25. Sonia wants to write a positive five-digit integer whose digits are $1,2,3,4$ and 5 in some order. The first digit of the integer is to be divisible by 1 , the first two digits are to form a two-digit integer divisible by 2 , the first three digits are to form a three-digit integer divisible by 3 , the first four digits are to form a four-digit integer divisible by 4 and the five-digit integer itself is to be divisible by 5 . How many such five-digit integers could Sonia write?
A 10
B 5
C 2
D 1
E 0
Solution
E

Suppose such a five-digit integer exists. Since it is to be divisible by 5 , its last digit must be 5 . The two-digit integer consisting of the first two digits and the four-digit integer consisting of the first four digits are to be divisible by 2 and by 4 respectively. Therefore the second and the fourth digits of the five-digit integer are both even. Therefore the five-digit integer is of the form $* 2 * 45$ or $* 4 * 25$, with the missing digits being 1 and 3 in some order. The three-digit integer made up of the first three digits is to be divisible by 3 and hence the sum of the first three digits should be divisible by 3 . However neither 143 nor 341 is divisible by 3 and hence the five-digit integer is of the form $* 2 * 45$. But neither 1234 nor 3214 is divisible by 4 . Therefore there are no possible five-digit integers that Sonia could write.