Difference between revisions of "1968 IMO Problems/Problem 2"
(→Solution 2) |
Mathboy100 (talk | contribs) |
||
Line 7: | Line 7: | ||
Let the decimal expansion of <math>x</math> be <math>\overline{d_1d_2d_3\dots d_n}</math>, where <math>d_i</math> are base-10 digits. We then have that <math>x\geq d_1\cdot 10^{n-1}</math>. However, the product of the digits of <math>x</math> is <math>d_1d_2d_3\dots d_n\leq d_1\cdot 10\cdot 10\dots 10=d_1\cdot 10^{n-1}</math>, with equality only when <math>x</math> is a one-digit integer. Therefore the product of the digits of <math>x</math> is always at most <math>x</math>, with equality only when <math>x</math> is a base-10 digit. This implies that <math>x^2-10x-22\leq x</math>, so <math>x^2-11x-22\leq 0</math>. Every natural number from 1 to 12 satisfies this inequality, so we only need to check these possibilities. It is easy to rule out 1 through 11, since <math>x^2-10x-22<0</math> for those values. However, <math>12^2-10\cdot 12-22=2</math>, which is the product of the digits of 12. Therefore <math>\boxed{12}</math> is the only natural number with the desired properties. <math>\blacksquare</math> | Let the decimal expansion of <math>x</math> be <math>\overline{d_1d_2d_3\dots d_n}</math>, where <math>d_i</math> are base-10 digits. We then have that <math>x\geq d_1\cdot 10^{n-1}</math>. However, the product of the digits of <math>x</math> is <math>d_1d_2d_3\dots d_n\leq d_1\cdot 10\cdot 10\dots 10=d_1\cdot 10^{n-1}</math>, with equality only when <math>x</math> is a one-digit integer. Therefore the product of the digits of <math>x</math> is always at most <math>x</math>, with equality only when <math>x</math> is a base-10 digit. This implies that <math>x^2-10x-22\leq x</math>, so <math>x^2-11x-22\leq 0</math>. Every natural number from 1 to 12 satisfies this inequality, so we only need to check these possibilities. It is easy to rule out 1 through 11, since <math>x^2-10x-22<0</math> for those values. However, <math>12^2-10\cdot 12-22=2</math>, which is the product of the digits of 12. Therefore <math>\boxed{12}</math> is the only natural number with the desired properties. <math>\blacksquare</math> | ||
− | ==Solution 2== | + | ==Solution 2(SFFT)== |
+ | It is pretty obvious that <math>x</math> cannot be three digits or more, because then <math>x^2 - 10x - 22</math> is way too big. | ||
+ | |||
+ | Write <math>x = 10a + b</math> where <math>a</math> and <math>b</math> are digits satisfying <math>0 \leq a, b < 10</math>. Then, we can use SFFT: | ||
+ | <cmath>(10a + b)^2 - 10(10a + b) - 22 = ab</cmath> | ||
+ | <cmath>(10a + b)^2 - 10(10a + b) - 24 = ab - 2</cmath> | ||
+ | <cmath>(10a + b + 2)(10a + b - 12) = ab - 2.</cmath> | ||
+ | We have | ||
+ | <cmath>(10a + b + 2)(10a + b - 12) \geq (10a + 2)(10a - 12) = 100a^2 - 100a + 24 = 100(a^2 - a) + 24.</cmath> | ||
+ | It is therefore clear that <math>a</math> must be either <math>0</math> or <math>1</math>. We can then split into two cases: | ||
+ | |||
+ | <math>\mathbf{a = 0:}</math> | ||
+ | |||
+ | We have <math>(b + 2)(b - 12) = -2</math> or <math>b^2 - 10b - 22 = 0</math>, which is only satisfied when <math>b = -2</math> or <math>12</math>. | ||
+ | |||
+ | <math>\mathbf{a = 1:}</math> | ||
+ | |||
+ | We have <math>(b + 12)(b - 2) = b - 2</math>. This is only satisfied when <math>b = 2</math>, or <math>b + 12 = 0</math>. Therefore, <math>b = 2</math>, and so <math>x = \boxed{12}.\square</math> | ||
+ | |||
+ | ~mathboy100 | ||
+ | |||
+ | ==Solution 3== | ||
Revision as of 23:51, 6 December 2022
Problem
Find all natural numbers such that the product of their digits (in decimal notation) is equal to
.
Solution 1
Let the decimal expansion of be
, where
are base-10 digits. We then have that
. However, the product of the digits of
is
, with equality only when
is a one-digit integer. Therefore the product of the digits of
is always at most
, with equality only when
is a base-10 digit. This implies that
, so
. Every natural number from 1 to 12 satisfies this inequality, so we only need to check these possibilities. It is easy to rule out 1 through 11, since
for those values. However,
, which is the product of the digits of 12. Therefore
is the only natural number with the desired properties.
Solution 2(SFFT)
It is pretty obvious that cannot be three digits or more, because then
is way too big.
Write where
and
are digits satisfying
. Then, we can use SFFT:
We have
It is therefore clear that
must be either
or
. We can then split into two cases:
We have or
, which is only satisfied when
or
.
We have . This is only satisfied when
, or
. Therefore,
, and so
~mathboy100
Solution 3
Let,
Now note that, if is a prime such that
then
.
That means,
But, which means
don't divivde
So, and
It is easy to see that has one solution and that is
( Prove it by contradiction)
So,
See Also
1968 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |