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How many $3$ digit numbers can be formed using digits $1,2,3,4$ and $5$ without repetition such that the number is divisible by $6$

First Approach:

A number is divisible by $6$ if it is divisible by $2$ and $3$.

Now the possible combinations I found are

$(1,3,2)$

$(3,1,2)$

$(2,3,4)$

$(3,2,4)$

$(4,3,2)$

$(3,4,2)$

$(3,5,4)$

$(5,3,4)$

total $8$ ways.

Second Approach

Case1: unit digit can be filled in only two ways $(2,4)$ for nos $(3,2,4)$

Tens digit can be filled in $2$ ways

Hundred digit can be filled in $1$ ways

the required number is 2*1*2=4 ways

Case2: unit digit can be filled in only one ways $(2)$ for nos $(1,2,3)$

Tens digit can be filled in $1$ ways

Hundred digit can be filled in $2$ ways

the required number is 2*1*1=2 ways

Case3: unit digit can be filled in only one ways $(4)$ for nos $(3,4,5)$

Tens digit can be filled in $1$ ways

Hundred digit can be filled in $2$ ways

the required number is 2*1*1=2 ways

So, Total ways=8

Is there still a better way to solve this problem?

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    $\begingroup$ your second argument makes no sense. you need to ensure that the digit sum is multiple of $3$. $\endgroup$
    – user251257
    Aug 26, 2015 at 12:58
  • $\begingroup$ okay i am trying again and will edit the post soon i think we need to make cases.Thankx for the hint $\endgroup$
    – justin takro
    Aug 26, 2015 at 13:01
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    $\begingroup$ After choosing the units digit, there is a little symmetry left in that the hundreds digit and the tens digit can be swapped ("without repetition"). So it suffices to count the leading pairs that work with unit place 2 and those that work with unit place 4, then double the count because of swapping those leading digits. $\endgroup$
    – hardmath
    Aug 26, 2015 at 13:04
  • $\begingroup$ Hint: Say the unit digit is 2. Denote the tens and hundred digit by $p$ and $q$. Then, $2+p+q$ is a multiple of $3$ if and only if $p+q \equiv 1 \pmod 3$. $\endgroup$
    – user251257
    Aug 26, 2015 at 13:11
  • $\begingroup$ @user251257 Edited the code. $\endgroup$
    – justin takro
    Aug 26, 2015 at 13:12

3 Answers 3

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A refinement of your first approach:

In numbers which end with $2$, the sum of the other two digits must be $4$ or $7$:

  • $132$ the sum of the other two digits is 4
  • $312$ the sum of the other two digits is 4
  • $342$ the sum of the other two digits is 7
  • $432$ the sum of the other two digits is 7

In numbers which end with $4$, the sum of the other two digits must be $5$ or $8$:

  • $234$ the sum of the other two digits is 5
  • $324$ the sum of the other two digits is 5
  • $354$ the sum of the other two digits is 8
  • $534$ the sum of the other two digits is 8
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Your second method does not take in to account that the numbers must be divisible by 3. If you had consecutive numbers you could divide by 3, which also seems to work here, but I suspect the best approach is to determine how many triples of {1,2,3,4,5} sum to a multiple of 3, and use that as your set instead of using the entire set, without 2 or 4.

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Hint:-Form even numbers such that their sum is divisible by 3. Divisibility rule by 6 says that the number must be divisible by both 2 and 3.Your second try does not take into account that the number must be divisible by 3.The possible ways are ,as you quoted-(1,3,2) (3,1,2) (2,3,4) (3,2,4) (4,3,2) (3,4,2) (3,5,4) (5,3,4).There is possibly no better way to do it without using divisibility.

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