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@EddieEC: You can get them from books like "The Most Enormous Book of Brainteasers Ever!".

More Riddles

MSD wants math related more than lingo related riddles, so here it goes:

1) Complete the sequence: 34 7 29 11 23 16 16 22 ?

2) Complete the matrix:

6 3 4 6
5 5 7 4
8 3 4 8
3 9 7 ?

Hint: A single-digit number should be filled.

3) 2-1-7-3-8-9-5 is to 9-7-2-5-3-8-1 as 9-6-7-4-8-1-2 is to ? (fill in the proper sequence)

4) Find the missing matrix element:

1 7 5 9 5 7
6 4 8 1 4 4
2 3 2 ? 9 2
9 1 2 3 3 5
2 6 5 4 3 7

5) Little Tommy Tomkins riddle:

A teacher at a local school decides to improve the young children's arithmetic by teaching them how to measure. He explains that most things are measurable, and demonstrates with a number of different measuring instruments. He groups the students into teams and asks them to measure different things like areas, change in temperature, weight of different objects, etc... For homework, he asks them to use a measuring instrument at home to measure and calculate whatever they choose. The next day, he receives the students' assignments. While correcting them, he's shocked by the results that Tommy Tomkins wrote in his notebook:

7 + 10 = 5
8 + 5 = 1
4 + 11 = 3
9 + 7 = 4
6 + 8 = 2
5 + 7 = 12

He immediately asks to see Tommy. "This is unacceptable work" he says, "only one right out of six simple additions".

"But they are all the right answers" objects Tommy. "I checked them carefully, every one of the sums is right".

The teacher allows Tommy to explain further and is amazed to find out that Little Tommy was actually correct.

So, what was Tommy measuring?
1. 8
2.
3. 1-7-9-2-4-8-6
4.
5. time
By using the hint (I don't know how I would have solved it otherwise)
8. 924 - 1716
@EddieEC:

E1: 4+5=3^2

E2: Send 1&2 (2min), bring back 1 (1 min), send 5&10 (10min), bring back 2 (2 min), send 1&2 (2 min). You're done in exactly 17min

E3: She opened the box labeled "Apple and Oranges". As it is labeled incorrectly, it will have only one type of fruit.

If she found apples:
The box labeled "Apples and Oranges" contains Apples
The box labeled Oranges contains Apples and Oranges
The box labeled Apples Oranges.

If she found oranges:
The box labeled "Apples and Oranges" contains oranges
The box labeled Oranges contains Apples
The box labeled Apples contains apples and oranges
@karim
E1: 4+5=3^2 where did the ^ come from?
@geek: 1, 3 and 5 are correct :)

@MSD: I think karim meant:4 + 5 = 3²
karim wrote@EddieEC:

E1: 4+5=3^2

E2: Send 1&2 (2min), bring back 1 (1 min), send 5&10 (10min), bring back 2 (2 min), send 1&2 (2 min). You're done in exactly 17min

E3: She opened the box labeled "Apple and Oranges". As it is labeled incorrectly, it will have only one type of fruit.

If she found apples:
The box labeled "Apples and Oranges" contains Apples
The box labeled Oranges contains Apples and Oranges
The box labeled Apples Oranges.

If she found oranges:
The box labeled "Apples and Oranges" contains oranges
The box labeled Oranges contains Apples
The box labeled Apples contains apples and oranges
Correct :)
Ra8 wrote10. Difficulty 1/5
http://funshots.in/images/13986005235273888330.png
The image is not on the scale
The side of the square and the circumference of the circle are equal.
We are rolling the circle on the square (like a wheel) how many complete tours will the circle make f it has rolled all over the square and came back to its original place?
5
Indeed, by 3^2 i meant 3 squared. Which doesn't need any extra operator to write on a piece of paper.
@unforgiven correct some people don't see that on the corner the wheel rotates 90 degrees

@mesa #2 and #4 answer is 6
#2: the sum of the horizontal lines increases 2 by 2 (1st line:19, 2nd:21, 3rd:23, and 4th should be 25 a six completes it)
#4: the sum of the vertical lines increases 1 by 1 (1st:20 2nd:21 3rd:22 5th:24 6th:25 so the 4th should be 23 and a six completes it)
#14
Find the odd number (explain why): 3;7;31;73;127
#14

127? because its the only prime number in the set over 100? haha.
xterm wrote#14

127? because its the only prime number in the set over 100? haha.
no haha :P
14. just shooting in the dark: 127 since it is the only one that if flipped would not be prime, since 721 is not prime?
#14
127 since there is 2 number starting w 3 (3;31) and 7 (7;73) ......
@MSD and az09za90za I think these answers could be considered right. But that's not what i was expecting.
#14: Why 73 is the odd number?
Ra8 wrote11. Difficulty 2/5
for any x different from 0 => 0^x=0
for any x different from 0 => x^0=1

what is lim x->0 (x^x) ??
show all steps
#15 difficulty:2/5
what is lim x->infinity of (1+1/x)^x
the answer is not 1 .. so show all steps and find the answer
Ra8 wrote@mesa #2 and #4 answer is 6
#2: the sum of the horizontal lines increases 2 by 2 (1st line:19, 2nd:21, 3rd:23, and 4th should be 25 a six completes it)
#4: the sum of the vertical lines increases 1 by 1 (1st:20 2nd:21 3rd:22 5th:24 6th:25 so the 4th should be 23 and a six completes it)
yep :)
15) lim x->infinity of (1+1/x)^x = e (it's actually one of the definitions of e)

I'm finishing the proof, just need a couple of minutes

[Edit] Proof:

Let n = lim(x->infinity) of (1+1/x)^x and function f(x) = ln(x)

f(n) = f(lim(x->infinity) of (1+1/x)^x) = lim(x->infinity) of f((1+1/x)^x)

since for any a and any x>0, f(x^a) = af(x) when f(x) = ln(x):
proof:

if g(x) = f(x^a) and h(x) = af(x), and f(1) = 0, then g(1) = h(1) = 0.
By chain rule: g'(x) = f'(x^a).a.x^(a-1) = (1/x^a).a.x^(a-1) = a/x
and h'(x) = a.f'(x) = a/x

then g'(x) = h'(x) for all x>0 and g(1) = h(1) => g(x) = h(x) for all x>0

so f(n) = lim(x->infinity) of f((1+1/x)^x)
= lim(x->infinity) of xf((1+1/x))
= lim(x->infinity) of f((1+1/x))/(1/x)
= lim(x->infinity) of [f((1+1/x))-f(1)]/(1/x) {f(1) = 0 for f(x) = e^x)}
= lim(h->0) of [f(1+h)-f(1)]/h
= f'(1)
= 1/1
=> f(n) = 1 = ln(e) => n = e
[/Edit]
This is the only way I could think of it, but not sure if it's right: [edit]I'm positive it's right[/edit]

11. Solution:

lim(x->0) of x^x
= lim(x->0) of e^(ln (x^x)) [e^(ln(x)) = x]
= lim(x->0) of e^(xln(x)) [ln(a^b) = bln(a)]
= lim(x->0) of e^x [x > ln(x)]
= 1

Edit: The second proof was based on an equation that I mixed up between ln(ab) = ln a + ln b and e^(ab). My bad.
@mesa #11 and 15 correct :D

Edit:
[e^(ab) = e^a + e^b]
since when?
try e^(6) - (e^2 + e^3) on your calculator it doesn't give you 0
it either (e^a)^b =e^ab or e^(a+b) =e^a * e^b

the first proof is correct