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:
The answer is 13
Step-by-step explanation:
Hope this helps :)
Answer:
No
Step-by-step explanation:
Given that:
Amount earned when a student does comminty service = $5
Number of students in club = 12
Amount earned E, is a function of number of members (n) who does community service.
Hence,
E = n * amount earned per member
Is 24 a possible output?
To test, then E = 24
24 = n * 5
24 = 5n
24/5 = 5n/5
4.8 = 5
Hence, 24 is not a possible output, be use number of membwra cannot be a decimal
Answer:
3 * 14+x
Step-by-step explanation:
* is an multiplication sign
Answer:
ANSWER IS 12
Step-by-step explanation: