Difference between revisions of "2001 IMO Shortlist Problems/N1"
(New page: == Problem == Prove that there is no positive integer <math>n</math> such that, for <math>k = 1,2,\ldots,9</math>, the leftmost digit (in decimal notation) of <math>(n + k)!</math> equals ...) |
(Added a solution) |
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== Solution == | == Solution == | ||
− | {{ | + | Suppose that there is such a number <math>n</math>. Let <math>a</math> be the number of digits of <math>(n + 8)!</math>, and let <math>b</math> be the number of digits of <math>n + 9</math>. Then we have: |
+ | |||
+ | <math>8 \cdot 10^{a-1} \leqslant (n + 8)! < 9 \cdot 10^{a-1}</math> | ||
+ | |||
+ | and | ||
+ | |||
+ | <math>10^{b-1} \leqslant n + 9 < 10^b</math> | ||
+ | |||
+ | Combining these inequalities via multiplication we get: | ||
+ | |||
+ | <math>8 \cdot 10^{a+b-2} \leqslant (n+9)! < 9 \cdot 10^{a+b-1}</math> | ||
+ | |||
+ | From this inequality, it can be seen that <math>(n+9)!</math> has at least <math>a+b-1</math> and at most <math>a+b</math> digits. However, if it has <math>a+b</math> digits then the first digit is less than <math>9</math>, which contradicts with how <math>n</math> is defined. Thus, <math>(n+9)!</math> must have <math>a+b-1</math> digits. Expressing this as an inequality, we get: | ||
+ | |||
+ | <math>9 \cdot 10^{a+b-2} \leqslant (n+9)! < 10^{a+b-1}</math> | ||
+ | |||
+ | Combining these values with the earlier bounds for <math>(n+8)!</math> via division, we get: | ||
+ | |||
+ | <math>10^{b-1} < n+9 < \frac{5}4 10^{b - 1}</math> | ||
+ | |||
+ | Suppose that <math>n+1 < 10^{b-1}</math>. Then <math>10^{b-1}</math> is between <math>n+1</math> and <math>n+9</math>. Thus, one of the values <math>n+2,n+2,\ldots n+8</math> must be <math>10^{b-1}</math>. Call that value <math>n+l (1 < l < 9)</math>. This means that the digits of <math>(n+l)!</math> are just the digits of <math>(n+l-1)!</math>, followed by <math>b-1</math> zeroes. Thus the first digit of the two numbers are equal, which contradicts with how <math>n</math> is defined. | ||
+ | This means that: | ||
+ | |||
+ | <math>10^{b-1} \leqslant n+1 < n+2 < n+3 < n+4 < n+9 < \frac{5}4 10^{b - 1}</math> | ||
+ | |||
+ | Let <math>c</math> be the number of digits of <math>(n+1)!</math>. Then we have: | ||
+ | |||
+ | <math>10^{c-1} \leqslant (n+1)! < 2 \cdot 10^{c-1}</math> | ||
+ | |||
+ | Combining these with the bounds for <math>n+2</math>, <math>n+3</math> and <math>n+4</math> via multiplication we get: | ||
+ | |||
+ | <math>10^{c+3b-4} \leqslant (n+4)! < \frac{125}{32}10^{c+3b-4} < 4 \cdot 10^{c+3b-4}</math> | ||
+ | |||
+ | So <math>(n + 4)!</math> has <math>c+3b-3</math> digits, but is smaller than the smallest <math>c+3b-3</math> digit number with the first digit 4, which means that its first digit can't be 4. This contradicts with how <math>n</math> is defined, thus no such value of <math>n</math> exists. | ||
== Resources == | == Resources == |
Revision as of 02:41, 1 April 2022
Problem
Prove that there is no positive integer such that, for
, the leftmost digit (in decimal notation) of
equals
.
Solution
Suppose that there is such a number . Let
be the number of digits of
, and let
be the number of digits of
. Then we have:
and
Combining these inequalities via multiplication we get:
From this inequality, it can be seen that has at least
and at most
digits. However, if it has
digits then the first digit is less than
, which contradicts with how
is defined. Thus,
must have
digits. Expressing this as an inequality, we get:
Combining these values with the earlier bounds for via division, we get:
Suppose that . Then
is between
and
. Thus, one of the values
must be
. Call that value
. This means that the digits of
are just the digits of
, followed by
zeroes. Thus the first digit of the two numbers are equal, which contradicts with how
is defined.
This means that:
Let be the number of digits of
. Then we have:
Combining these with the bounds for ,
and
via multiplication we get:
So has
digits, but is smaller than the smallest
digit number with the first digit 4, which means that its first digit can't be 4. This contradicts with how
is defined, thus no such value of
exists.