Reaman Numeral

Copyright 1997 by Paul Niquette. All rights reserved.

The sophisticated solver of this puzzle would surely start out by letting x represent the unknown number.
    Here is a picture of the problem:
x =
~
~
~
~
~
~
#
9x =
#
~
~
~
~
~
~

At this point we do not know much about x -- not even how many digits x has. For now, let's just assign n to represent the number of digits in x.

Like all the rest of the cyphers in x, the least significant digit is unknown. Better give it a name: d.

    About all we really know is that d is also the most significant digit of 9x. That means d cannot be zero, otherwise 9x would be one digit shorter than x -- BLAGH! -- and 9x would then be smaller than x. Can't have that. {Scenario}
x =
~
~
~
~
~
~
d
9x =
d
~
~
~
~
~
~

With so little known, a problem like this can give you a brain blister.

Before clicking off to some other website, though, let's write an equation.

    From the statement of the problem, we have Using a few algebraic steps we are able to rearrange our equation so that x winds up all by itself over on the left hand side.
      x = (10n d - d) / 89
Here is merely a fraction. As for its denominator, 89, we know everything there is to know about it.
    How about that numerator, though?
The expression 10nd is nothing but a digit d followed by a string of zeroes.
    There are exactly n of them, as a matter of fact. It looks like this:
10n d =
d
0
0
0
...
0
0
    Now, you subtract d from that and what do you get? Mostly a string of nines, for sure. The rightmost digit, however, is 10 - d. The leftmost digit is d - 1.
10n d - d =
d-1
9
9
9
...
9
10-d
    If you're keeping track, you will have observed that the numerator we're working on has an extra digit: It is n + 1 digits long. That's a good thing, as we shall see.
Here is our fraction, then.
 
x =
d-1
9
9
9
...
9
10-d
/ 89
    That fraction is not really a fraction, but a whole number. The quotient is, after all, x.
Looks like all we need to do now is...
Step 1. Pick a value for digit d,
Step 2. Put digit d - 1 at the leftmost end of a string of nines,
Step 3. Put digit 10 - d at the rightmost end of that string of nines, and
Step 4. Carry out the indicated long division by 89.
That Step 4 is a bit tricky. During the long division, we have to keep sticking in nines as we go along until the quotient comes out even (zero remainder, as they used to teach fourth graders before calculators).

There is just this one more thing we know: That we have to get n digits in the quotient.

    Fortunately the numerator has one extra digit: It is n + 1 digits long. As already mentioned, that will come in handy.
The first step of the long division has us looking for a non-zero quotient digit when the value of the leftmost two digits of the dividend gets compared to 89.
    The value of the leftmost two digits in the numerator is given by...
      10 (d -1) + 9
    ...and that dang well better be at least as big as 89.
The expression 10 (d -1) + 9 will be equal to or greater than 89 if d is at least 9 -- which is as big as d can possibly be in decimal-land.
    Time for a small hoo-hah, since d is no longer unknown.
Oh, and the first quotient digit -- the leftmost digit of x -- is also now known to be 1, since the value of the leftmost two digits of the numerator is known to be 89.
    That's something to get excited about. The leftmost digit 1 in the quotient is the smallest possible, which meets the first condition in the problem statement ("Find the smallest decimal integer...")
So much for Step 1. Here are the results of Step 2 and Step 3.
 
x =
8
9
9
9
...
9
1
/ 89

All we really don't know about the numerator at this point is the length of that numerator -- how many 9s there are between the 8 over on the left and that 1 there on the right.

    We'll just have to let our long division take care of that.
Ho-hum. After Step 4, we get the following solution.
 
x =
10,112,359,550,561,797,752,808,988,764,044,943,820,224,719
9x =
91,011,235,955,056,179,775,280,898,876,404,494,382,022,471


Historical Note

You might imagine my own astonishment when, in 1953, hunched over a desk in my room, dividing and dividing, watching in wonder while unruly digits marched across a quadrilled page -- not especially confident that my algebraic methods were valid, not knowing that a discovery was at hand, only hoping my elementary procedure would eventually end, and then finally gasping in triumph at the zero remainder. You might imagine how I gaped wide-eyed at that magnificent 44-digit number.

The exhilaration of that moment gave at least a transient significance to an otherwise undistinguished life. It was truly a time to cry out, "Hoo-hah."

    "Why 'Reaman Numeral'?" a student asked years later.

    "It's my father's first name," I answered. "He inspired me to study numbers."

Epilog: For a sequel to this puzzle see Reaman Numerals (plural).  Also, in 2007, the author found out that another name has come along over the intervening 54 years for integers with this feature: "parasite numbers."
 


Scenario

Intuition plays a speaking role in solving problems. Thus, the word 'seems' belongs in the script and gets uttered by a whole troupe of repertory actors on the mathematical stage. After a solution is found, though, analytic purists get embarrassed about any seeming that might have gone on when the problem was not yet solved (Eeoo, the s-word).

When the curtain first parted on this problem back in 1953, an empty theatre echoed with an opening soliloquy from an offstage narrator, "It seems like the least significant digit in x would have to be 9." The sentiments in that line derived, no doubt, from a venerable number play of about the same vintage (see Secret Message).

"Hark!" cried the Voice of Reason from the front line of the chorus. "The leftmost digit of x cannot be a zero." The chorus made sounds of agreement. "Otherwise the expression 'n-digit number' makes no sense: Why, if zero counted as a leading digit, any number of them might then be stuck on so that even the smallest integer would say of itself, 'I have an infinite number of digits'!"

The audience laughed at the absurdity.

The 'great unraveler,' George Polya (How to Solve It), strolled onto the stage, "So what if you cannot immediately conquer the general problem!" exclaimed he. "Be not ashamed to attack a few specific cases."

Out stepped a real ham, Mr. Empirical, grinning. He commenced one of his 'trial-and-error' schticks...
 

19 : 91 = 1 : 4.79
109 : 910 = 1 : 8.35
1009 : 9100 = 1 : 9.02

"There, you see," commented Polya. "The ratio 9 is indeed feasible. Mathematicians call that an 'existence theorem' and -- "

"Watch this," interrupted Mr. Empirical...
 

18 : 81 = 1 : 1.50 
108 : 810 = 1 : 7.50 
1008 : 8100 = 1 : 8.04 
10008 : 81000 = 1 : 8.09 
10000008 : 81000000 = 1 : 8.10

"Seems like d has to be 9, doesn't it," mused the narrator, emphasizing the s-word. {Return}
 


Derivation

In slow motion, then...

  • (x - d) represents the elementary act of taking away the least significant digit from x.
  • (x - d) / 10 is the arithmetical equivalent of shifting all the rest of the digits in x to the right one place.
  • + 10n-1 d is math-speak for taking d and adding it back in at the left-most position of the shifted number. {Return}


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