Difference between revisions of "1987 AIME Problems/Problem 3"
(→Solution 2) |
Whitephoenix (talk | contribs) m (→Solution 1) |
||
Line 14: | Line 14: | ||
In the latter case, then <math>p(n) = p \cdot p^2 \cdots p^{(e-1)} = p^{(e-1)e/2}</math>. | In the latter case, then <math>p(n) = p \cdot p^2 \cdots p^{(e-1)} = p^{(e-1)e/2}</math>. | ||
− | For <math>p(n) = n</math>, we need <math>p^{(e-1)e/2} = p^e \Longrightarrow e^2 - e = 2e \Longrightarrow </math> <math>e = 0 or e = 3</math>. | + | For <math>p(n) = n</math>, we need <math>p^{(e-1)e/2} = p^e \Longrightarrow e^2 - e = 2e \Longrightarrow </math> <math>e = 0</math> or <math>e = 3</math>. |
Since <math>e \neq 3</math>, in the case <math>e = 0 \Longrightarrow n = 1</math> does not work. | Since <math>e \neq 3</math>, in the case <math>e = 0 \Longrightarrow n = 1</math> does not work. | ||
Line 20: | Line 20: | ||
Thus, listing out the first ten numbers to fit this form, <math>2 \cdot 3 = 6,\ 2^3 = 8,\ 2 \cdot 5 = 10,</math> <math>\ 2 \cdot 7 = 14,\ 3 \cdot 5 = 15,\ 3 \cdot 7 = 21,</math> <math>\ 2 \cdot 11 = 22,\ 2 \cdot 13 = 26,</math> <math>\ 3^3 = 27,\ 3 \cdot 11 = 33</math>. | Thus, listing out the first ten numbers to fit this form, <math>2 \cdot 3 = 6,\ 2^3 = 8,\ 2 \cdot 5 = 10,</math> <math>\ 2 \cdot 7 = 14,\ 3 \cdot 5 = 15,\ 3 \cdot 7 = 21,</math> <math>\ 2 \cdot 11 = 22,\ 2 \cdot 13 = 26,</math> <math>\ 3^3 = 27,\ 3 \cdot 11 = 33</math>. | ||
[[Sum]]ming these yields <math>\boxed{182}</math>. | [[Sum]]ming these yields <math>\boxed{182}</math>. | ||
− | |||
== Solution 2== | == Solution 2== |
Latest revision as of 09:29, 9 January 2023
Contents
Problem
By a proper divisor of a natural number we mean a positive integral divisor other than 1 and the number itself. A natural number greater than 1 will be called nice if it is equal to the product of its distinct proper divisors. What is the sum of the first ten nice numbers?
Solution 1
Let denote the product of the distinct proper divisors of
. A number
is nice in one of two instances:
- It has exactly two distinct prime divisors.
- If we let
, where
are the prime factors, then its proper divisors are
and
, and
.
- If we let
- It is the cube of a prime number.
- If we let
with
prime, then its proper divisors are
and
, and
.
- If we let
We now show that the above are the only two cases. Suppose that another nice number existed that does not fall into one of these two categories. Then we can either express it in the form (with
prime and
) or
(with
).
In the former case, it suffices to note that
.
In the latter case, then .
For , we need
or
.
Since , in the case
does not work.
Thus, listing out the first ten numbers to fit this form,
.
Summing these yields
.
Solution 2
Alternatively, we could note that is only nice when it only has two proper divisors, which, when multiplied, clearly yield
. We know that when the prime factorization of
, the number of factors
of
is
Since is nice, it may only have
factors (
,
,
, and
). This means that
. The number
can only be factored into
or
, which means that either
and
, or
. Therefore the only two cases are
, or
.
And then continue.
See also
1987 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.