Math Puzzles *UPDATED*

Vertimus

Banned
Apr 2, 2004
1,441
0
0
April 29th:

1. [easy] Find a 10-digit number so that the first digit is divisible by 1, the first two digits are divisible by 2, the first three digits are divisible by 3, and so on. Solved by JujuFish
2. [medium] Find the smallest integer k such that 1^2+2^2+3^2+......+k^2 is divisible by 200. Solved by chuckywang
3. [hard] Two positive integers differ by 60. The sum of their square roots is the square root of an integer that is not a perfect square. What is the maximum possible sum of the two integers? Solved by chuckywang
4. [bonus] The equation 26+62=126 is FALSE. However, if one moves one of the digits to another place, the equation becomes true. Find the move. (Note: Rotation, Translating, is allowed. Reflection, i.e. flipping, is NOT allowed).Solved by dullard
5. [easy] Let N be the greatest integer multiple of 8, no two of whose digits are the same. Find N. Solved by JujuFish
*6. [medium] Circles A, B, and C are externally tangent to each other and internally tangent to circle D. Circles B and C are congruent. Circle A has radius 1 and passes through the center of D. What is the radius of circle B?
7. [hard] Given that log(sin x)+log(cos x)=-1 and that log(sin x+cos x)=1/2*(log(n) -1), find n. (We are assuming base 10.) Solved by dullard
8. [easy] Let N be the largest positive integer with the following property: reading from left to right, each pair of consecutive digits of N forms a perfect square. What are the leftmost three digits of N? Solved by chuckywang
9. [medium] Two distinct, real, infinite geometric series each have a sum of 1 and have the same second term. The third term of one of the series is 1/8. Find the second term of both series. Solved by chuckywang
*10. [hard] How many positive integer divisors of 2004^(2004) are divisible by exactly 2004 positive integers?
11. [hard] How many positive integers less than 10,000 have at most two different digits? Solved by sao123
*12. [easy] Find the sum of all positive two-digit integers that are divisible by each of their digits.
*13. [medium] Jane is 25 years old. Dick is older than Jane. In n years, where n is a positive integer, Dick's age and Jane's age will both be two-digit numbers and will have the property that Jane's age is obtained by interchanging the digits of Dick's age. Let d be Dick's present age. How many ordered pairs of positive integers (d,n) are possible?
*14. [hard] The perimeter of triangle APM is 152, and angle PAM is a right angle. A circle of radius 19 with center O on AP is drawn so that it is tangent to AM and PM. Find the length of OP.
*15. [hard] Given that 1/(2!17!)+1/(3!16!)+1/(4!15!)+1/(5!14!)+1/(6!13!)+1/(7!12!)+1/(8!11!)+1/(9!10!) =N/(18!), find N.
*16. [hard x3] A long thin strip of paper is 1024 units in length, 1 unit in width, and is divided into 1024 unit squares. The paper is folded in half repeatedly. For the first fold, the right end of hte paper is folded over to coincide with and lie on top of the left end. The result is a 512 by 1 strip of double thickness. Next, the right end of the strip is folded over to coincide with and lie on top of hte left end, resulting in a 256 by 1 strip of quadruple thickness. This process is repeated 8 more times. How many of these squares lie below the square that was originally the 942nd square counting from the left?
*17. [hard x10] Find the least positive integer n such that 1/(sin45*sin46) + 1/(sin47*sin48) + .... + 1/(sin133*sin134)=1/sin(n) (Note: all the angles are in degrees.)


*=unclaimed problem

The first person to provide the answer to any of these problems will have their s/n edited in.

If you need clairification with a question, please speak up.

Good Luck! :p:p:p:p:beer:

Edit: From now on, any solution must have a short description along with it. I dont' need a proof, but a short explanation would work. Of course, hard problems should have longer explanations than easy problems.

Edit2: All these problems can be done without electronics, and should be done without electronics.. So, in your explanation, I shouldn't see methods that require massive amounts of computation.
 

Whitecloak

Diamond Member
May 4, 2001
6,074
2
0
Originally posted by: Vertimus
April 29th:



Medium:
Find the smallest integer k such that 1^2+2^2+3^2+......+k^2 is divisible by 200.



Edit: This is NOT my homework. I'm currently taking multi-var calc, which is nothing related to math puzzles.

400?
 

Vertimus

Banned
Apr 2, 2004
1,441
0
0
Originally posted by: whitecloak
Originally posted by: Vertimus
April 29th:



Medium:
Find the smallest integer k such that 1^2+2^2+3^2+......+k^2 is divisible by 200.



Edit: This is NOT my homework. I'm currently taking multi-var calc, which is nothing related to math puzzles.

400?

Nope.
 

