Math CentralQuandaries & Queries


Question from Jerry, a student:

Does there exist a positive integer such that when it is written in base 10 and its leftmost digit is crossed out, the new number is 56 times less than the original number?

We have two responses for you

Hi Jerry,

I am not sure what you mean by "the new number is 56 times less than the original number". I am going to assume that this means that "the original number is 56 times the new number".

Let x be the original number with the leftmost digit removed (this is the new number) and suppose that x has k digits. Let the leftmost digit of the original number be d then the original number is

x + 10k d

The requirement is then that

x + 10k d = 56 x


10k d = 55 x

What are the prime divisors of the left side? What are the prime divisors of the right side?



Hi Jerry.

Let's say that such a number does exist and we'll call it z.

Let z = i + b such that i is the value of the leftmost digit. For example, if z = 2078, then i = 2000 and b = 78.

Then we have z - i = z/56, because removing i leaves a number 56 times less than the original. "56 times less than" is an awkward phrase which should be avoided, but if you contrast this with the phrase "56
times", you can interpret it to mean "1/56th".

z - i = z/56
(i+b) - i = (i+b)/56
56b = i+b
55b = i.

Now we know that i can be expressed as a single digit times some power of ten, so let's say i = D*10T.

55b = D * 10T
(5)(11)(b) = (D)(5T)(2T)

Thus, D must be a multiple of 11, due to the fundamental theorem of arithmetic.

However, we know that D is a single digit from 1 to 9. There are no such multiple of 11 here.

So this is not possible.

Stephen La Rocque.

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