The only way 3 digits can have product 24 is
1 x 3 x 8 = 241 x 4 x 6 = 242 x 2 x 6 = 242 x 3 x 4 = 24
So the digits comprises of 1,3,8 or 1,4,6, or 2,2,6, or 2,3,4
To be divisible by 3 the sum of the digits must be divisible by 3.
1+ 3+ 8=12, 1+ 4+ 6= 11, 2 +2 + 6=10, 2 +3 + 4=9Of those sums of digits, only 12 and 9 are divisible by 3.
So we have ruled out all but integers whose digits consist of1,3,8, and 2,3,4.
Meanwhile they must be odd they either must end in 1 or 3.
The only ones which can end in 1 are 381 and 831.
The others must end in 3.
They must be greater than 152 which is 225. So the
First digit cannot be 1. So the only way its digits can contain of1,3,8 and close in 3 is to be 813.
The rest must contain of the digits 2,3,4, and the only way they can end in 3 is to be 243 or 423.
So there are precisely five such three-digit integers: 381, 831, 813, 243, and 423.
The primes are 2, 3, 5, 7, 11, 13, 17, 19. There are 8 of them, so they make up 40% of the numbers 1–20.
Simplify 6b2+7bc+2c2.
Hope this helps
Answer:
2
Step-by-step explanation:
