Ganita Prakash · Chapter 5
Prime Time
Complete, step-by-step solutions to every in-text and exercise question — multiples, factors, primes, co-primes, prime factorisation and divisibility tests.
Common Multiples & Common Factors
‘Idli’ is said for multiples of the smaller number (here 4) and ‘vada’ for multiples of the larger number. Nobody said just ‘vada’, which means every multiple of the larger number is also a multiple of 4 — so it always becomes ‘idli-vada’.
This happens only when the larger number is itself a multiple of 4. Among 2, 3, 5, 8, 10 the only multiple of 4 is 8.
Check: multiples of 8 are \(8, 16, 24, 32,\dots\) — each one is also a multiple of 4, so ‘vada’ alone is never said.
A jump size lands on a number only if it is a factor of that number. To land on both, the jump size must be a common factor of 15 and 30.
Factors of 15: \(1, 3, 5, 15\)
Factors of 30: \(1, 2, 3, 5, 6, 10, 15, 30\)
Common factors: \(1, 3, 5, 15\)
1. Shaded numbers: the shaded numbers (33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69) are all multiples of 3.
2. Circled numbers (32, 36, 40, 44, 48, 52, 56, 60, 64, 68) are all multiples of 4.
3. Both shaded and circled: \(36, 48, 60\). These are the common multiples of 3 and 4 — every one is a multiple of \(3\times4=12\).
‘Idli-vada’ is said at numbers that are multiples of both 3 and 5 — that is, the multiples of 15.
\(15, 30, 45, 60, 75, 90, 105, 120, 135, \mathbf{150}\)
The 10th multiple of 15 is \(15\times10\).
- How many times would the children say ‘idli’ (including ‘idli-vada’)?
- How many times would they say ‘vada’ (including ‘idli-vada’)?
- How many times would they say ‘idli-vada’?
a. ‘Idli’ = every multiple of 3 up to 90:
\(\left\lfloor \dfrac{90}{3}\right\rfloor = 30\)
b. ‘Vada’ = every multiple of 5 up to 90:
\(\left\lfloor \dfrac{90}{5}\right\rfloor = 18\)
c. ‘Idli-vada’ = every multiple of 15 up to 90:
\(\left\lfloor \dfrac{90}{15}\right\rfloor = 6\)
Use the same idea, now up to 900:
‘Idli’ (multiples of 3): \(\dfrac{900}{3} = 300\)
‘Vada’ (multiples of 5): \(\dfrac{900}{5} = 180\)
‘Idli-vada’ (multiples of 15): \(\dfrac{900}{15} = 60\)
Yes. The left circle holds the ‘idli’ numbers (multiples of 3), the right circle holds the ‘vada’ numbers (multiples of 5), and the overlap holds the numbers that are multiples of both — exactly the ‘idli-vada’ numbers (multiples of 15).
Playing the game up to 60, the diagram looks like this:
List multiples of 40 and keep those strictly between 310 and 410:
\(40\times7 = 280\;(\text{too small})\)
\(40\times8 = 320,\quad 40\times9 = 360,\quad 40\times10 = 400\)
\(40\times11 = 440\;(\text{too big})\)
- I am less than 40. One of my factors is 7. The sum of my digits is 8.
- I am less than 100. Two of my factors are 3 and 5. One of my digits is 1 more than the other.
a. Multiples of 7 below 40: \(7, 14, 21, 28, 35\). Their digit sums are \(7, 5, 3, 10, 8\). Only \(35\) gives a digit sum of \(3+5=8\).
b. Factors 3 and 5 ⇒ a multiple of 15: \(15, 30, 45, 60, 75, 90\). We need digits differing by 1 → \(45\) (since \(5 = 4+1\)).
Try \(6\). Its factors are \(1, 2, 3, 6\).
