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## The year 2011

• Posted by a hidden member.
Take the year you were born (87 for me) and add that to the age you will become in 2011. EVERYONE born before 2000 will add up to 111...I can make no sense of it!!
• TheAlchemixt

Posts: 2294
I was born in 84 in 2011 I will be 27. 84 + 27= 111.

So kanaka in 2011 you should be 24 because

87 + 24 = 111. I hope that made sense. Your buddy - Ryan
• Posted by a hidden member.
Lol Ryan. Your addition made perfect sense. What I don't understand is how everyone in the world, 12 and older, share 111.
• Posted by a hidden member.
it's perfectly logical. the numbers added "fills" up last century (100) plus the eleven years you've lived in this century.
next year the "mystery" number will be 112 for everyone who was born last century.
edit: All it describes is that from 1900 - 2011 is a span of 111 years. any point (birth year) where you divide that timeline the number of years on each side added will be 111, easy.

(.... whoever came up with this "weird fact" must have a lot of time on his hands or have strange interests LOL)
• TheAlchemixt

Posts: 2294
Oh okay Kanaka. I misunderstood you then lol but I think judo guy knows what he is talking about.v
• Posted by a hidden member.
hahah perfect! I was studying all night and drifted a bit lol.
• Posted by a hidden member.
Kanaka24 saidhahah perfect! I was studying all night and drifted a bit lol.

Get some rest and then back to studying. We need someone to solve world peace, cure health problems, solve world finance issues, etc.. and have somebody hot like yourself do it.

Do it!
• Posted by a hidden member.

That was a bit of a math fail Kanaka
• turtleneckjoc...

Posts: 4685
Worked for me too.

Born in '54 and I will be 57 this year. Comes to 111.
• Posted by a hidden member.
a total fail! too many line integrals and diverging parametric equations "skewing" my addition skills
• Posted by a hidden member.
92+19=111
• alphatop

Posts: 1955
75+36=111
• Posted by a hidden member.
You're welcome....
1 110 111
2 109 111
3 108 111
4 107 111
5 106 111
6 105 111
7 104 111
8 103 111
9 102 111
10 101 111
11 100 111
12 99 111
13 98 111
14 97 111
15 96 111
16 95 111
17 94 111
18 93 111
19 92 111
20 91 111
21 90 111
22 89 111
23 88 111
24 87 111
25 86 111
26 85 111
27 84 111
28 83 111
29 82 111
30 81 111
31 80 111
32 79 111
33 78 111
34 77 111
35 76 111
36 75 111
37 74 111
38 73 111
39 72 111
40 71 111
41 70 111
42 69 111
43 68 111
44 67 111
45 66 111
46 65 111
47 64 111
48 63 111
49 62 111
50 61 111
51 60 111
52 59 111
53 58 111
54 57 111
55 56 111
56 55 111
57 54 111
58 53 111
59 52 111
60 51 111
61 50 111
62 49 111
63 48 111
64 47 111
65 46 111
66 45 111
67 44 111
68 43 111
69 42 111
70 41 111
71 40 111
72 39 111
73 38 111
74 37 111
75 36 111
76 35 111
77 34 111
78 33 111
79 32 111
80 31 111
81 30 111
82 29 111
83 28 111
84 27 111
85 26 111
86 25 111
87 24 111
88 23 111
89 22 111
90 21 111
91 20 111
92 19 111
93 18 111
94 17 111
95 16 111
96 15 111
97 14 111
98 13 111
99 12 111
• Posted by a hidden member.
Kanaka24 saidTake the year you were born (87 for me) and add that to the age you will become in 2011. EVERYONE born before 2000 will add up to 111...I can make no sense of it!!

Let 1900 + x = the year you were born, where 0 < or = x < or = 99.

Then your age in 2011 is 2011 - (1900 + x) = 111 - x.

Therefore, the year you were born (in the 20th century) + your age
= x + (111 - x)
= 111

• Posted by a hidden member.
To add onto Judo guys observation... it has to do with our numeral system, which is based on the number 10.. that is... after 9 you begin with the first numeral (1) and add a 0..... after 99 you add two zeros to the 1 etc.

Its the same reason why any multiple of 9 (18, 27, 36, 108 etc) always adds up to 9... its the highest number in our system

If you were to use another numeral system, say the Mayan one, with cycles of 20 being the base... you would get different patterns.. any multiple of 19 would add up to 19 there

.....

Interestingly enough, the old german peoples used a system that went up to twelve.... which is why german languages have different names for "eleven" and "twelve" than the following "ten" or "teen"-based numbers... (thirteen = three-ten... fourteen = four-ten etc.)
• Posted by a hidden member.
Pato_Rico saidTo add onto Judo guys observation... it has to do with our numeral system, which is based on the number 10.. that is... after 9 you begin with the first numeral (1) and add a 0..... after 99 you add two zeros to the 1 etc.

