Sum of even factors

What is the sum of all even divisors of 1000? #math #maths #mathematics #shorts

Source

Send me suggestions by email (address at end of many videos). I may not reply but I do consider all ideas!

If you purchase through these links, I may be compensated for purchases made on Amazon. As an Amazon Associate I earn from qualifying purchases. This does not affect the price you pay.

Book ratings are from January 2022.

My Books (worldwide links)

My Books (US links)
Mind Your Decisions: Five Book Compilation
A collection of 5 books:
"The Joy of Game Theory" rated 4.2/5 stars on 224 reviews
"The Irrationality Illusion: How To Make Smart Decisions And Overcome Bias" rated 4/5 stars on 24 reviews
"40 Paradoxes in Logic, Probability, and Game Theory" rated 4.1/5 stars on 38 reviews
"The Best Mental Math Tricks" rated 4.2/5 stars on 76 reviews
"Multiply Numbers By Drawing Lines" rated 4.3/5 stars on 30 reviews

Mind Your Puzzles: Collection Of Volumes 1 To 3
A collection of 3 books:
"Math Puzzles Volume 1" rated 4.4/5 stars on 87 reviews
"Math Puzzles Volume 2" rated 4.1/5 stars on 24 reviews
"Math Puzzles Volume 3" rated 4.2/5 stars on 22 reviews

2017 Shorty Awards Nominee. Mind Your Decisions was nominated in the STEM category (Science, Technology, Engineering, and Math) along with eventual winner Bill Nye; finalists Adam Savage, Dr. Sandra Lee, Simone Giertz, Tim Peake, Unbox Therapy; and other nominees Elon Musk, Gizmoslip, Hope Jahren, Life Noggin, and Nerdwriter.

My Blog

Twitter

Instagram

Merch

Patreon

Press

[racheline_nya]

2184 is also a great year to be born, because your 3rd birthday will be in the year 3^7, and your 13th birthday will be in the year 13^3

[paulgreen]

Nice! Things like this lead to very interesting topics.

[SmokeyBear]

Didn’t get the answer but trying made me discover. The sum of all factors of any number that is an answer to 2^(all positive integers) (1, 2, 4, 8, 16, 32, etc) is equal to “(that number) x 2 - 1”

Sum of all factors of y = 2y - 1 if y is an answer to 2^(all positive integers)

[sabe]

alternatively, note that the susm of the odd factors is 1/14 the sum of the even factors, so the sum of the even factors is 14/15 the ssum of all factors
sum of all factors = (2^4-1)/1 * (5^4-1)/4 = 15*156
14/15 * 15 * 156 = 14*156=2184.

[deandelvin]

Glad this channel has some problems I can solve in my head

[SASA_maxillo]

It is
Counter = 0
For i in range(1000):
If i % 2 == 0:
Counter += i
Print(counter)

[mtlmathstutoring]

I got this the second time around! Very satisfying argument!

[AliBaBaSR]

This is the simplest of simplest thing I have ever seen

[arjunraja]

I'm so happy I got this one on my own

[_man_of_culture]

He should use arithmetic progression it would be easy

[jucom]

I solved it by the simple method of 2×sum(2²5³)=2×(2³-1)/(2-1)×(5⁴-1)/(5-1), without having to work out all the powers

[irfaniskandar]

ChatGPT answer:
To find the sum of all even factors of 1000, we need to identify all the even factors of 1000 and then add them up.

The prime factorization of 1000 is: 2^3 * 5^3.

The even factors of 1000 will have one or more powers of 2 in their factorization, but not more than three powers of 2 since we are only considering even factors. The powers of 5 do not affect the evenness of the factors.

The possible combinations of powers of 2 are:

2^0 = 1 (odd factor)
2^1 = 2
2^2 = 4
2^3 = 8
Now we add up all the even factors:
2 + 4 + 8 = 14

Therefore, the sum of all even factors of 1000 is 14.

[Muhammed_Basheer]

The prime factorization of 1000 is $2^3 \cdot 5^3$. An even factor of 1000 will be of the form $2^a \cdot 5^b$, where $0 \leq a \leq 3$ and $0 \leq b \leq 3$. Thus, the even factors of 1000 are:

\begin{align*}
2^0 \cdot 5^0 &= 1\
2^1 \cdot 5^0 &= 2\
2^2 \cdot 5^0 &= 4\
2^3 \cdot 5^0 &= 8\
2^0 \cdot 5^1 &= 5\
2^1 \cdot 5^1 &= 10\
2^2 \cdot 5^1 &= 20\
2^3 \cdot 5^1 &= 40\
2^0 \cdot 5^2 &= 25\
2^1 \cdot 5^2 &= 50\
2^2 \cdot 5^2 &= 100\
2^3 \cdot 5^2 &= 200\
2^0 \cdot 5^3 &= 125\
2^1 \cdot 5^3 &= 250\
2^2 \cdot 5^3 &= 500\
2^3 \cdot 5^3 &= 1000\
\end{align*}

We need to find the sum of all these even factors. To do so, we can group the factors as follows:

\begin{align*}
&(1 + 2 + 4 + 8)(1 + 5 + 25 + 125) \
&= 15 \cdot 156 \
&= 2340
\end{align*}

Therefore, the sum of all the even factors of 1000 is 2340.

[thidasvinnath]

You could use primorials I just learnt about them and I'm pretty tired to explain so look it up for yourself

[KIKU_]

There is also a difficult method called Brute force😆

[prasadadke]

Can we draw it's factors by permutation method?

[rantarothegreat]

Now what if the question was not sum but product.
Is there a way you could have done it then?

[firelow]

cool i did not expect it to seem so simple

[kickthacan]

There is a direction formula for this JEE aspirants will know it

[krzysztofmazurkiewicz]

Hmm now im curious. As even factors i would take only those who are actually even and that wouls be only all powers of 2 except 2^0. 5 would be an odd number. But i may be missing something here