Prime Time: questions with answers

Add every factor, including 1 and 28: 1 + 2 + 4 + 7 + 14 + 28 = 56. Also, 2 x 28 = 56, so this matches the definition of a perfect number.

A number that has the sum of all its factors equal to twice the number is called a perfect number.

The factors of 8 are 1, 2, 4 and 8. The factors of 12 are 1, 2, 3, 4, 6 and 12. The factors of 16 are 1, 2, 4, 8 and 16. The common factors of all three sums are 1, 2 and 4, so the greatest is 4. The number 2 is a common factor, but it is not the greatest. The number 8 is a factor of 8 and 16, but not of 12.

Add some pairs: $1+3=4$, $3+5=8$, $5+7=12$, $7+9=16$. Every sum is a multiple of 4. To see why, look at the even number between the two odd numbers, such as 6 between 5 and 7. The first odd number is 1 less than it and the second is 1 more. So the sum is twice the even number: $5+7=6+6=12$. An even number is 2 times some number, so twice it is 4 times that number. So 4 always divides the sum. The sum $1+3=4$ is not divisible by 6 or 8. The number 2 also always divides the sum, but 4 is larger. So the largest number that always divides the sum is 4.

The thread is tied to every peg only when the number of pegs and the gap are co-prime. Here 9 = 3 × 3 and 4 = 2 × 2. No prime appears in both, so the only common factor of 9 and 4 is 1. So the answer is Yes. Being composite does not matter, and checking only the factor 2 is not enough. The gap also does not need to divide the number of pegs.

Any pair of consecutive numbers has exactly 1 common factor, and that factor is 1. For example, the factors of 14 are 1, 2, 7 and 14, and the factors of 15 are 1, 3, 5 and 15. The only common factor is 1. Two multiples of a number greater than 1 can never be just 1 apart. So consecutive numbers are always co-prime.

For a product, the count of each prime factor is the count in the first factor plus the count in the second factor. For prime 2, the count is 2 + 2 = 4.

Prime factorisation means writing 84 as a product of prime numbers only. The split 84 = 12 × 7 is not complete if 12 or 7 is composite. Continuing gives 2 × 2 × 3 × 7.

The first rectangle uses 1 row and 10 columns. The second rectangle uses 2 rows and 5 columns, so 2 x 5 = 10. Because 2 and 5 are both greater than 1, 10 has factors other than 1 and itself. Therefore, 10 is composite.

The circled numbers in the completed sieve are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37. Counting these circled numbers gives 12 primes.

The first common multiple is the smallest positive multiple shared by all the given numbers. A later common multiple does not meet the requirement.

Every entry must be a prime number, and each product must match the number shown. Check the entries: 4 = 2 × 2, so 4 is not prime. The other three entries are prime. The row products are 3 × 4 = 12 and 7 × 5 = 35, and both match. The column products are 3 × 7 = 21 and 4 × 5 = 20. The grid shows 24 for the first column, so that product is wrong. So there are two faults. Checking only the entries and the rows misses the column fault. Checking only the products misses the entry that is not prime.

All entries are prime. The matching direction is the row products, but 2 x 5 = 10, not 14. A correct grid must satisfy both row products and column products.

Add the two odd numbers: 13 + 21 = 34. Since 34 lies between 4 × 8 = 32 and 4 × 9 = 36, the sum is not a multiple of 4. The last digit 4 alone cannot decide divisibility by 4. Check the number formed by the last two digits instead. The sum of two odd numbers is sometimes a multiple of 4 and sometimes not, so 'always' and 'never' are both wrong.

Every number divisible by 8 is divisible by 4, because 8 = 2 × 4. The reverse is not always true. 16 is divisible by 4 and also by 8. But 12 is divisible by 4 and not divisible by 8. So some numbers divisible by 4 are also divisible by 8, and some are not.