Its the same reason why any multiple of 9 (18, 27, 36, 108 etc) always adds up to 9... its the highest number in our system

If you were to use another numeral system, say the Mayan one, with cycles of 20 being the base... you would get different patterns.. any multiple of 19 would add up to 19 there

.....

Interestingly enough, the old german peoples used a system that went up to twelve.... which is why german languages have different names for "eleven" and "twelve" than the following "ten" or "teen"-based numbers... (thirteen = three-ten... fourteen = four-ten etc.)

Actually this doesn't have anything to do with our numeral system. It just has everything to do with the fact that if you have two numbers and you add 1 to one of the numbers and subtract 1 from the other that the result of adding the two numbers will be the same as it was before

• Posted by a hidden member.
Pato_Rico saidIts the same reason why any multiple of 9 (18, 27, 36, 108 etc) always adds up to 9... its the highest number in our system

Take any 4-digit number ABCD.

ABCD
= 1000A + 100B + 10C + D
= 999A + 99B + 9C + (A + B + C + D)
= 9 (111A + 11B + C) + (A + B + C + D)

Therefore ABCD is divisibe by 9 if and only if A + B + C + D is divisible by 9.

The same argument can easily be extended to a number consisting of any number of digits. Just use 10^3 instead of 1000, etc.

Pato_Rico saidIf you were to use another numeral system, say the Mayan one, with cycles of 20 being the base... you would get different patterns.. any multiple of 19 would add up to 19 there

Take any 4-digit number ABCD in the 20-numeral system.

ABCD
= 20^3*A + 20^2*B + 20C + D
= (20^3 - 1)A + (20^2 -1)B + 19C + (A + B + C + D)
= 19 (421A + 21B + C) + (A + B + C + D)

Therefore ABCD is divisibe by 19 if and only if A + B + C + D is divisible by 19.

Note that 20^n - 1 = 19(20^(n-1) + 20^(n-2) + ... + 20 + 1) is divisible by 19 for all positive integers n.

• Posted by a hidden member.
Ummmm....k
• Posted by a hidden member.
Try this one:

REAL
+ JOCK
GAGA

Where: L=C+K, O=E+R, C=7, G=6, and R=2. There is a unique solution.
• Posted by a hidden member.
*sigh* I'm so old- thanks for reminding me.
• Posted by a hidden member.
okaaayyyy....last night i was waaayyy spaced out off your addy, scotty :-( crazy shit would not leave my mind!!!
• Posted by a hidden member.
chocmandms saidTry this one:

REAL
+ JOCK
GAGA

Where: L=C+K, O=E+R, C=7, G=6, and R=2. There is a unique solution.

I randomly guessed A=9 and worked from there.
2398
+4571
6969

• Posted by a hidden member.
math.... eh.... i can add by 45 lbs really easily... but other than that.... meh
• Posted by a hidden member.
While we're on the subject of 2011, what is the sum of all integers from 1 to 2011? A fun thing to do Saturday night.
1 + 2 + 3 + ..... + 2011

Well the good news is you don't have to spend your whole evening and then some figuring it out. First write it forwards then backwards (system not letting me line up the columns - imagine 1 in the first row is directly above 2011 in the second row, 2 directly above 2010, etc):
1 + 2 + 3 + ..... + 2011
2011 + 2010 + 2009 + ..... + 1

Now if we add up each column we get:
2012 + 2012 + 2012 + ..... + 2012

There are 2011 columns, each with a sum of 2012. So our answer is
2012 x 2011 / 2 = 2,023,066.
• Posted by a hidden member.
monochrome said
Pato_Rico saidIts the same reason why any multiple of 9 (18, 27, 36, 108 etc) always adds up to 9... its the highest number in our system

Take any 4-digit number ABCD.

ABCD
= 1000A + 100B + 10C + D
= 999A + 99B + 9C + (A + B + C + D)
= 9 (111A + 11B + C) + (A + B + C + D)

Therefore ABCD is divisibe by 9 if and only if A + B + C + D is divisible by 9.

The same argument can easily be extended to a number consisting of any number of digits. Just use 10^3 instead of 1000, etc.

Pato_Rico saidIf you were to use another numeral system, say the Mayan one, with cycles of 20 being the base... you would get different patterns.. any multiple of 19 would add up to 19 there

Take any 4-digit number ABCD in the 20-numeral system.

ABCD
= 20^3*A + 20^2*B + 20C + D
= (20^3 - 1)A + (20^2 -1)B + 19C + (A + B + C + D)
= 19 (421A + 21B + C) + (A + B + C + D)

Therefore ABCD is divisibe by 19 if and only if A + B + C + D is divisible by 19.

Note that 20^n - 1 = 19(20^(n-1) + 20^(n-2) + ... + 20 + 1) is divisible by 19 for all positive integers n.

Note also that the more generalized result is that any number written in the k-numeral system is divisible by k-1 if and only if the sum of its digits is divisible by k-1.
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