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OK, here we have 2 puzzle submissions. The first, Matchtricks, comes from Suineg :
Correct the equations by moving only one match:
NOTE: THE EQUAL SIGNS AND THE DIVISION OPERATOR ARE NOT MADE FROM MATCHES.
The 2nd comes from the Chairman of an engineering company, and I must warn: is very difficult:
Find the missing number in the following series
10,11,12,13,14,15,16,17,20,22,24,31,100,—–,10000.
Feel free to enter your answers below in the comment section; will give it some time before unmasking. By the way, the picture riddle below still remains unsolved!
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for the number puzzles:
1) by moving the last match in the denominator to the far left of the numerator, one can form the equation IXIII/VI = II (12/6 = 2). The numerator is not technically well-formed, but would equal the amount stated.
2) one solution (which also works for number 1) is to place any match diagonally across the equal sign to form a not-equal sign. another solution is to move the second match from the left to the right and slightly above the left-hand side of the equation to turn it into an exponent, so that it reads: II^I = II (2^1 = 2). The very first match is still slanted, but it is legible. The exponent could similarly be placed on the other side of the equation.
Don’t have the number series yet, I’ll write again if I get it.
Not sure if I’ve got the first matchtrick right, but I believe you simply tilt the last match of XIII so the equation becomes XIV/VII=II.
For the second, I may have cheated: Simply cross the matches that make up the V, and think outside the box. It becomes 2)X I=II (or 2 times 1 equals 2
)
matchstick 1 : move the last match in “XIII” so that it joins with the adjacent match, forming a V as in “XIV”. XIV/VII=II
matchstick 2 : move any match so that it crosses the = sign perpendicularly, forming a “not-equals” sign. Cheap, but effective.
series : you have to get to the base of the problem. If I’m right, the series that defines the answer is 1,2,3,11,12,20,21,22,100,101,102,110,111,112,120,121, so the answer is 121.
Matchtricks #1
XIV/VII = II
II to V on left side, top.
Matchtricks #2
XI = 11
V to X on left side. The left side of the equation is roman numerals and the right is numbers, both are eleven.
121 is the missing number in the series (16 in base 3:all the numbers are 16 in decreasing base)
finally! i actually figured s.t. out!! okay, for equation #1 move the last matchstick in XIII to form a V with the one in front of it so the equation reads: XIV/VII=II (14/7=2)
equation #2 move one of the matchstick in the V to form an X so the equation reads XI=11!
For the Roman numerals
1) move a matchstick from the denominator and place it horizontally over the second I to produce:
XI + I
—— = II
V I
2) All I could think of was removing one match from the V to leave a lobsided I = I
OR to place one match from the right on top of the I on the left thus looking like a square root sign where the Sq root of 1 = 1
so: __
V I = I (bit difficult to draw but..)
matchsticks
1. XII/V+I=II
2.XI=11(in arabic)
The missing number is 121. 16 is 121 in base 3
wow that series is hard!!!, is pure aritmetic RK???
It is tough, Suineg. Only a few people have gotten it so far. It kind of goes beyond arithmetic in a way….
will unmask submitted answers Thursday afternoon so you can see
for the series it looks like it may have something to do with computers, like how everything is in base 16 or somehting but not sure on the calculation of it…
i think 121 is the answer to the sequence, but hexagesimal comon RK I am not a machine man, jajajajajajajajajajajaja, it was hard, I try with the base 8 because of the jump from 17 to 20, but then it came the light, not sure yet!
i will be a good sport and put a good explanation because this series is hard:
Number 16 expresed in every base from hexagesimal to base 1 so: 16 will be according to his base: note(base,16 in that base)
(16,10), (15,11), (14,12), (13,13), (12,14), (11,15), (10,16), (9,17), (8,20), (7,22), (6,24), (5,31), (4,100), (3,121), (2,10000)
Let explain the harder position: 10.000
base 2 mean that you have 2 numbers(binary), 1 and 0;
so: 10000= every position in binary counts from right to left: 2^0… 2^4=16 wowowowowowowowow!!!!!!!!!!!!
This puzzle is the trickiest puzzle I have seen in my life because it depends of a fixed number 16 on different numeric bases, the binary was the one that gave me the answer, but it was too hard for someone with regular mathematical knowlegde, I thought i had a lot of mathematical background in that matter, but cool!!!
wow, great alternative answers to the matchtrick puzzle, great ones, the answers I had in mind pop in the second equation in two cases, not so for the first puzzle, but I like to see many alternative great answers indeed, I send the answers I had in mind to RK, cool
Thank you Suineg for submitting the match puzzle.
While there were a couple of interesting ways to solve (as shown above), here’s what Suineg had in mind:
1) (X-III)/VII = I
2) SQUARE ROOT(I)= I
Lots of lateral thinking used for the 2nd one I see
And great job to Shawn, bizarette18, bvan23, and Suineg on the series question; that thing was just brutal
arrr…
121
(quick cheat: http://www.research.att.com/~n.....;go=search)
suineg, now I know where my mathematical hemisphere resides, awesome!