\(1 + 2 + 3 + 6 = 12 = 2\times 6\) ✓
- 20 and 28
- 35 and 50
- 4, 8 and 12
- 5, 15 and 25
a. \(20\!:\,1,2,4,5,10,20\) · \(28\!:\,1,2,4,7,14,28\) → common 124
b. \(35\!:\,1,5,7,35\) · \(50\!:\,1,2,5,10,25,50\) → common 15
c. \(4,8,12\) → common 124
d. \(5,15,25\) → common 15
Multiples of 50 are the even multiples of 25 (50, 100, 150…). So the odd multiples of 25 work:
\(25,\;75,\;125\quad(\text{also }175, 225,\dots)\)
The first ‘idli-vada’ happens at the first common multiple (the LCM) of the two numbers. We need two numbers below 10 whose first common multiple is more than 50.
\(7 \text{ and } 8:\ \text{LCM} = 56 > 50\) ✓
\(8 \text{ and } 9:\ \text{LCM} = 72 > 50\) ✓
The jump sizes are the common factors of 28 and 70.
\(28\!:\,1,2,4,7,14,28\)
\(70\!:\,1,2,5,7,10,14,35,70\)
Common: \(1, 2, 7, 14\)
The common multiples shown are \(24, 48, 72\) — these are consecutive multiples of \(24\). So the two numbers must have a first common multiple (LCM) of 24.
A neat choice is 3 and 8 (since \(\text{LCM}(3,8)=24\)). Then:
- Multiples of 3 only: 3, 6, 9, 12, 15, 18, 21, 27, 30, …
- Multiples of 8 only: 8, 16, 32, 40, 56, 64, …
- Common (centre): 24, 48, 72
Other pairs also work, e.g. 6 and 8 or 8 and 24 — any pair whose LCM is 24.
Take the highest power of each prime needed by 1–10 (leaving out 7):
\(2^3\) (from 8), \(3^2\) (from 9), \(5\) (from 5, 10)
\(2^3\times 3^2\times 5 = 8\times 9\times 5 = 360\)
Now we also include the prime 7:
\(2^3\times 3^2\times 5\times 7 = 360\times 7 = 2520\)
Prime Numbers
From 21 to 30 there are 10 numbers.
Primes: \(23, 29\) → 2 primes
Composites: \(21, 22, 24, 25, 26, 27, 28, 30\) → 8 composites
No. Every even number other than 2 has 2 as a factor, so it has at least three factors (1, 2 and itself) — making it composite.
Smallest: \(3 – 2 = 1\) (only between 2 and 3).
Largest: \(97 – 89 = 8\).
No — the count varies.
Most primes: 1–10 and 11–20 have 4 each (2, 3, 5, 7 and 11, 13, 17, 19).
Fewest primes: 91–100 has just 1 (only 97).
\(23\) → prime ✓
\(51 = 3\times 17\) → composite
\(37\) → prime ✓
\(26 = 2\times 13\) → composite
\(2 + 3 = 5\)
\(3 + 7 = 10\)
\(2 + 13 = 15\)
Other pairs also work, such as \((7, 13)\) since \(7+13 = 20\).
Look for two-digit primes whose reversal is also prime:
\(17 \leftrightarrow 71,\quad 37 \leftrightarrow 73,\quad 79 \leftrightarrow 97\)
The stretch just before 97 has a long run with no primes:
\(90, 91, 92, 93, 94, 95, 96\)
(Here \(91 = 7\times 13\) — none of these seven is prime.)
(3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), (71, 73)
- There is no prime number whose units digit is 4.
- A product of primes can also be prime.
- Prime numbers do not have any factors.
- All even numbers are composite numbers.
- 2 is a prime and so is 3. For every other prime, the next number is composite.
a. True. A units digit of 4 makes the number even, so it is divisible by 2 and cannot be prime.
b. False. A product of two or more primes has those primes as factors, so it is composite (e.g. \(2\times3 = 6\)).
c. False. Primes have exactly two factors — 1 and the number itself.
d. False. 2 is even but prime; only even numbers greater than 2 are composite.
e. True. Any prime greater than 2 is odd, so the next number is even and greater than 2 — hence composite. Only 2 & 3 are consecutive primes.
