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Hi,
To reconsideryouranswer: The reasons for your solutions are fine.
But these are >>> mathematical problems! <<<[/math]
They're *all* mathematical problems, so that isn't new information.
That doesn't change anything. Not only is it a "mathematical
question," it is simultaneously a real-world question.
You simply admit you had a flawed question. If you want a
proper mathematical answer to a mathematical problem,
then please word it with that in mind.
You avoid a problem with this, for example, by asking what is
the longest straight-line distance (line segment) inside of a
rectangular box with those dimensions.
Then the problem or question is clear/precise and unambiguous.
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Here is another one:
There are two rectangular boxes, each having dimensions of
If the two boxes touch each other in at least at one shared point,
then what is the longest possible straight-line distance in feet
spanning from the furthest section of one box to the furthest
section of the other box?
#2973. What is the maximum length of a pencil that can be kept in
a rectangular box of dimensions 8 cm x 6 cm x 2 cm?#2974. Find the length of the longest rod that can be placed in a
m high.
room 16 m long, 12 m broad, and
ganesh,
I understand the intent of these questions, #2973 and #2974, but the answers
do not work.
#2973
The pencil, whether sharpened (at one end) or not sharpened at all,
would not fit along a longest diagonal of the box. It would have to
be infinitely thin, such as a line segment.
Note: This can be checked "in real life" with a pencil whose length is as
close to your answer (given in the hidden answer as a reduced radical).
Or, it may be easier to tell/check if all of the dimensions of the rectangular
box and the alleged (length of the) pencil are multiplied, by say,
a factor of 5.
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#2974
The same is true for the rod. It would have to be infinitely thin, too.
Anakin, you typed steps where you have this variable constant, a,
but you did not state where it came from and/or what classification
of numbers it belongs to.
It just shows up without being introduced.
Hi reconsideryouranswer,
These questions are to be attempted to be answered orally,
and attempted within a minute, without the use of paper, pen/pencil and calculator.
After sampling many questions here, I would propose the above alteration to be realistic. And my
involvement/effort to my solution to #1049 would be representative of that change.
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Edit:
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ashwil
Full Member
Re: Oral puzzles
mathsy, 2 easy ways of getting approximate answers to 20 without a calculator:-
a) every power of 10 requires just over 3 powers of 2. Therefore 50 powers of 2
suggests between 15 and 16 powers of 10.
b) If you know that log 2 = 0.3010 (which you should!), then log 2^50 = 15.05.
Again, about 15 powers of 10
By both methods, this gives me 16 digits in the answer.
_______________________________________________________________________
But how would a person know that they would be correct with either of those methods?
Hi bobbym,
The solution 1048 is perfect. Good work!
#1049. Find the sum to n terms of the series
.
bobbym and anonimnystefy,
for #1049, neither of your answers *were* correct.
Notice if n = 1 in your formulas, neither produces 2/3, which is the 1st term of the series.
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bobbym,
Then, the summation on the right-hand side of the subtraction sign is equal to:
The whole infinite series (**) sums to 1/2.
Edit:
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I will take the credit for the solution.
L # 4
I don't want a method that will rely on defining certain functions, taking derivatives,
noting concavity, etc.
Change of base:
Each side is positive, and multiplying by the positive denominator
keeps whatever direction of the alleged inequality the same direction:
On the right-hand side, the first factor is equal to a positive number less than 1,
while the second factor is equal to a positive number greater than 1. These
facts are by inspection combined with the nature of exponents/logarithms.
Because of (log A)B = B(log A) = log(A^B), I may turn this into:
I need to show that
Then
Then 1 (on the left-hand side) will be greater than the value on the
right-hand side, and the truth of the original inequality will be established.
I want to show
Raise a base of 3 to each side:
Each side is positive, and I can square each side:
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Then I want to show that when 2 is raised to a number equal to
(or less than) 1.5, then it is less than 3.
Each side is positive, and I can square each side:
(3) How many times would you have to write the number 6 when you write from 1 to 100?
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6, 16, 26, 36, 46, 56, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 76, 86, 96
Definition of Prime number: A number which has EXACTLY TWO factors (divisors) is called a Prime number!
e. g. 2 has factors 1, 2 (exactly two factors) hence is a Prime number.
3 has factors 1, 3 (exactly two factors) hence is a Prime number.
4 has factors 1, 2, 4 (more than two factors) hence is Not Prime. Such numbers are called Composite numbers.
13 has factors 1, 13 (exactly two factors) hence is a Prime number.Note: We consider the positive natural number divisors only.
Use this instead:
Prime Number: A positive integer which has exactly two distinct positive integer factors (divisors).
[I made up the following (or rediscovered it).]
Without estimating any square roots (but allowing repeated squaring
of sides and using the four main operations), in theory, is it possible
to work out which of the following sides is larger?
Break up the original sum into the sum of these separate parts:
sum (1 to oo) [(n^2)/(2^n)]:
Let S = 1/2 + 4/4 + 9/8 + 16/16 + ...
Then (1/2)S = 1/4 + 4/8 + 9/16 + ...
S - (1/2)S = 1/2 + 3/4 + 5/8 + 7/16 + ...
(1/2)S = 1/2 + 3/4 + 5/8 + 7/16 + ...
(1/4)S = 1/4 + 3/8 + 5/16 + 7/32 + ...
(1/2)S - (1/4)S = 1/2 + 2/4 + 2/8 + 2/16 + ...
(1/4)S = 1/2 + 2(1/4 + 1/8 + 1/16 + 1/32 + ...)
(1/4)S = 1/2 + 2(1/2)
(1/4)S = 1/2 + 2/2
(1/4)S = 3/2
4(1/4)S = 4(3/2)
S = 6
---------------------------------------------------------
sum (1 to oo) [(2n)/(2^n)]:
= 2[sum (1 to oo) (n)/(2^n)]:
= 2[1/2 + 2/4 + 3/8 + 4/16 + ...] **
Let S = 1/2 + 2/4 + 3/8 + 4/16 + ...
(1/2)S = 1/4 + 2/8 + 3/16 + ...
S - (1/2)S = 1/2 + 1/4 + 1/8 + 1/16 + ...
(1/2)S = 1
2(1/2)S = 2(1)
S = 2
Then ** = 2[2]
= 4
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sum (1 to oo) [3/(2^n)]:
= 3[sum (1 to oo) 1/(2^n)]
= 3[1/2 + 1/4 + 1/8 + ...]
= 3[1]
= 3
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Then the sum equals
6 + 4 + 3 =
13
Hi all;
I do not think there is much to show. You only have to sum the first 2 to see the >> sum is greater than 3. <<
The above amendment comes right out of the first two consecutive terms, as well as
it shows 3 followed by an excess positive number.
This exemplifies the highlighted comment in the quote box.
http://www.mathsisfun.com/puzzles/24-from-8-8-3-3-solution.html
Any of them with 8^(1/3), for example, do not count, because there are 1s being used
to make these fractional exponents, but the problem is to have only two 8s
and two 3s. No other digits are to be used.
phrontister's "CONCATENATE" one is not a solution.
What amounts to (1, 0, 0, 0) is not 1000, so it does
not count.