NCERT MCQ Solutions for Class 6 Maths Chapter 5 Prime Time

The positive common multiples of 3 and 5 are 15, 30, 45, … .
The tenth such number = 15 × 10 = 150.
Each new “idli-vada” occurs 15 numbers after the previous one.

The multiples of 3 are 3, 6, 9, …, 90.
Number of these multiples = 90 ÷ 3 = 30
The common multiples of 3 and 5 are included because the question also counts “idli-vada”.

Common factors of 28 and 70 are 1, 2, 7 and 14.
28 = 14 × 2 and 70 = 14 × 5
So jumps of 14 land on both treasures, and no larger common factor is available.

40 × 8 = 320, 40 × 9 = 360 and 40 × 10 = 400.
The previous multiple, 280, is too small; the next multiple, 440, is too large.
Therefore, the required multiples are 320, 360 and 400.

Positive factors of 6 are 1, 2, 3 and 6.
Their sum = 1 + 2 + 3 + 6 = 12
Twice the number = 2 × 6 = 12
The two values are equal, so 6 is perfect.

The smallest prime number is 2.

The player says Idli for 3 and it’s multiple and Vada for 5 and it’s multiple, the first common multiple of 3 and 5 will be Idli-Vada.

A prime number has exactly two positive factors: 1 and the number itself.
For example, the factors of 7 are 1 and 7.
Having any additional positive factor would make a number greater than 1 composite.

The only positive factor of 1 is 1 itself.
A prime needs exactly two positive factors, while a composite number needs more than two.
Since 1 has just one positive factor, it belongs to neither group.

The factors of 2 are 1 and 2, so it is prime.
Every even number greater than 2 has 1, 2 and itself as different factors.
Those numbers have more than two factors and are composite.

The numbers 22, 24, 26, 28 and 30 are even and greater than 2.
Also, 21 = 3 × 7, 25 = 5 × 5 and 27 = 3 × 9.
Only 23 and 29 have no positive factors other than 1 and themselves.

41 and 43 are both prime, and 43 − 41 = 2.
The other pairs each include a composite number: 21 = 3 × 7, 33 = 3 × 11 and 51 = 3 × 17.
Therefore, only 41 and 43 meet both conditions.

9 is not a factor of 56.

Prime factorization of 72 = 2 × 2 × 2 × 3 × 3
Or 2³ × 3²

Factors of 4 are 1, 2 and 4.
Factors of 9 are 1, 3 and 9.
Their only common factor is 1, so they are co-prime. Both are composite because each has more than two positive factors.

Multiples of 4 include 4, 8, 12, …, 36.
Multiples of 9 are 9, 18, 27, 36, … .
Their first positive common multiple is 36. Since 4 and 9 are co-prime, it also equals 4 × 9.

The visited pegs are 1, 8, 5, 2, 9, 6, 3, 10, 7, 4 and then 1 again.
All 10 pegs are visited before the return. The peg count 10 and thread-gap 7 have only 1 as a common factor.

56 = 7 × 8
= 7 × 2 × 2 × 2
= 2 × 2 × 2 × 7
Every factor in the final product is prime. Expressions containing 4, 8 or 14 are not yet complete prime factorisations.

Required number = 2 × 3 × 3 × 11
= 2 × 9 × 11
= 18 × 11 = 198
Both copies of the factor 3 must be included; using only one would give a different number.

The only even prime number is 2.

There are 6 factors of 28 of which 28 itself is the biggest factor and its positive factors are 1, 2, 4, 7, 14 and 28.

1955 ÷ 5 = 391
391 = 17 × 23
Therefore, 1955 = 5 × 17 × 23.
The numbers 5, 17 and 23 are all prime, so this gives the complete prime factorisation of 1955.

The four smallest different primes are 2, 3, 5 and 7.
Their product = 2 × 3 × 5 × 7
= 210
Replacing any of them with a larger prime or repeating an extra factor would make the product larger.

56 = 2 × 2 × 2 × 7 and 25 = 5 × 5.
Combining these factors gives 56 × 25 = 2 × 2 × 2 × 5 × 5 × 7.
The factors can be rearranged without changing their product.

42 = 2 × 3 × 7
12 = 2 × 2 × 3
The factorisation of 12 needs two copies of 2, but 42 contains only one.
So the complete prime factorisation of 12 is not included in that of 42.

57 = 3 × 19 and 85 = 5 × 17.
There is no common prime factor, so their only common positive factor is 1.
The other pairs have common factors 3, 11 and 7 respectively, so they are not co-prime.

Perfect number: A positive integer that is equal to the sum of its proper divisors. The smallest perfect number is 6, which is the sum of 1, 2, and 3.

A composite number is one that has more than two factors.

A number divisible by 5 ends in 0 or 5.
The largest such number below 399 is 395.
395 = 5 × 79
The next multiple is 400, which is greater than 399 and therefore cannot be used.

The divisibility test for 4 uses the last two digits.
36 ÷ 4 = 9
Therefore, 8536 is divisible by 4. An even last digit alone is not sufficient: for example, 22 is even but is not divisible by 4.

For divisibility by 8, the number formed by the last three digits is checked.
552 = 8 × 69
So 8552 is divisible by 8. Looking only at whether the last digit is even cannot establish divisibility by 8.

173 = 10 × 17 + 3
The multiple 170 can be divided exactly by 10, leaving 3.
Since 3 is less than 10, it is the remainder. The units digit gives the remainder when a whole number is divided by 10.

Divisibility by 8 includes divisibility by 2 and 4.
If the number is also divisible by 5, its factorisation includes both 2 and 5, so it is divisible by 10.
Thus, checking 5 and 8 covers all five requirements.

The leap years are 2024, 2028, …, 2096. There is no century year in this interval.
Number of four-year gaps = (2096 − 2024) ÷ 4 = 18
Including the first leap year gives 18 + 1 = 19.

A four-digit palindrome has matching first and last digits. Divisibility by 4 requires an even last digit, so the first digit is at least 2.
2002 is not divisible by 4, but 2112 is because 12 ÷ 4 = 3.
Thus, 2112 is the smallest.

2 + 6 = 8, which is divisible by 4.
2 + 4 = 6, which is not divisible by 4.
Both examples add two even numbers, so divisibility by 4 holds sometimes, but not for every such sum.

The units digits are 6 and 5, so neither number ends in 0.
The other pairs also have product 10,000, but at least one number in each pair ends in 0 and fails the condition.

Product of the known entries = 5 × 5 = 25
Missing entry = 75 ÷ 25 = 3
Check: 5 × 5 × 3 = 75.
The number 3 is prime, so it satisfies both the product and prime-entry conditions.

A jump size of 15 will allow Jumpy to land on both 15 and 30, as 15 is a factor of both numbers; meaning you can jump 15 steps to reach both from a starting point.

There are 4 prime numbers between 1 and 10 are 2, 3, 5, 7.