\(45 = 3\times 3\times 5\) — only two distinct primes
\(60 = 2\times 2\times 3\times 5\) — a prime repeats
\(91 = 7\times 13\) — only two primes
\(105 = 3\times 5\times 7\) — exactly three distinct primes ✓
\(330 = 2\times 3\times 5\times 11\) — four distinct primes
The possible numbers are \(245, 254, 425, 452, 524, 542\).
Every one ends in \(2, 4\) or \(5\), so each is divisible by 2 or 5 — none can be prime.
\(2\times 2+1 = 5\) ✓
\(2\times 5+1 = 11\) ✓
\(2\times 11+1 = 23\) ✓
\(2\times 23+1 = 47\) ✓
\(2\times 29+1 = 59\) ✓
Co-prime Numbers for Safekeeping Treasures
A pair is “safe” when the two numbers share no common factor other than 1.
a. \(15\) & \(39\) share the factor 3 → not safe.
b. \(4\) & \(15\): factors \(\{1,2,4\}\) and \(\{1,3,5,15\}\) → only 1 in common → safe.
c. \(18\) & \(29\): 29 is prime and doesn’t divide 18 → only 1 in common → safe.
d. \(20\) & \(55\) share the factor 5 → not safe.
Two numbers are co-prime if their only common factor is 1.
a. \(18 = 2\times 3^2,\ 35 = 5\times 7\) → no common prime → co-prime ✓
b. \(15 = 3\times 5,\ 37\) is prime → co-prime ✓
c. \(30 = 2\times 3\times 5,\ 415 = 5\times 83\) → share 5 → not co-prime
d. \(17\) prime, \(69 = 3\times 23\) → co-prime ✓
e. \(81 = 3^4,\ 18 = 2\times 3^2\) → share 3 → not co-prime
First common multiple = product (the numbers are co-prime):
\((3,5)\to 15 = 3\times 5\); \((3,7)\to 21\); \((4,9)\to 36\)
First common multiple < product (they share a factor):
\((3,6)\to 6 < 18\); \((3,12)\to 12 < 36\); \((6,15)\to 30 < 90\)
Relation: the first common multiple equals the product exactly when the two numbers are co-prime. If they share a factor, the first common multiple is smaller.
Prime Factorisation
Every composite number can be broken down into a product of primes — its prime factorisation. A factor tree makes this easy:
\(64 = 2^6\)
\(104 = 2^3\times 13\)
\(105 = 3\times 5\times 7\)
\(243 = 3^5\)
\(320 = 2^6\times 5\)
\(141 = 3\times 47\)
\(1728 = 2^6\times 3^3\)
\(729 = 3^6\)
\(1024 = 2^{10}\)
\(1331 = 11^3\)
\(1000 = 2^3\times 5^3\)
\(2\times 3\times 3\times 11 = 2\times 9\times 11\)
1955 ends in 5, so divide by 5:
\(1955 = 5\times 391\)
\(391 = 17\times 23\)
- 56 × 25
- 108 × 75
- 1000 × 81
Factorise each part, then combine:
a. \(56\times 25 = (2^3\times 7)\times(5^2) = 2^3\times 5^2\times 7\)
b. \(108\times 75 = (2^2\times 3^3)\times(3\times 5^2) = 2^2\times 3^4\times 5^2\)
c. \(1000\times 81 = (2^3\times 5^3)\times(3^4) = 2^3\times 3^4\times 5^3\)
Use the smallest primes, once each:
a. \(2\times 3\times 5 = 30\)
b. \(2\times 3\times 5\times 7 = 210\)
a. No. \(30 = 2\times 3\times 5,\ 45 = 3^2\times 5\) → share 3 and 5.
b. Yes. \(57 = 3\times 19,\ 85 = 5\times 17\) → no common prime.
c. No. \(121 = 11^2,\ 1331 = 11^3\) → share 11.
d. Yes. \(343 = 7^3,\ 216 = 2^3\times 3^3\) → no common prime.
The second divides the first only if its whole prime factorisation fits inside the first’s.
a. No. \(225 = 3^2\times 5^2,\ 27 = 3^3\). 27 needs three 3s but 225 has only two.
b. Yes. \(96 = 2^5\times 3,\ 24 = 2^3\times 3\) fits inside → \(96 = 24\times 4\).
c. No. \(343 = 7^3\); 17 is a prime that isn’t a factor.
d. No. \(999 = 3^3\times 37,\ 99 = 3^2\times 11\); 11 isn’t a factor of 999.
