Removing digits

N cannot have 1 digit, because it won't be divisible by 25.

With 2 digits $\rightarrow$ Multiples from 25 are 25, 50 and 75 (not useful)

With 3 digits $\rightarrow$ The first digit of N is X:

$\frac {N}{25}$ = $N - 100 \times X$

$N = 25N - 25 \times 100 \times X$

$24 N = 25 \times 100 \times X$

$2^3 \times 3 N = 5^2 \times (2^2 \times 5^2) \times X$

So X must be multiple of 2 and 3 $\rightarrow$ X = 6

With 4 digits:

$\frac {N}{25} = N - 1000 \times X$

$N = 25N - 25 \times 1000 \times X$

$2^3 \times 3 N = 5^2 \times (2^3 \times 5^3) \times X$

X must be a multiple of 3 $\rightarrow$ X can be 3, 6 or 9.

Repeating this until N has 10 digits, we can see there are 3 numbers for each amount of digit:

5 digits $\rightarrow$ 3 numbers ... 9 digits $\rightarrow$ 3 numbers

So, there are $3 \times 6 + 1 = 19$ numbers.

Let N=m (10^t)+N/25 Solving the equation for N yields N=(25 m 10^t)/(2^3 3) We know that m is between [1,9] and t is between [1,8]. Since neither 25 or 10^t are divisible by 3, so m must be a multiple of 3. Casework: 1. m=3 t=[3,8] 2.m=6 t=[2,8] 3.m=9 t=[3,8] Total 19 values of m and t.

Let $N$ be any positive integer such that crossing out the first digit results in $\frac{N}{25}$ . Let $a$ denote the first digit of $N$ and let $M$ denote the number obtained from $N$ by crossing out this digit. Thus $N=a\cdot 10^k+M$ , where the total number of digits of $N$ equals $k+1$ (so $M$ has at most $k$ digits, but could have fewer, in case the second digit of $N$ is $0$ ).

By assumption, $N=25\cdot M$ , which yields $24\cdot M=a\cdot 10^k$ . In particular, since $3$ and $10$ are relatively prime, $a$ must be divisible by $3$ , so the only choices for $a$ are $3$ , $6$ and $9$ .

If $a=3$ , we obtain $M=\frac{10^k}{8}$ ; in particular, we must have $k\geq 3$ in order for $10^k$ to be divisible by $8$ . Conversely, if $k\geq 3$ is an integer, then taking $M=\frac{10^k}{8}=125\cdot 10^{k-3}$ leads to a valid solution of the problem: $N=3125\cdot 10^{k-3}$ .

Very similar considerations show that taking $a=6$ leads to the family of solutions of the form $N=625\cdot 10^{k-2}$ , where $k\geq 2$ is an integer, and taking $a=9$ leads to the family of solutions of the form $N=9375\cdot 10^{k-3}$ , where $k\geq 3$ is an integer.

It remains to count the total number of solutions under $10^9$ . We get $6$ from the first family, $7$ from the second family and $6$ from the third one, so the answer is $6+7+6=19$ .

Let $A$ be the first digit of $N$ and $x$ be the number of digits that $N$ has.

From our condition, we have: $N-\frac {N}{25}=A*10^{x-1}$

$N=25\frac {A*10^{x-1}}{24}$ .where the fraction has to be an integer.

Since $3$ is a factor of $24$ but not $10$ , hence $3\mid A\Rightarrow A=3,6,9$ . Since $N$ has $x$ digits, therefore $\frac {25A}{24}$ has to be less than 10, which holds true when $A$ is max at $9$ .

Now there's no solution for $x=1,2$ . When $x=3$ , $2^2\mid 10^2$ so only $6$ works. When $4\le x\le 9$ , $2^3\mid 10^{x-1}$ so $A$ can be $3,6,9$ .

Hence this gives us $3*6+1=\boxed {19}$ values of $N$ .