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?
Best Answer
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.