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Class 12 Tuition > Learn Class XI-XII Tuition (PUC) > How many six digit positive integers that are divisible...
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How many six digit positive integers that are divisible by 3 can be formed using the digits 0,1,2,3,4,5 and 6 without any of the digits getting repeating?

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Answer is 216. The sum of the digits of any number that is divisible by '3' is divisible by 3. For instance, take the number 54372. Sum of its digits is 5+4+3+7+2=21 As 21 is divisible by '3', 54372 is also divisible by 3. There are six digits viz., 0,1,2,3,4 and 5. To form 5-digit numbers...
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Answer is 216. The sum of the digits of any number that is divisible by '3' is divisible by 3. For instance, take the number 54372. Sum of its digits is 5+4+3+7+2=21 As 21 is divisible by '3', 54372 is also divisible by 3. There are six digits viz., 0,1,2,3,4 and 5. To form 5-digit numbers we need exactly 5 digits. So we should not be using one of the digits. The sum of all the six digits 0,1,2,3,4 and 5 is 15. We know that any number is divisible by 3 if and only if the sum of its digits is divisible by '3'. Combining the two criteria that we use only 5 of the 6 digits and pick them in such a way that the sum is divisible by 3, we should not use either '0' or '3' while forming the five digit numbers. Case 1 If we do not use '0', then the remaining 5 digits can be arranged in: 5! ways=120 numbers. Case 2 If we do not use '3', then the arrangements should take into account that '0' cannot be the first digit as a 5-digit number will not start with '0'. The first digit from the left can be any of the 4 digits 1,2,4 or 5 Then the remaining 4 digits including '0' can be arranged in the other 4 places in 4! ways. So, there will be 4×4! numbers =4×24=96 numbers. Combining Case 1 and Case 2, there are a total of 120+96= 216, 5 digit numbers divisible by '3' that can be formed using the digits 0 to 5. read less
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B.tech

Last two digits should be 30,03,12,21,36,63,45,54 to divisible by three... so total possible arrangements could be = 4x(4x3x2x1)x2 = 192 digits
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Answer is 1320 Explaination: To form 6 digit number which is divisilble by 3. first condition is sum of number is divisible by 3i.e., 0+1+2+3+4+5+6=21 which is divisible by 3. since out of 7 digit , only 6 digit has to be placed, one digit has to be removed. (which should be divisible by 3) So, by...
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Answer is 1320 Explaination: To form 6 digit number which is divisilble by 3. first condition is sum of number is divisible by 3i.e., 0+1+2+3+4+5+6=21 which is divisible by 3. since out of 7 digit , only 6 digit has to be placed, one digit has to be removed. (which should be divisible by 3) So, by removing 0 and 3 from series the total number will always be divisible by 3. 1.)) by removing 0 from series(1,2,3,4,5,6):-we get 6 digits can be arranged in 6! times = 720 2.)) by removing 3 from series(0,1,2,4,5,6):- 0 cannot take first place from left so without it, we have remaining 5 number and now including 0 in other 5 places we can arrange 5 digits in 5! ways. = 5*5!= 5*120=600 3)) In total 720+600=1320 read less
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here i have posted the answer.
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1. If '0' is not used then it can be arranged in 6 possible ways 6!=720 numbers 2.if we don't use '3' and '0' cannot be the first digit, hence we have only 5 digits as 1st number they are 1,2,4,5,6 Then remaining 5 digits can be arranged in other 5 places hence we will have 5 digits in 5 ways 5*5!=600...
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1. If '0' is not used then it can be arranged in 6 possible ways 6!=720 numbers 2.if we don't use '3' and '0' cannot be the first digit, hence we have only 5 digits as 1st number they are 1,2,4,5,6 Then remaining 5 digits can be arranged in other 5 places hence we will have 5 digits in 5 ways 5*5!=600 numbers By adding 1 and 2 we will have 720+600=1320 numbers read less
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ans is 1920 divisibility test of 3 is the sum of digits should be divisible by 3. So sum of 0,1,2,3,4,5,6 is 21. case 1: when 1,2,3,4,5,6 numbers are taken possible number will be factorial 6=120 Case 2: when 0,1,2,4,5,6 are taken possible numbers will be 5*5! (as 0 cannot come as first digit otherwise...
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ans is 1920 divisibility test of 3 is the sum of digits should be divisible by 3. So sum of 0,1,2,3,4,5,6 is 21. case 1: when 1,2,3,4,5,6 numbers are taken possible number will be factorial 6=120 Case 2: when 0,1,2,4,5,6 are taken possible numbers will be 5*5! (as 0 cannot come as first digit otherwise that would not be an 6 digit no. Case3 : when 0,1,2,3,4,5 numbers are taken possible number will be 5*5! Ans is 120+(2*5*5!) read less
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C++ Tutor

ones place can be filled in 6 different ways tens place can be filled in 5 different ways hundredth place can also be filled in 5 different ways ans should be 6!*5!*5!
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