How many positive three-digit integers with each digit greater than 4 are divisible by 6?
So whats the solution
Hey, just count the options, the answer is 16.
there's 558, 576, 588, 666, 678, 696, 756, 768, 786, 798, 858, 876, 888, 966, 978, and 996.
make sure to check your answer before calling it simple ;)
You did it all wrong, Damon. Only the last digit has to be even (to be divisible by 2).
6 = 2*3
Last digit has two choices: 6 and 8.
For it to be divisible by 3, sum of all three digits must be a multiple of 3.
There are only so many choices, since the digits are limited to 5, 6, 7, 8, 9.
The largest possible sum is 9+9+8 = 26, and the greatest MULTIPLE OF 3 below 26 is 24.
We find the minimum this time, and its 16, and the smallest MULTIPLE OF 3 larger than 16 is 18.
Cases:
18: 556
21: 696, 786, 876, 966, 588, 678, 768, 858
24: 998
Quite a simple casework problem, and count for yourself.
Final answer: 10
To solve this problem, we need to find the number of positive three-digit integers where each digit is greater than 4 and is divisible by 6.
Step 1: Find the range of three-digit integers
The range of three-digit integers is from 100 to 999.
Step 2: Determine the divisibility rule for 6
A number is divisible by 6 if it is divisible by both 2 and 3.
Step 3: Determine the divisibility rule for 2
A number is divisible by 2 if its last digit is even, i.e., 0, 2, 4, 6, or 8.
Step 4: Determine the divisibility rule for 3
A number is divisible by 3 if the sum of its digits is divisible by 3.
Step 5: Find the count of three-digit integers divisible by 6
To find the number of three-digit integers divisible by 6, we need to consider two conditions:
- All digits are greater than 4.
- The sum of the digits is divisible by 3.
Condition 1: All digits are greater than 4
For the hundreds place, we have 5, 6, 7, 8, or 9 (total of 5 options).
For the tens and units place, we have 5, 6, 7, 8, or 9 (total of 5 options).
Therefore, there are 5 * 5 * 5 = 125 three-digit integers with all digits greater than 4.
Condition 2: The sum of the digits is divisible by 3
The smallest possible sum of three digits greater than 4 is 5 + 5 + 5 = 15.
The largest possible sum is 9 + 9 + 9 = 27.
To find the count of three-digit integers with a sum divisible by 3, we count the number of multiples of 3 in the range from 15 to 27.
15, 18, 21, 24, 27
There are 5 multiples of 3 within this range.
Step 6: Find the count of three-digit integers satisfying both conditions
To find the count of three-digit integers satisfying both conditions, we multiply the counts from Condition 1 and Condition 2.
Total count = 125 * 5 = 625
Therefore, there are 625 positive three-digit integers where each digit is greater than 4 and divisible by 6.
6 = 2 * 3
x y z
all must be divisible by 2 (therefore even) and all bigger than 4
so each is 6 or 8
eg
6 6 6 yes
6 6 8 NO
6 8 6 NO
6 8 8 NO
8 6 6 NO
8 6 8 NO
8 8 6 NO
8 8 8 YES