Suppose that, we have a set that contains {1,2,3,4,5,6,7,8} and we want to create a 3-digit number.

a. How many number of multiple three, but the digits are not repeated, can be created?

b. If the digits are repeated, how many number of multiple three can be created?

c. How many prime number from this case?

Can anybody show me the way to solve it? Thanks

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## Comments

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TopNewesti think a. Multiple three and different. So, you can write 8x7x6 = 336 b. Same number. So, you can write 8x8x8 = 512 c. The prime number is 2,3,5,7. There are 4 prime number terima kasih, semoga benar dan bermanfaat

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same as llham.A apply permutation and combination

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b.) this is the same as a three digit base 8 number, the largest of which is 8^3 =512. since 510 is the highest multiple of three lower than this we know there are a total of 170 multiples of three included in this set.

a.) to find the the multiples of three that have repeating digits use n=a

64+b8+c = a(63+1)+b(6+2)+c = a(63)+a(1)+b(6)+b(2)+cSince 63 and 6 are both multiples of three they will not affect whether the final number (n) is divisible by three so they drop out of the equation leaving a+2b+c=n' where if n' is divisible by 3 so will n. we then look at the case where the first and second number are the same (a=b)

a+2b+c=n' 3a+c=n' this will be divisible by 3 for c=3,c=6 or a total of 2

8=16 cases for having the same second and third number, (b=c) we have8=16 cases-2 repeating cases for 14 new cases for having the same first and third number, (a=c) we havea+2b+c=n' a+3b=n' this will be divisible by 3 for a=3,a=6 or a total of 2

a+2b+c=n' 2a+2b=n' a+b=n' here n' will be between 2 and 16 so will be divisible by three for n'=3,6,9,12,15 n'=3 will have 2 cases, n'=6 will have 5, n'=9 will have 8 cases, n'=12 will have 5 and n'=15 will have 2 for a total of 22 cases -2 repeated cases=20 new cases this means there are a total of 50 cases that are divisible by three and have repeating numbers which mean the number of non repeating numbers divisible by three are: 170-50=120

c.) there are 97 primes less than 512

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With letter a, I think this can be solved by cases. If is a number of multiple three, then the sum of the digits must give us a total of three. The possibilities are from 6 to 21 (6, 9, 12, 15, 18, 21). Now, lets organize the possibilities:

6 = 1 + 2 + 3

9 = 2 + 3 + 4 = 1 + 2 + 6 = 1 + 3 + 5

12 = 3 + 4 + 5 = 2 + 3 + 7 = 2 + 4 + 6 = 1 + 2 + 9 = 1 + 3 + 8 = 1 + 4 + 7 = 1 + 5 + 6

15 = 4 + 5 + 6 = 3 + 4 + 8 = 3 + 5 + 7 = 2 + 5 + 8 = 2 + 6 + 7 = 1 + 6 + 8

18 = 5 + 6 + 7 = 4 + 6 + 8 = 3 + 7 + 8

21 = 6 + 7 + 8

If we count, we have 21 different sums. For each sum we have 6 different numbers. That means that we have 6 x 21 = 126 different 3-digit numbers that are of multiple three.

If any mistake, advise me :P

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