Difference between revisions of "2019 AMC 10B Problems/Problem 19"
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==Solution== | ==Solution== | ||
− | The prime factorization of 100,000 is <math>2^5 \cdot 5^5</math>. Thus, we choose two numbers <math>2^a5^b</math> and <math>2^c5^d</math> where <math>0 \le a,b,c,d \le 5</math> and <math>(a,b) \neq (c,d)</math>, whose product is <math>2^{a+c}5^{b+d}</math>, where <math>0 \le a+c \le 10</math> and <math>0 \le b+d \le 10</math>. | + | The prime factorization of <math>100,000</math> is <math>2^5 \cdot 5^5</math>. Thus, we choose two numbers <math>2^a5^b</math> and <math>2^c5^d</math> where <math>0 \le a,b,c,d \le 5</math> and <math>(a,b) \neq (c,d)</math>, whose product is <math>2^{a+c}5^{b+d}</math>, where <math>0 \le a+c \le 10</math> and <math>0 \le b+d \le 10</math>. |
− | Consider <math>100000^2 = 2^{10}5^{10}</math> | + | Consider <math>100000^2 = 2^{10}5^{10}</math>, whose number of divisors is <math>(10+1)(10+1) = 121</math>. However, some of the divisors of <math>2^{10}5^{10}</math> cannot be written as a product of two distinct divisors of <math>2^5 \cdot 5^5</math>, namely: <math>1 = 2^05^0</math>, <math>2^{10}5^{10}</math>, <math>2^{10}</math>, and <math>5^{10}</math>. The last two can not be created because the maximum factor of <math>100,000</math> involving only <math>2</math>s or <math>5</math>s is only <math>2^5</math> or <math>5^5</math>. Since the factors chosen must be distinct, the last two numbers cannot be created because those require <math>2^5 \cdot 2^5</math> or <math>5^5 \cdot 5^5</math>. This gives <math>121-4 = 117</math> candidate numbers. It is not too hard to show that every number of the form <math>2^p5^q</math> where <math>0 \le p, q \le 10</math>, and <math>p,q</math> are not both <math>0</math> or <math>10</math>, can be written as a product of two distinct elements in <math>S</math>. Hence the answer is <math>\boxed{\textbf{(C) } 117}</math>. |
==See Also== | ==See Also== |
Revision as of 23:25, 17 February 2019
- The following problem is from both the 2019 AMC 10B #19 and 2019 AMC 12B #14, so both problems redirect to this page.
Problem
Let be the set of all positive integer divisors of
How many numbers are the product of two distinct elements of
Solution
The prime factorization of is
. Thus, we choose two numbers
and
where
and
, whose product is
, where
and
.
Consider , whose number of divisors is
. However, some of the divisors of
cannot be written as a product of two distinct divisors of
, namely:
,
,
, and
. The last two can not be created because the maximum factor of
involving only
s or
s is only
or
. Since the factors chosen must be distinct, the last two numbers cannot be created because those require
or
. This gives
candidate numbers. It is not too hard to show that every number of the form
where
, and
are not both
or
, can be written as a product of two distinct elements in
. Hence the answer is
.
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 | ||
All AMC 10 Problems and Solutions |
2019 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.