If digits are not to be used with repetition, find the number of four- digit odd numbers divisible by 15 which can be formed by using 0,2,3,5,6,7, and 9?
Answer: 30
The answer is 30, but I don't know how to get 30, please help me how to solve it🙏
divisible by 15 means divisible by 5 and by 3
... the number must end in 5 (an odd number)
... if a number is divisible by 3, the sum of its digits is divisible by 3
so the three digits before the five must sum to
... 10 or 13 or 16 or 19 or 22
... the digits can be in any order
307 , 370 , 703 , 730
etc.
I still don't understand🙏 why must sum to 10 or 13 or 16 etc.?
Can we use combination to solve it?
Sure, I'll be happy to help you solve the problem step by step.
To find the number of four-digit odd numbers that are divisible by 15 using the given digits without repetition, we can break down the problem into smaller steps.
Step 1: Determine all possible combinations of the given digits without repetition.
In this case, we are given the digits 0, 2, 3, 5, 6, 7, and 9. We need to find all possible combinations of these digits without repetition for a four-digit number. To do this, we can use the concept of permutations.
The formula to find the number of permutations of n objects taken r at a time is given by:
P(n, r) = n! / (n - r)!
In our case, n = 7 (the number of digits) and r = 4 (the number of positions in the four-digit number). So, the number of possible combinations without repetition is:
P(7, 4) = 7! / (7 - 4)! = 7! / 3! = 7 x 6 x 5 = 210
Therefore, there are 210 possible combinations of the given digits without repetition.
Step 2: Determine the conditions for the four-digit numbers to be divisible by 15.
To find out if a number is divisible by 15, it means that it should be divisible by both 3 and 5.
- Divisible by 3: A number is divisible by 3 if the sum of its digits is divisible by 3.
- Divisible by 5: A number is divisible by 5 if its last digit is either 0 or 5.
Step 3: Apply the conditions to reduce the possible combinations.
We can apply the conditions mentioned in Step 2 to reduce the number of possible combinations.
Condition 1: Divisible by 3
We need to find combinations that have a sum divisible by 3. From the given digits (0, 2, 3, 5, 6, 7, and 9), the possible combinations with a sum divisible by 3 are:
(3, 6, 9)
(2, 5, 8)
(0, 3, 6, 9)
(0, 3, 6)
(0, 6, 9)
So, we have reduced the possibilities to 5 combinations.
Condition 2: Divisible by 5
We need to find combinations that have a last digit of either 0 or 5. From the reduced combinations in the previous step, the possible combinations with a last digit of 0 or 5 are:
(3, 6, 9)
(0, 3, 6, 9)
(0, 3, 6)
(0, 6, 9)
Note: These combinations are the same as the previous step as all of them end with 0 or 5.
Step 4: Determine the number of odd numbers.
Since we are looking for odd numbers, the last digit cannot be 0. Therefore, we need to exclude combinations that end with 0.
From the possible combinations above, we have 3 combinations remaining:
(3, 6, 9)
(0, 3, 6, 9)
(0, 3, 6)
Step 5: Determine the number of four-digit numbers.
To form a four-digit number using these combinations, we place them in different positions.
For the first combination (3, 6, 9), we have 3 choices for the first position (thousands digit). For the remaining three digits, we have 6 permutations (P(6, 3)).
For the second combination (0, 3, 6, 9), we have 4 choices for the first position (thousands digit). For the remaining three digits, we have 6 permutations (P(6, 3)).
For the third combination (0, 3, 6), we have 4 choices for the first position (thousands digit). For the remaining three digits, we have 6 permutations (P(6, 3)).
So, the total number of four-digit odd numbers that are divisible by 15 and can be formed by using the given digits without repetition is:
(3 x 6) + (4 x 6) + (4 x 6) = 18 + 24 + 24 = 66
However, it's important to note that we are only looking for numbers divisible by 15. So, the answer should be smaller.
Step 6: Consider only the numbers divisible by 15.
Among the 66 possible four-digit odd numbers, we need to identify the ones that are divisible by 15. From the condition of divisibility by 15, we know that the numbers divisible by 15 should also be divisible by 3 and 5.
From the possible combinations, we can see that only the combination (0, 3, 6, 9) is divisible by 15.
Step 7: Determine the number of final combinations.
For the combination (0, 3, 6, 9), we have 4 choices for the first position (thousands digit). For the remaining three digits, we have 6 permutations (P(6, 3)).
Therefore, the total number of four-digit odd numbers divisible by 15 and can be formed using the given digits without repetition is:
4 x 6 = 24.
So, the answer is 24, not 30 as you mentioned before. I apologize for any confusion caused.