25. How many four-digit numbers have the following property? "For each of its digits, when this digit is deleted the resulting three-digit number is a factor of the original number."
A 5
B 9
C 14
D 19
E 23
25. C Let the 4-digit integer be ' $a b c d$ '. When it is divided by ' $a b c$ ', we get ' $a b c d$ ' $=10 \times$ ' $a b c$ ' $+d$. Since ' $a b c$ ' is a factor of ' $a b c d$ ', we must have $d=0$, and the integer is ' $a b c 0$ '. Similarly, when ' $a b c 0$ ' is divided by ' $a b 0$ ', we get ' $a b c 0$ ' $=10 \times$ ' $a b 0$ ' $+{ }^{\prime} c 0$ '. But ' $c 0$ ' can only be divisible by ' $a b 0$ ' if $c=0$. Thus the integer is ' $a b 00$ '. This is divisible by ' $b 00$ ', so we can't have $b=0$. When we divide ' $a b 00$ ' by ' $a 00$ ', we get ' $a b 00$ ' $=10 \times$ ' $a 00$ ' + ' $b 00$ ', hence ' $b 00$ ' must be a multiple of ' $a 00$ ', and therefore $b$ is a multiple of $a$.
Since ' $b 00$ ' is a factor of ' $a b 00$ ' and ' $a b 00$ ' $=$ ' $a 000$ ' + ' $b 00$ ', it must be that ' $b 00$ ' divides ' $a 000$ ', hence ' $a 0$ ' is a multiple of $b$. For $b$ to be multiple of $a$ and a factor of $10 \times a, b$ must be $a, 2 a, 5 a$ or $10 a$. But $10 a$ is more than one digit long. When $b=a$ we get ' $a b 00$ ' is 1100 , $2200,3300,4400,5500,6600,7700,8800,9900$. When $b=2 a$, we get $1200,2400,3600,4800$. When $b=5 a$, we get 1500 . This is $9+5=14$ possibilities.