he
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.
efore
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.
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 10^{n}d
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:
10^{n} 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.
10^{n} d  d =

d1

9

9

9

...

9

10d

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 =

d1

9

9

9

...

9

10d

/ 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
nonzero 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...
...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 decimalland.
Time for a small hoohah, 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.
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.
Hohum. 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
ou
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 wideeyed
at that magnificent 44digit 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, "Hoohah."
"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
ntuition
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 sword).
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 'ndigit
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 'trialanderror' 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 sword. {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.

+ 10^{n1} d is mathspeak for taking d and adding
it back in at the leftmost position of the shifted number. {Return}
