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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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7! = 5040
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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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The basic divisibility rule of 3 is that the sum of digits should be divisible by 3, Since they are asking for a 6 digit number and have provided 7 digits, Sum = 0+1+2+3+4+5+6 = 21, So either 0 or 3 or 6 can be removed and the remaining numbers can be rearranged to form a multiple of 3. When 0 is removed,...
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The basic divisibility rule of 3 is that the sum of digits should be divisible by 3, Since they are asking for a 6 digit number and have provided 7 digits, Sum = 0+1+2+3+4+5+6 = 21, So either 0 or 3 or 6 can be removed and the remaining numbers can be rearranged to form a multiple of 3. When 0 is removed, 6 numbers can arrange among themselves in 6! ways. When 3 and 6 removed, 0 cannot occupy the first position and so 6 numbers will arrange among themselves in 5*5! ways each for 3 and 6. Total = 6! + 5*5! + 5*5! = 1920 read less
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The sum of all the digits = 0+1+2+3+4+5+6 = 21 Hance any six digit number created with these digit is divisible by 3. At the leftmost position '0' can not be there. So total six digits numbers without repeatation of digits : 6 × 6 × 5 × 4 × 3 × 2 = 4320 If repeatation is allowed than this question...
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The sum of all the digits = 0+1+2+3+4+5+6 = 21 Hance any six digit number created with these digit is divisible by 3. At the leftmost position '0' can not be there. So total six digits numbers without repeatation of digits : 6 × 6 × 5 × 4 × 3 × 2 = 4320 If repeatation is allowed than this question become more complicated and complexity will be more then the time :-) read less
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now don't consider zero) the summation of 3 consecutive number is always divisible bye 3. so on this note take e.g. 1,2 and 3. we can make 6(3!) three digit number by using this 3 number. 2 more series will w there that is (2,3,4) and (3,4,5) so now total = 6*3=18. one more series will be there i.e....
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now don't consider zero) the summation of 3 consecutive number is always divisible bye 3. so on this note take e.g. 1,2 and 3. we can make 6(3!) three digit number by using this 3 number. 2 more series will w there that is (2,3,4) and (3,4,5) so now total = 6*3=18. one more series will be there i.e. (1,3,5)=6 now consider zero. by considering 0 we will get 16 three digit number which is divisible by 3. so total=18+6+16=40 read less
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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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