4: 2, 3, 5 and 7. 1 is not a prime.
42
How many two digit number are divisible by 5
6 x 7 = 42 (it helps to know the multiplication tables), so the prime factors are 2 x 3 x 7, and the combination is 237
Yes
997 is the largest 3-digit prime number.
2, 3, 5, and 7 are the only one-digit prime numbers.
The prime factors of 42 are 2, 3, and 7. These prime numbers multiplied create 42.2*3*7 = 422 x 3 x 7 = 42.
One way to solve this is to multiply the lowest prime numbers together, starting from 2: 2x3=6. 6x5=30. Multiplying by 7 would yield a 3-digit number, so we are looking for all 2-digit numbers with 3 distinct prime factors. We already have 30. 2x3x7=42. 2x3x11=66. 2x3x13=78. 2x5x7=70. So, to answer the original question, the 2-digit numbers that have the most factors in their prime factorization are, in ascending order, 30, 42, 66, 70, and 78.
There are only two prime numbers that are consecutive numbers, 2 and 3. Their product is 2 x 3 = 6. The first prime numbers are 2, 3, 5, and 7 and the only two consecutive prime numbers whose product is a single digit are 2 and 3. (The next two consecutive prime numbers, 3 and 5, have a two-digit product.)
1, 2, 3, 7 are the prime factors of 42.
The prime factors of 42 are 2, 3 and 7
The prime factors of 42 are 2, 3 and 7.
42 = 2 * 3 * 7the prime factors of 42 are 1,2,3,7the product of these numbers is... 1x2x3x7 = 42
7, 3, 2
It is: 2*3*7 = 42
2 x 3 x 7 = 42