They share the primes 3 and 7, so they are not co-prime.
Neither factorisation fits completely inside the other (the first has a 2 the second lacks; the second has an 11 the first lacks), so neither divides the other.
Yes. Two different primes have no factors except 1 and themselves, so their only common factor is 1.
e.g. \(2\) & \(3\), or \(3\) & \(11\) — always co-prime.
Divisibility Tests
For divisibility by 4, only the last two digits matter.
Last two digits: \(36\), and \(36\div 4 = 9\) — exact.
All three are true. Every hundred (100, 200, 300…) is a multiple of 4, so the hundreds/thousands part never affects divisibility by 4 — only the last two digits decide it.
e.g. \(124, 364, 4028\) are all divisible by 4, matching their last two digits (24, 64, 28).
120–140: \(128, 136\)
1120–1140: \(1128, 1136\)
3120–3140: \(3128, 3136\)
Observation: the same endings (…128, …136) repeat — for divisibility by 8 only the last three digits matter.
Keep 85 and pick last two digits making the last three digits divisible by 8:
\(8552\) → last three digits \(552 = 8\times 69\) ✓
a. This depends on your birth year — list the multiples of 4 from then until now (e.g. 2012, 2016, 2020, 2024 …), skipping any century year not divisible by 400.
b. Leap years from 2024 to 2099 are the multiples of 4: \(2024, 2028, \dots, 2096\) (2100 is beyond the range).
Count \(= \dfrac{2096 – 2024}{4} + 1 = \dfrac{72}{4} + 1 = 18 + 1 = 19\)
A 4-digit palindrome looks like \(\overline{abba}\). It is divisible by 4 only when its last two digits \(\overline{ba}\) are divisible by 4.
Largest: if \(a = 9\), the ending \(\overline{b9}\) is odd — never divisible by 4. Trying \(a = 8\) gives \(\overline{b8}\); the largest works at \(b = 8\):
\(8888\) → \(88 \div 4 = 22\) ✓
Smallest: \(a = 1\) gives an odd ending. With \(a = 2\), \(\overline{b2}\) is divisible by 4 at \(b = 1\):
\(2112\) → \(12 \div 4 = 3\) ✓
a. Sometimes true.
\(2 + 6 = 8\) ✓ but \(2 + 4 = 6\) ✗
b. Sometimes true.
\(1 + 3 = 4\) ✓ but \(1 + 5 = 6\) ✗
The last digit alone gives all three remainders.
| Number | ÷ 10 | ÷ 5 | ÷ 2 |
|---|---|---|---|
| 78 | 8 | 3 | 0 |
| 99 | 9 | 4 | 1 |
| 173 | 3 | 3 | 1 |
| 572 | 2 | 2 | 0 |
| 980 | 0 | 0 | 0 |
| 1111 | 1 | 1 | 1 |
| 2345 | 5 | 0 | 1 |
Checking 8 automatically covers 2 and 4 (any multiple of 8 is a multiple of 2 and 4). Checking 5 together with divisibility by 2 covers 10.
Divisible by 8 ⇒ divisible by 2 and 4; divisible by 5 and 2 ⇒ divisible by 10.
The number must end in 0 (for 10) and have its last three digits divisible by 8.
\(572\) — doesn’t end in 0 ✗
\(2352\) — doesn’t end in 0 ✗
\(5600\) — ends in 0, and \(600 \div 8 = 75\) ✓
\(6000\) — ends in 0, and \(000 \div 8 = 0\) ✓
\(77622160\) — ends in 0, and \(160 \div 8 = 20\) ✓
Factorise: \(10000 = 2^4\times 5^4\). A trailing 0 appears only when a number has both a 2 and a 5. So put all the 2s in one number and all the 5s in the other:
\(2^4 = 16\) and \(5^4 = 625\)
\(16\times 625 = 10000\) ✓ (units digits 6 and 5)