Same question.Labib wrote:What does "find all their sums" mean? I mean, do I have to find the sum of the 3 divisors??Masum wrote:$4.$ Write it as a product of $3$ numbers in as many ways you can, and find all their sums. It turned out that one sum was twice one of the others and a perfect square.
Search found 8 matches
- Tue Dec 27, 2011 6:15 pm
- Forum: Number Theory
- Topic: Preparation Marathon
- Replies: 112
- Views: 48762
Re: Preparation Marathon
- Mon Oct 31, 2011 11:05 am
- Forum: National Math Camp
- Topic: Exercise 1.13 [Section 1, BOMC-2011]
- Replies: 26
- Views: 13289
Re: Exercise 1.13 [Section 1, BOMC-2011]
Hmmm... Okay...
I have a knack for forgetting to review a problem for further details, very often causing me to make (devastating) silly mistakes. I guess I should try to be more cautious from now on
I have a knack for forgetting to review a problem for further details, very often causing me to make (devastating) silly mistakes. I guess I should try to be more cautious from now on
- Mon Oct 31, 2011 10:56 am
- Forum: National Math Camp
- Topic: Exercise 1.13 [Section 1, BOMC-2011]
- Replies: 26
- Views: 13289
Re: Exercise 1.13 [Section 1, BOMC-2011]
In THIS case, this might actually have a value, you know $\frac{0}{0}$, which is smaller than $\frac{45}{8}$... Like $\frac{sinx}{x}$ tends to one as $x$ tends to zero...
- Mon Oct 31, 2011 10:35 am
- Forum: National Math Camp
- Topic: Exercise 1.13 [Section 1, BOMC-2011]
- Replies: 26
- Views: 13289
Re: Exercise 1.13 [Section 1, BOMC-2011]
You mean $ -\frac{1}{2}$? Doesn't $x=0$ fall between the two limits anyway?
- Mon Oct 31, 2011 10:27 am
- Forum: National Math Camp
- Topic: Exercise 1.13 [Section 1, BOMC-2011]
- Replies: 26
- Views: 13289
Re: Exercise 1.13 [Section 1, BOMC-2011]
@Jini you have included the case when $x=0$ and $x \geq - \frac{1}{2}$ And for the IMO longlisted problem: (i) If you're a student under class -9 "Well done" (ii)If you're a student of upper class-9 then.... The answer comes out to be imaginary otherwise... Should I add the lower limit, then? About...
- Mon Oct 31, 2011 3:20 am
- Forum: National Math Camp
- Topic: Exercise 1.13 [Section 1, BOMC-2011]
- Replies: 26
- Views: 13289
Re: Exercise 1.13 [Section 1, BOMC-2011]
@Sourav:
$\frac{1}{log_{a}abc}+\frac{1}{log_{b}abc}+\frac{1}{log_{c}abc}$
$a;b;c>0$
I wonder if this is right:
$\frac{1}{log_{a}abc}+\frac{1}{log_{b}abc}+\frac{1}{log_{c}abc}$
$a;b;c>0$
I wonder if this is right:
- Mon Oct 31, 2011 3:05 am
- Forum: National Math Camp
- Topic: Exercise 1.13 [Section 1, BOMC-2011]
- Replies: 26
- Views: 13289
Re: Exercise 1.13 [Section 1, BOMC-2011]
This is how I did it: $\frac{(4x^2)}{(1-\sqrt{1+2x})^2}$ $= \frac{(4x^2)}{(1-\sqrt{1+2x})^2} . \frac{(1+\sqrt{1+2x})^2}{(1+\sqrt{1+2x})^2}$ $= \frac{(4x^2)(1+\sqrt{1+2x})^2}{(1^2-(1+2x))^2}$ $= \frac{(4x^2)(1+\sqrt{1+2x})^2}{(-2x)^2}$ $= (1+\sqrt{1+2x})^2$ Now, we are given: $\frac{(4x^2)}{(1-\sqrt{...
- Fri Oct 21, 2011 8:33 pm
- Forum: National Math Camp
- Topic: বাংলাদেশ অনলাইন ক্যাম্প - ২০১১
- Replies: 105
- Views: 58066
Re: বাংলাদেশ অনলাইন ক্যাম্প - ২০১১
What is or was this "extension camp" exactly? Does it have anything to do with the "Lilaboti Camp" that was held for girls? :? According to a few of the earlier posts,there will be an evaluation exam on the first day, but in the final decision about the structure and rules, there was no mention of a...