They are prime numbers, divisible only by one and themselves.
How many subsets are there in 2 3 5 7 11 13 17 19 23?
They are consecutive prime numbers 2 3 5 7 11 13 17 19 23 29 31 ..... etc
2 3 5 7 11 13 17 19 23 29 31 They are prime numbers.
5/6 of 13/5 = 5/6 of 8/5 = 8/6 = 4/3 or 11/3
There are 10 possible combinations so 60 possible permutations. The combinations are:(1, 3, 21), (1, 5, 19), (1, 7, 17), (1, 9, 15), (1, 11, 13),(3, 5, 17), (3, 7, 15), (3, 9, 13), (5, 7, 13), (5, 9, 11).
equivalent fractions for 11/13 = 22/26, 33/39, 44/52, 55/65,... To get the equivalent fractions of 11/13: Multiply 11/13 by 2/2, 3/3, 4/4, 5/5,... 11/13 * 2/2 = 22/26 11/13 * 3/3 = 33/39 11/13 * 4/4 = 44/52 11/13 * 5/5 = 55/65
3 + 3 + 5 = 11 3 + 3 + 7 = 13 3 + 5 + 7 = 15 3 + 3 + 11 = 17 3 + 5 + 11 = 19 3 + 7 + 11 = 21 3 + 7 + 13 = 23 3 + 11 + 11 = 25 5 + 11 + 11 = 27 7 + 11 + 11 = 29
Primes: 3, 5, 11, 13 Others are combinations of the primes, i.e. 3 x 5 = 15 3 x 11 = 33 3 x 13 = 39 5 x 11 = 55 5 x 13 = 65 11 x 13 = 143 3 x 5 x 11 = 165 2 x 5 x 13 = 195 3 x 11 x 13 = 429 5 x 11 x 13 = 715
To get the fraction equivalents for 11/13:Multiply 11/13 by 2/2, 3/3, 4/4, 5/5,...3 Fraction equivalents of 11/13: 22/26, 33/39, 44/52
11 is to 13 and 18 is to 25
5, 3, 11, 13 are the prime factors 32 * 5 * 11 * 13 = 6435
1, 3, 5, 7, 9, 11, 13, 15, 17, 19. 1, 3, 5, 7, 9, 11, 13, 15, 17, 19. 1, 3, 5, 7, 9, 11, 13, 15, 17, 19. 1, 3, 5, 7, 9, 11, 13, 15, 17, 19.
2, 3, 5, 7, 11, and 13
The prime factors of 30,030 are: 2, 3, 5, 7, 11 and 13.
2, 3, 5, 7, 11 and 13 are prime.
The factors of 33 are:1, 3, 11, 33The factors of 55 are:1, 5, 11, 55The HCF of both numbers is 11
16 = 1+15, 3+13, 5+11, 7+92,3,5,7,11,13 are primes < 1616 = 3+13 or 5+11