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?
Answer
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.
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