Deleting digits How many four digit numbers have the following property? 
‘For each of its digits, when this digit is deleted the resulting $3$ digit number is a factor of the original number’. There are multiple alternatives:
$$A)5$$$$ B)9 $$$$C)14 $$$$D)19 $$$$E)23$$
Any ideas on how to tackle this problem?

 A: Suppose the number is $xyzw$ then since $xyz \mid xyzw$ so we must have $$100x+10y+z \mid 1000x+100y+10z+w.$$ Since $100x+10y+z \mid 1000x+100y+10z$ So we must have $$100x+10y+z \mid w.$$ Which is possible only if $w=0.$ Now using similar argument for $xyw$ we get $$10z-9w=0 \implies z=0$$ since $w=0.$
Once you got this the number looks like a very easy form which is $xy 00.$ Now use the facts that $x00|xy00$ and $y00|xy00$ to see that $x \mid y$ and $y \mid 10 x$ whenever $y \neq 0.$ This shows that for $y \neq 0$ we have $y=x$ or $2x$ or $5x.$ So the values of $(x,y)$ can be listed as $$(1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(7,7),(8,8),(9,9),(1,2),(2,4),(3,6),(4,8),(1,5).$$
So the four digits numbers are precisely $$1100,2200,3300,4400,5500,6600,7700,8800,9900,1200,2400,3600,4800,1500.$$
Also for $y=0$ we have $9$ four digits numbers which are $$1000,2000,3000,4000,5000,6000,7000,8000,9000.$$ But they don't satisfy the given property if the first digit of each of these numbers is deleted. 
So in total there are precisely $14$ four digits numbers having the given property.
A: This is the official answer:
Let the 4-digit integer be ‘abcd’. When it is divided by ‘abc’, we get ‘abcd’ = 10 × ‘abc’ + d. Since ‘abc’ is a factor of ‘abcd’, we must have d = 0, and the integer is ‘abc0’. Similarly, when ‘abc0’ is divided by ‘ab0’, we get ‘abc0’ = 10 × ‘ab0’ + ‘c0’. But ‘c0’ can only be divisible by ‘ab0’ if c = 0. Thus the integer is ‘ab00’. This is divisible by ‘b00’, so we can’t have b = 0. When we divide ‘ab00’ by ‘a00’, we get ‘ab00’ = 10 × ‘a00’ + ‘b00’, hence ‘b00’ must be a
multiple of ‘a00’, and therefore b is a multiple of a.
Since ‘b00’ is a factor of ‘ab00’ and ‘ab00’ = ‘a000’ + ‘b00’, it must be that ‘b00’ divides ‘a000’,hence‘a0’is a multiple of b For b to be a multiple of a and a factor of 10×a, b must be a, 2a, 5a or 10a. But 10a is more than one digit long. When b = a we get ‘ab00’ is 1100,
2200, 3300, 4400, 5500, 6600, 7700, 8800, 9900. When b = 2a, we get 1200, 2400, 3600, 4800. When b = 5a, we get 1500. This is 9 + 5 = 14 possibilities.
A: All multiples of 5 with only the last two digits as zero should be included. Along with all the multiples of 12 and 11 with only the last two digits as zero.
1100,2200,3300,4400,5500,6600,7700,8800,9900,1200,2400,3600,4800,7200,8400,9600,1500,2500,3500,4500,6500,7500,8500,9500.
