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Q How many 4 digit numbers can be formed using numbers 2,3,4,5,6,7 such that the number is only once divisible by 25?

My approach:

Case1: Unit digit is 5.Ten's digit will be 2 or 7.Taking here 2 first

No of ways=4*3*1*1=12

Case2: Unit digit is 5.Ten's digit will 7.

No of ways=4*3*1*1=12

Total no of ways=12+12=24

Is my approach correct?Is there any other approach through which u solve the question?

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    $\begingroup$ Where have you used the condition about being "only once divisible by 25"? $\endgroup$ – Gerry Myerson Aug 27 '15 at 13:14
  • $\begingroup$ The number has to end with $25$, so just choose the other two digits freely. Your question does not state whether or not each digit can be used more than once. So with replacement you have $6^2=36$ options, and without replacement you have $\binom42=6$ options. $\endgroup$ – barak manos Aug 27 '15 at 13:18
  • $\begingroup$ @barakmanos1st mistake u did i think is that number can end with 25 or 75 both to be divisible by 25.2nd mistake each digit should be used only once. $\endgroup$ – justin takro Aug 27 '15 at 13:28
  • $\begingroup$ @justintakro: First mistake is indeed my mistake. Second mistake, as I've already mentioned in the comment is yours, as you did not specify this fact in the question. $\endgroup$ – barak manos Aug 27 '15 at 15:35
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The condition that the number is only once divisible by $25$ means that it is a multiple of $25$ but not $25^2 = 625$. Since we are restricted to using the digits $2, 3, 4, 5, 6, 7$, the four-digit number will only be divisible by $25$ if the last two digits are $25$ or $75$. From these, we must remove those multiples of $625$ between $2000$ and $8000$ in which the only digits are $2, 3, 4, 5, 6, 7$. Those multiples are:

$$2500, 3125, 3750, \color{blue}{4325}, 5000, 5625, 6250, 6875, 7500$$

Assuming digits cannot be repeated, you correctly found the number of four-digit numbers that are divisible by $25$. However, $4325$ is also a multiple of $625$ in which the only digits used are in the set $\{2, 3, 4, 5, 6, 7\}$ and no digits are repeated. Therefore, there are $24 - 1 = 23$ numbers that satisfy the given condition.

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  • $\begingroup$ Why you need to care of 625 because we need to find all the numbers divisible by 25 only.It is just like added functionality.Give me an example to explain this.A bigger number will always be divisible i think. $\endgroup$ – justin takro Aug 27 '15 at 14:18
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    $\begingroup$ The problem asked for numbers that were divisible by $25$ only once. If a number is divisible by $625$, then it is divisible by $25$ at least twice. $\endgroup$ – N. F. Taussig Aug 27 '15 at 14:19
  • $\begingroup$ @N.FTaussig Okay but there are 9 numbers you mentioned which are divisible by 25 and 625. Why these cannot be deleted.It should be 24-9 then $\endgroup$ – justin takro Aug 27 '15 at 14:21
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    $\begingroup$ Note that the only digits we are allowed to use are $2, 3, 4, 5, 6, 7$. This eliminates $2500, 3125, 3750, 6250, 6875, 7500$. As for $5625$, it has a repeated $5$. $\endgroup$ – N. F. Taussig Aug 27 '15 at 14:22
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i think your approach is correct but the number 625*7=4375 should not be counted then the answer is 24-1=23

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  • $\begingroup$ @Rupasa Yes you are right.+1 for the sharp observation :) $\endgroup$ – justin takro Oct 10 '15 at 14:31

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