three digit numbers are to be formed from digits 0,1,2,3...9 if repetitionis allowed.
(a) how many of such numbers can be formed?
(b)how manyof the numbers in(a)are divisible by 3 and how many of 9?
(c) find the sum of all the numbers divisible by 3 but not divisible by 9
Effectively, you have all the numbers form 100 to 999 , (can't start with 0 or else we have a 2 digit number)
There would be 999-100 + 1 = 900 of them
or
using the standard permutation method
choices for 1 digit = 9
choices for 2nd digit = 10
choices for 3rd digit = 10
number of cases = 9x10x10 = 900
those divisible by 3 are:
102, 105, 108 , ... 999
they form an arithmetic sequence with
a = 102, d = 3
what term number is 999?
102 + (n-1)(3) = 999
3n - 3 = 897
3n = 900
n = 300 , makes sense that 1/3 of them would be divisible by 3
those divisible by 9:
108, 117, 126,...,999
108 + (n-1)(9) = 999
9n - 9 = 891
9n = 900
n = 100 , or 1/9 of the 900 are divisible by 9
using the formula:
sum of n terms of an AS = (n/2)(first + last)
sum of the multiples of 3
= 150(102 + 999) = 165150
sum of multiples of 9
= 50(108 + 999) = 55350
sum of all divisible by 3 but not by 9
= 165150 - 55350 = 109800
To solve this problem, we can break it down into smaller steps and use some basic concepts of combinatorics and divisibility.
(a) To find the total number of three-digit numbers that can be formed using the digits 0 to 9 with repetition allowed, we can think of each digit as having 10 options (since 0 to 9 are 10 digits). So, for each of the three digits, there are 10 options. Since repetition is allowed, the total number of three-digit numbers is calculated by multiplying these options together: 10 * 10 * 10 = 1000. Therefore, there are 1000 such numbers that can be formed.
(b) To find the number of three-digit numbers that are divisible by 3 or 9, we have to consider the divisibility rules of these numbers.
- Divisibility by 3: A number is divisible by 3 if the sum of its digits is divisible by 3. Since we have numbers from 0 to 9, the possible combinations that sum up to a number divisible by 3 are:
- (0, 0, 0)
- (1, 2, 3)
- (2, 1, 3)
- (3, 0, 3)
- (3, 3, 0)
- (4, 2, 3)
- (5, 1, 3)
- (6, 0, 3)
- (6, 3, 0)
- (7, 2, 3)
- (8, 1, 3)
- (9, 0, 3)
So, there are 12 combinations that sum up to a number divisible by 3.
- Divisibility by 9: A number is divisible by 9 if the sum of its digits is divisible by 9. Following the same approach as above, we can see that the only combination that sum up to a number divisible by 9 is (0, 0, 0). So, there is only 1 combination that sums up to a number divisible by 9.
(c) To find the sum of all three-digit numbers that are divisible by 3 but not divisible by 9, we need to consider the numbers that have a sum of digits divisible by 3 but not a sum divisible by 9. From the previous calculations, we know that there are 12 combinations that sum to a number divisible by 3. Among these 12 combinations, only one combination sums up to a number divisible by 9. Thus, there are 12 - 1 = 11 combinations that are divisible by 3 but not divisible by 9.
To find their sum, we can calculate the sum for each combination and add them up:
(1 + 2 + 3) + (2 + 1 + 3) + (3 + 0 + 3) + ... + (9 + 0 + 3)
In each set of parentheses, the first two digits have 10 options each, and the last digit has 3 options. So, the sum of each set of parentheses is (10+10+3) = 23. Multiplying this by the number of combinations, we get: 11 * 23 = 253.
Therefore, the sum of all three-digit numbers that are divisible by 3 but not divisible by 9 is 253.