Sorry for the ambiguity in the title, I had no idea what else to put.

I am studying for a test I have to take for a Japanese scholarship and I have been able to do all of the other sample questions except this one:

The total number of subtractions that result in 11111 remaining after a four-digit number has been subtracted from a five-digit number and the digits 1 through 9 have all been used is...

When I first saw this question I immedietly thought that this was a simple question and only took a little bit of logical guesswork. Soon, I found that 19753 - 8642 = 11111 exactly and figured the number of subtractions must be 1. The answer is not 1, it is 24. This makes no sence to me because the numbers I chose are within the requirements and therefore valid, so even if you could find a 5-digit number that could subtract a 4-digit number 24 times to get 11111, it would not make my answer wrong.

In my opinion this problem can be shown by the equation y - mx = 11111, with y being the 5-digit number, m being the number of subtractions, and x being the 4-digit number.

This test includes topics ranging from algebra 1 to integral calculus, so this question could be an algebra question like I was thinking, but it might be something else.

Rather than showing me the work that results in the answer, 24, I would greatly appreciate it if someone could nudge me in the right direction.

If you need clarification I will be glad to do so. Thank you for your time.

  • 1
    $\begingroup$ If I'm not mistaken, the question asks for the number of pairs $(x,y)$ such that $x$ is a 5-digit positive integer, $y$ is a 4-digit positive integer, $x$ and $y$ contain all digits from 1 to 9 and $x-y=11111$. $\endgroup$ – lattice Jun 4 '16 at 15:45
  • $\begingroup$ I recommend you quote the problem exactly. It is possible you misunderstood the problem. $\endgroup$ – hardmath Jun 4 '16 at 15:45

I would say when thinking about the number of possible solutions, consider how many times you could arrange the pairs of numbers to get 1. you have 4 pairs. i.e. 4 factorial "number of subtractions."

To clarify, yes youre solution is correct. But I think the question is asking how many different solutions there are. That would be 24 ways you can get 11111.

  • $\begingroup$ Thank you, i understand the question now $\endgroup$ – Richard Watson Jun 4 '16 at 16:06

Lattice's comment about the interpretation of the question yields 24 solutions.

Rephrasing it different the question could be thought of as: How many possible subtractions exist between a five digit and four digit number using only the digits 1 to 9 once each which gives an answer of 11111.

The answer of 24 comes from the following pairs of five and four digit numbers:

$13579-2468=11111$, $13597-2486=11111$, $13759-2648=11111$ $13795-2684=11111$, $13957-2846=11111$, $13975-2864=11111$ $15379-4268=11111$, $15397-4286=11111$, $15739-4628=11111$ $15793-4682=11111$, $15937-4826=11111$, $15973-4862=11111$ $17359-6248=11111$, $17395-6284=11111$, $17539-6428=11111$ $17593-6482=11111$, $17935-6824=11111$, $17953-6842=11111$ $19357-8246=11111$, $19375-8264=11111$, $19537-8426=11111$ $19573-8462=11111$, $19735-8624=11111$, $19753-8642=11111$

You will notice in the answer (and in the one that you found) that the five digit number must consist of odd numbers and the four digit number must consider of the even numbers. This is because for each of the subtractions there needs to be no carry.


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