome divisible by 4: 1001. — Check palindromes starting from smallest and largest 4-digit numbers and test divisibility by 4.

Question 13

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.

Accepted Answer

a. Sometimes true. Example: 2 + 6 = 8 (multiple of 4), but 2 + 4 = 6 (not multiple of 4).
b. Sometimes true. Example: 1 + 3 = 4 (multiple of 4), but 1 + 5 = 6 (not multiple of 4). — Sum of two even or two odd numbers can be multiple of 4 depending on the numbers chosen.

Question 14

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

Accepted Answer

Remainders:
78 → 8 (mod 10), 3 (mod 5), 0 (mod 2)
99 → 9, 4, 1
173 → 3, 3, 1
572 → 2, 2, 0
980 → 0, 0, 0
1111 → 1, 1, 1
2345 → 5, 0, 1 — Divide each number by 10, 5, and 2 and note the remainder.

Question 15

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?

Accepted Answer

The two numbers are 5 and 8. If a number is divisible by 5 and 8, it is divisible by 2, 4, 5, 8, and 10. — Divisibility by 8 ensures divisibility by 2 and 4; divisibility by 5 ensures divisibility by 5 and 10.

Prime Time

Prime Time — Study Notes

Introduction to Prime Numbers

The Sieve of Eratosthenes

Prime Factorization

Practice Questions — Prime Time

All 10 Chapters in Ganita Prakash