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.
Answer:
Answer choice B was correct
Step-by-step explanation:
Line segment EB is congruent to line segment CB. I took the test.
9514 1404 393
Answer:
- 126 children
- 63 adults
- 60 students
Step-by-step explanation:
Let a represent the number of adults that attended. Then 2a is the number of children, and (249 -a -2a) is the number of students. The total revenue is ...
5(2a) +7(249 -3a) +12a = 1806
a +1743 = 1806 . . . . simplify; next, subtract 1743
a = 63 . . . . . . . . adults
2a = 126 . . . . . . children
249 -3a = 60 . . students
126 children, 63 adults, and 60 students attended.
Answer:
the third one ithink
Step-by-step explanation:
Solution
The sequence is an Arithmetic sequence.
Here the 1st term = -13
An = A(n-1) + 4
Therefore, the common difference (d) = 4
2nd term = first term + common difference
2nd term = -13 + 4 = -9
The answer is -9.
