Difference between revisions of "1996 AHSME Problems/Problem 29"
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<math> \text{(A)}\ 32\qquad\text{(B)}\ 34\qquad\text{(C)}\ 35\qquad\text{(D)}\ 36\qquad\text{(E)}\ 38 </math> | <math> \text{(A)}\ 32\qquad\text{(B)}\ 34\qquad\text{(C)}\ 35\qquad\text{(D)}\ 36\qquad\text{(E)}\ 38 </math> | ||
− | ==Solution== | + | ==Solution 1== |
Working with the second part of the problem first, we know that <math>3n</math> has <math>30</math> divisors. We try to find the various possible prime factorizations of <math>3n</math> by splitting <math>30</math> into various products of <math>1, 2</math> or <math>3</math> integers. | Working with the second part of the problem first, we know that <math>3n</math> has <math>30</math> divisors. We try to find the various possible prime factorizations of <math>3n</math> by splitting <math>30</math> into various products of <math>1, 2</math> or <math>3</math> integers. | ||
Line 15: | Line 15: | ||
<math>3\cdot 10 \rightarrow p^2q^9</math> | <math>3\cdot 10 \rightarrow p^2q^9</math> | ||
− | <math>5\cdot 6 \ | + | <math>5\cdot 6 \rightarrow p^4q^5</math> |
<math>2\cdot 3\cdot 5 \rightarrow pq^2r^4</math> | <math>2\cdot 3\cdot 5 \rightarrow pq^2r^4</math> | ||
Line 31: | Line 31: | ||
In the third case, <math>q=3</math> gives <math>2n = 2\cdot p^2\cdot 3^8</math>. If <math>p=2</math>, then there are <math>4\cdot 9</math> factors, while if <math>p \neq 2</math>, there are <math>2\cdot 3\cdot 9</math> factors. | In the third case, <math>q=3</math> gives <math>2n = 2\cdot p^2\cdot 3^8</math>. If <math>p=2</math>, then there are <math>4\cdot 9</math> factors, while if <math>p \neq 2</math>, there are <math>2\cdot 3\cdot 9</math> factors. | ||
− | In the fourth case, <math>p=3</math> gives <math>2n = 2\cdot 3^3\cdot q^5</math>. If <math>q=2</math>, then there | + | In the fourth case, <math>p=3</math> gives <math>2n = 2\cdot 3^3\cdot q^5</math>. If <math>q=2</math>, then there are <math>7\cdot 4= 28</math> factors. This is the factorization we want. |
Thus, <math>3n = 3^4 \cdot 2^5</math>, which has <math>5\cdot 6 = 30</math> factors, and <math>2n = 3^3 \cdot 2^6</math>, which has <math>4\cdot 7 = 28</math> factors. | Thus, <math>3n = 3^4 \cdot 2^5</math>, which has <math>5\cdot 6 = 30</math> factors, and <math>2n = 3^3 \cdot 2^6</math>, which has <math>4\cdot 7 = 28</math> factors. | ||
In this case, <math>6n = 3^4\cdot 2^6</math>, which has <math>5\cdot 7 = 35</math> factors, and the answer is <math>\boxed{C}</math> | In this case, <math>6n = 3^4\cdot 2^6</math>, which has <math>5\cdot 7 = 35</math> factors, and the answer is <math>\boxed{C}</math> | ||
+ | |||
+ | |||
+ | == Solution 2 == | ||
+ | Because <math>2n</math> has <math>28</math> factors and <math>3n</math> has <math>30</math> factors, we should rewrite the number <math>n = 2^{e_1}3^{e_2}... p_n^{e_n}</math> | ||
+ | As the formula for the number of divisors for such a number gives: | ||
+ | <math>(e_1+1)(e_2+1)... (e_n+1)</math> | ||
+ | We plug in the variations we need to make for the cases <math>2n</math> and <math>3n</math>. | ||
+ | <math>2n</math> has <math>(e_1+2)(e_2+1)(e_3+1)... (e_n+1) = 28</math> | ||
+ | <math>3n</math> has <math>(e_1+1)(e_2+2)(e_3+1)...(e_n+1) = 30</math> | ||
+ | |||
+ | If we take the top and divide by the bottom, we get the following equation: | ||
+ | <math>\frac{(e_1+2)(e_2+1)}{(e_1+1)(e_2+2)} = \frac{14}{15}</math>. Letting <math>e_1=x</math> and <math>e_2 = y</math> for convenience and expanding this out gives us: | ||
+ | <math>xy-13x+16y+2=0</math> | ||
+ | |||
+ | We can use Simon's Favorite Factoring Trick (SFFT) to turn this back into: | ||
+ | <math>(x+16)(y-13) +2 + 208 = 0</math> or | ||
+ | <math>(x+16)(y-13) = - 210</math> | ||
+ | |||
+ | As we want to be dealing with rather reasonable numbers for <math>x</math> and <math>y</math>, we try to make the <math>x+16</math> term the slightly larger term and the <math>y-13</math> term the slightly smaller term. This effect is achieved when <math>x+16 = 21</math> and <math>y-13 = -10</math>. Therefore, <math>x = 5, y = 3</math>. We get that this already satisfies the requirements for the number we are looking for, and we take <math>(5+2)(3+2) = \boxed{35}</math> | ||
