Question

\[2^4+10^3+3^3\times37\]

Answer 1

37*9 is special
adding 1032 to tht gives 2015 :)

Answer 2

37*3
37*6
37*9
...
a special pattern ;)

Answer 3

**adding 1016 :P

Answer 4

\[2^4+10^3+3^3\times37\\16+1000+3^2\times3\times37\\1016+9\times111\\1016+999\\\]

Answer 5

\(\Large 37\times 3 =111 \)
\(\Large 37\times 6 =222 \)
\(\Large 37\times 9 =333 \)
\(\Large 37\times 12 =444 \)
\(\Large 37\times 15=555 \)
\(\Large 37\times 18 =666 \)
\(\Large 37\times 21 =777 \)
\(\Large 37\times 24 =888 \)
\(\Large 37\times 27 =99 \)

Answer 6

\(\Large 37\times 27 =999\)
***

Answer 7

What is an expression for 2015 in only prime numbers

Answer 8

Something better than
\[2^4+2^35^3+3^3\times37\]

Answer 9

2015 is a square free integer

Answer 10

\(\Large 5\times 13 \times 31\)

Answer 11

is there a special number have same property as 37 ?

Answer 12

makes me remember this
37
373737
3737373737
37373737373737
.....

Answer 13

all composite

Answer 14

this is the rule (why its composite)
37(10101...).

Answer 15

a 3 digit number , say pqr.

\(\Large pqr \times 1001 = pqrpqr\)

Answer 16

\[\begin{vmatrix}5 & 2 & 7 \\ 0 & 13 & 11 \\ 0 & 0 & 31\end{vmatrix} = 2015\]

Answer 17

\(\begin{vmatrix}5 & \heartsuit & \clubsuit \\ 0 & 13 & \spadesuit \\ 0 & 0 & 31\end{vmatrix} = 2015\)
:P

Answer 18

neat

Answer 19

2014 + 1 = 2015 :D

Answer 20

omg, why didn't i think of that! :P
smart

Answer 21

Ik ik ~*bows*~

Answer 22

Clever ;)
by goldbach conjecture it can be represented as sum of 3 primes

Answer 23

What is the sum of primes that equals 2015?

Answer 24

Nope

Answer 25

659 + 673 + 683 ?

Answer 26

No, dang it. Lol I am trying

Answer 27

2011+2+2

Answer 28

Oh my goodness I feel like somebody just smacked me with a bat

Answer 29

659 + 673 + 683 is way better.

Answer 30

:|

Answer 31

661 + 643 + 3*2^3 + 673 + 7*2

I am getting bored as you can tell. I just created a few equations on how to get 2015. I am working on a more complex one right now

Answer 32

hey no wait there are like thousands of ways in which you can represent 2015 as sum of 3 different primes.. here is a more fun one :

represent 2015 as sum of squares of two integers

Answer 33

\[2015 = a^2 + b^2\]

find \(a\) andn \(b\)

Answer 34

I don't know but this is the closest I have gotten xD 35.2^2 + 27.87^2

Answer 35

To 2015. I am just guessing right now. I will probably try to work it out when I am fully awake

Answer 36

eh no integer solution
(face palm)

Answer 37

any way this circle gives all solution :P

Answer 38

lol does that mean all the points on that circle have irrational coordinates ?
it turns out that 2015 is a special number which cannot be represented as sum of squares of two integers and the reason for that goes back to 17th century
https://www.math.hmc.edu/funfacts/ffiles/20008.5.shtml

Answer 39

infact i check from 1 to 45

Answer 40

checked*

Answer 41

bruteforce is for programmers, not mathematicians :P

Answer 42

who cares

Answer 43

but yes i likes 4k+3 and 4k+1 idea
since 2015 is 3 mod 4
i completely forgot about that

Answer 44

but u seems like exited about 2015 :P @ganeshie8
note that its a happy number

Answer 45

2015 -> 2^2 + 0^2 + 1^2 + 5^2 = 4+0+1+25 = 30
-> 3^2 + 0^2 = 9
-> 9^2 = 81
-> 8^2 + 1^2 = 64+1 = 65
-> 6^2 + 5^2 = 36 + 25 = 61
-> 6^2+1^2 = 37
im tired :O

Answer 46

well 3 is a happy number i stopped when i got 30 :P

Answer 47

how do u know 3 is happy

Answer 48

but lets continue
37=9+49=58
58=25+64=89
89=64+81=145
145=1+16+25=42
42=16+4=20
20=4
4=16=1+36=37

seems like sad not happy

Answer 49

does that mean 3 is also not happy ? because all the numbers in the sequence of steps have to fall on same branch

Answer 50

yes

Answer 51

then i would say a happy number can be written as sum of 2 squares

Answer 52

counter example : 19

Answer 53

hmm or sum of squares can be a happy number :P