## Fall 2015, problem 2

Let $x,y$ be integers such that $10000 \leq x \lt 100000$, and $y$ is obtained from $x$ by removing the third digit. Determine all pairs $x,y$ as above such that $\frac{x}{y}$ is also an integer.

### Comments

Please check x=10001 We actually don't know any thing about the first two digits, just the third one

Elwardi, I believe we have different ideas of what it means to"remove the third digit". Your interpretation seems to be taking the number with digits $ abcde$ and subtracting $ c00$ to arrive at $ ab0de$, while I would toss away the $ c$ to get $ abde$. To me therefore $ x=10001$ is not a counterexample because 10001 is not a multiple of 1001. Staying with your interpretation though, the answer you give seems quite right.

It was meant for the third digit in $x$ to be removed in the manner Nelix wrote. However, Elwardi's interpretation of the problem is interesting as well as I had not considered that.

- If we intepret this question as removing the third non-zero digit left to right, when it exist: Since the third non-zero digit cannot appear before the third digit, Nelix's solution is still essentially right.
- If we allow "clock arithmetic" when counting the third non-zero digit L to R, then, trivially, all x with one or two non-zero digit and between 10000 and 99999(inclusive) will work.
- If we remove the second digit from L to R:

Things times 4n+1: 10125, 19125; 11250, 16250, 20250, 29250; 12375, 21375, 30375; 14625, 23625, 32625, 41625, 50625; 15750, 24750, 33750, 42750, 51750, 60750; 16875, 25875, 34875, 43875, 52875, 61875, 70875;

Times 5n+1: 10800,19800; 18700, 29700; 17600, 28600, 39600; 15400, 26400, 37400, 48400, 59400; 14300, 25300, 36300, 47300, 58300, 69300; 13200, 19200, 24200, 35200, 46200, 57200, 68200, 79200; 12100, 23100, 34100, 45100, 56100, 67100, 78100, 89100.

10500, 13500, 16500, 19500, 22500, 27500, 31500, 38500, 40500, 49500(don't want to split it between the two groups); Anything ending with 000. Is that all?

- Does such $\dfrac{\overline{abcde}}{\overline{acde}} \in \mathbb{Z}$ exist in any/all bases?

Jiazhen, what was your method to find the numbers listed in your third point?

Oh, that's an approch to it that i didn't think about, thanks Nelix and BenMC, it was my fault.It should be understood the way you did.

@Matthew By going through the possible choices for the last three digits. Since 1000 will divide into some integer multiple of them, the last three seems to be some multiple of 125 or 100.

@Jiazhen

> Does such $\frac{\overline{abcde}}{\overline{acde}} $ exist in any/all bases?

In any base $ p$ the numbers $ \overline {abc00}$ with $ a+c=b$ will work, since in this case $ \overline{ac00} × (p+1) = \overline {abc00}$.

If $ p$ is prime this will actually be the only class.

Oh, interesting observation! @Nelix

Hello to all,

The discussion may have moved beyond it, but for what it’s worth, here is my method:

X, expressed as abcde = 10000a + 1000b + 100c + 10d + e.

Y, omitting the 3rd digit is expressed as abde = 1000a + 100b + 10d +e.

Then X/Y = 10 with a remainder of R = 100c – 90d – 9e. For X/Y to be integer, R must have an integer solution. The only one (for 0 <= c, d, e <= 9) is c = d = e = 0.

Therefore, only X’s of the form ab000 will provide integer X/Y.

I understand your solution but I was having a problem that you did X=10Y+R, why you have taken 10 in particular, I mean you can take any other integer like 9,11 etc.

What I can understand from your solution is that if X/Y is an integer then that will be 10. Is this what you are doing in your solution? If yes then why it will be only 10?

@Singh I think joa is trying to prove that it can and can only be 10, and you seemed to understand though;

if X/Y is an integer then that will be 10

would you explain a bit more about your confusion?

What is important to observe in Jao's answer is that |R| < y obviously. So $\frac{x}{y} = 10 + \frac{R}{y} $ comes out integer only if $ R = 0$. Trying the same argument for the choice of 11 or 9 or so on then fails at some point, since $\frac{x}{y} $ has one definite value.