+ | |||
+ | ==Solution 3 (Alcumus Solution)== | ||
+ | Let <math>\, 2^{e_1} 3^{e_2} 5^{e_3} \cdots \,</math> be the prime factorization of <math>\, n</math>. Then the number of positive divisors of <math>\, n \,</math> is <math>\, (e_1 + 1)(e_2 + 1)(e_3 + 1) \cdots \; </math>. In view of the given information, we have <cmath> 28 = (e_1 + 2)(e_2 + 1)P </cmath>and <cmath> 30 = (e_1 + 1)(e_2 + 2)P, </cmath>where <math>\, P = (e_3 + 1)(e_4 + 1) \cdots \; </math>. Subtracting the first equation from the second, we obtain <math>\, 2 = (e_1 - e_2)P, \,</math> so either <math>\, e_1 - e_2 = 1 \,</math> and <math>\, P = 2, \,</math> or <math>\, e_1 - e_2 = 2 \,</math> and <math>\, P = 1</math>. The first case yields <math>\, 14 = (e_1 + 2)e_1 \,</math> and <math>\, (e_1 + 1)^2 = 15</math>; since <math>\, e_1 \,</math> is a nonnegative integer, this is impossible. In the second case, <math>\, e_2 = e_1 - 2 \,</math> and <math>\, 30 = (e_1 + 1)e_1, \,</math> from which we find <math>\, e_1 = 5 \,</math> and <math>\, e_2 = 3</math>. Thus <math>\, n = 2^5 3^3, \,</math> so <math>\, 6n = 2^6 3^4 \,</math> has <math>\, (6+1)(4+1) = \boxed{35} \,</math> positive divisors. | ||
==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=28|num-a=30}} | {{AHSME box|year=1996|num-b=28|num-a=30}} | ||
+ | {{MAA Notice}} |
Latest revision as of 05:52, 26 October 2023
Problem
If is a positive integer such that
has
positive divisors and
has
positive divisors, then how many positive divisors does
have?
Solution 1
Working with the second part of the problem first, we know that has
divisors. We try to find the various possible prime factorizations of
by splitting
into various products of
or
integers.
The variables are different prime factors, and one of them must be
. We now try to count the factors of
, to see which prime factorization is correct and has
factors.
In the first case, is the only possibility. This gives
, which has
factors, which is way too many.
In the second case, gives
. If
, then there are
factors, while if
, there are
factors.
In the second case, gives
. If
, then there are
factors, while if
, there are
factors.
In the third case, gives
. If
, then there are
factors, while if
, there are
factors.
In the third case, gives
. If
, then there are
factors, while if
, there are
factors.
In the fourth case, gives
. If
, then there are
factors. This is the factorization we want.
Thus, , which has
factors, and
, which has
factors.
In this case, , which has
factors, and the answer is
Solution 2
Because has
factors and
has
factors, we should rewrite the number
As the formula for the number of divisors for such a number gives:
We plug in the variations we need to make for the cases
and
.
has
has
If we take the top and divide by the bottom, we get the following equation:
. Letting
and
for convenience and expanding this out gives us:
We can use Simon's Favorite Factoring Trick (SFFT) to turn this back into:
or
As we want to be dealing with rather reasonable numbers for and
, we try to make the
term the slightly larger term and the
term the slightly smaller term. This effect is achieved when
and
. Therefore,
. We get that this already satisfies the requirements for the number we are looking for, and we take
Solution 3 (Alcumus Solution)
Let be the prime factorization of
. Then the number of positive divisors of
is
. In view of the given information, we have
and
where
. Subtracting the first equation from the second, we obtain
so either
and
or
and
. The first case yields
and
; since
is a nonnegative integer, this is impossible. In the second case,
and
from which we find
and
. Thus
so
has
positive divisors.
See also
1996 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 28 |
Followed by Problem 30 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.