Hi Singh and others,

You’re right. I could have expressed X/Y as 9 with remainder of (I call it) R9, or said another way, 9 + R9 / Y.

In this case R9 = X – 9Y = 1000a + 100b + 100c -80d -8

So, X / Y = 9 + (1000a + 100b + 100c -80d -8e) / Y =9 + ( [1000a + 100b + 10d + e] + 100c – 90d – 9e) / Y =9 + ( Y + 100c -90d – 9e) / Y =10 +(100c -90d – 9e) / Y = 10 + R10 / Y. This shows that 9 + R9 / Y is equivalent to 10 + R10 / Y.

I chose to express X / Y as 10 + R10 / Y because R10 ( = 100c – 90d – 9e) is more simply written and analyzed than R9. There are no terms containing a or b.

I did make a mistake in my original posting when I said that the remainder R10 had to be integer to result in integer X / Y. I meant to say that R10 / Y had to be integer. The maximum value of R10 is 100x9 = 900, the minimum value of R10 is -90x9 - 9x9 = -891 (maximum absolute value of 891), and the minimum value of Y is 1000x1 = 1000. This results in the conclusion that the only time R10 / Y can be integer is when R10 is 0. Therefore, c=d=e=0, and therefore X must be of form ab000.

Joa

x= abcde = 10000a + 1000b + 100c + 10d + e y = abde = 1000a + 100b + 10d + e

x/y = integer remember x is 5 digit number and y is 4 digit number min. value of x/y = 1 max value of x/y :: biggest 5 digit number divided by smallest 4 digit number. so, 99999/1000 = 99.999 but with a condition of middle digit removed: 99999/9999 = 10.00009

if we go and check any 5 digit number say 12355/1255 = 9.84

9 < x/y < 11

so only possible value we get is 10 (why is that , example below the solution)

x = 10 (1000a+ 100b) + 100c + 10d + e
= (1000a + 100b + 10d + e) + 9 (1000a + 100b) + 100c
=

let`s p = 1000a + 100b
and q =10d + e
x
x/y = (1000a + 100b + 10d + e) + 9 (1000a + 100b) + 100c / 1000a + 100b + 10d + e
= y + 9p + 100c / y
= 1 + (9p/y) +(100c/y)

if c=0 and d,e = 0 then q=0 then p/y=p/(p+q)=p thus x/y = 10 for c,d,e=0

so any number of ab000 will do the job....

Let's name a the third digit of x, so a=E(x/100)-10E(x/1000), where E(x) is the integer part of x; then y=x-100.a and x/y=x/(x-100.a)

if a=0 it's abvious that x=y and x/y=1 is an integer; if a=1, consider k=x/y to be an integer, with k can't be 0 (if so, x will be 0) or 1 (case of a=0) x/(x-100)=k gives us x=100.k/(k-1) and we now that for every integer k 1< k/(k-1)<=2 .....(1) x=100k/(k-1)<=200 (if we apply (1)) but we have x>10000, so a=1 is not an option. Now, if we consider in general 2<=a<=9 we get x=100ak/(k-1)<=200a<=1800 which is always <10000 but we know x sould be >10000. So the numbers that satisfie the problems' conditions are only wich have a null third digit (of the form ab0cd in decimal writing). Sorry for the bad writing,

This is yet another interpretation of the original problem - how should we interpret the meaning of "third" and "remove" from context? What's your first impression upon seeing these words?

That's probably why passing the Turing test is not easy...

About formatting: this site use MathJax. Enclose your LaTeX formula in a pair of $ and you should be fine.

In the nomenclature of the problem statement, let $ k:= \frac{x}{y} $. We know, that $x$ and $10y$ can differ at most in the last three digits. This implies: $$1000>|x - 10y| = |k - 10| y$$ But since already $y>=1000$, we conclude, that $ k=10$. Therefore the good pairs are exactly those of the form $ x=ab000$ and $ y=ab00$ with $ ab$ an arbitrary 2-digit number.