Vertimus

Banned
Apr 2, 2004
1,441
0
0
Originally posted by: Vertimus
Originally posted by: whitecloak
Originally posted by: Vertimus
April 29th:



Medium:
Find the smallest integer k such that 1^2+2^2+3^2+......+k^2 is divisible by 200.



Edit: This is NOT my homework. I'm currently taking multi-var calc, which is nothing related to math puzzles.

400?

Nope.

It's smaller than 400.
 

KnickNut3

Platinum Member
Oct 1, 2001
2,382
0
0
Originally posted by: Vertimus
Hard:
Two positive integers differ by 60. The sum of their square roots is the square root of an integer that is not a perfect square. What is the maximum possible sum of the two integers?

If the number is not a perfect square, its square root is nonwhole, meaning the sum of the square roots of the two integers is nonwhole.

How does that restrict us in any way? Is the answer infinity? If not, it's worded poorly.
 

Vertimus

Banned
Apr 2, 2004
1,441
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Originally posted by: KnickNut3
Originally posted by: Vertimus
Hard:
Two positive integers differ by 60. The sum of their square roots is the square root of an integer that is not a perfect square. What is the maximum possible sum of the two integers?

If the number is not a perfect square, its square root is nonwhole, meaning the sum of the square roots of the two integers is nonwhole.

How does that restrict us in any way? Is the answer infinity? If not, it's worded poorly.

You're missing the point of the question. Try testing a few numbers, and you'll see why.
 

Vertimus

Banned
Apr 2, 2004
1,441
0
0
Originally posted by: chuckywang
Originally posted by: Vertimus
Originally posted by: chuckywang
Originally posted by: Vertimus
Originally posted by: chuckywang
1) This is so easy, I'm not going to even attempt it.
2) k=399, I think.
3) 100

2 and 3 are wrong.

Whoops. No. 2) k=112.

DINGDINGDING!!! correct! :thumbsup:

Prob. 3) 156

Yep!

LOL... no wonder you got so many of these... you go to imsa!

On the other hand, I think you should probably stop answering for a while... maybe give others a chance :p
 

chuckywang

Lifer
Jan 12, 2004
20,133
1
0
Originally posted by: Vertimus
Originally posted by: chuckywang
Originally posted by: Vertimus
Originally posted by: chuckywang
Originally posted by: Vertimus
Originally posted by: chuckywang
1) This is so easy, I'm not going to even attempt it.
2) k=399, I think.
3) 100

2 and 3 are wrong.

Whoops. No. 2) k=112.

DINGDINGDING!!! correct! :thumbsup:

Prob. 3) 156

Yep!

LOL... no wonder you got so many of these... you go to imsa!

On the other hand, I think you should probably stop answering for a while... maybe give others a chance :p

How did you know about imsa? I didn't think many people knew about it outside of IL.
 

Vertimus

Banned
Apr 2, 2004
1,441
0
0
Originally posted by: chuckywang
Originally posted by: Vertimus
Originally posted by: chuckywang
Originally posted by: Vertimus
Originally posted by: chuckywang
Originally posted by: Vertimus
Originally posted by: chuckywang
1) This is so easy, I'm not going to even attempt it.
2) k=399, I think.
3) 100

2 and 3 are wrong.

Whoops. No. 2) k=112.

DINGDINGDING!!! correct! :thumbsup:

Prob. 3) 156

Yep!

LOL... no wonder you got so many of these... you go to imsa!

On the other hand, I think you should probably stop answering for a while... maybe give others a chance :p

How did you know about imsa? I didn't think many people knew about it outside of IL.

Teh Noah sheets :p
 

dullard

Elite Member
May 21, 2001
25,983
4,592
126
Originally posted by: Vertimus
The equation 26+62=126 is FALSE. However, if one moves one of the digits to another place, the equation becomes true. Find the move. (Note: Rotation, Translating, is allowed. Reflection, i.e. flipping, is NOT allowed).
Got here too late for the others.

The problem quoted above, I've seen before in a different form:

The equation 26 + 62 = 129 is FALSE. However, if one moves one of the digits to another place, the equation becomes true.

Just throwing this out there for added fun when people are solving Vertimus's problem.

 

dullard

Elite Member
May 21, 2001
25,983
4,592
126
Hmm thought people would still be around to solve these. Move the 1 to the left and rotate it a bit so that it overlaps and crosses out the +62 portion of the equation. Then it becomes 26 = 26.
 

Vertimus

Banned
Apr 2, 2004
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Originally posted by: dullard
Hmm thought people would still be around to solve these. Move the 1 to the left and rotate it a bit so that it overlaps and crosses out the +62 portion of the equation. Then it becomes 26 = 26.

Well, you can do that, but there's a better way, one that involves arithmtic and not crossing out. I think it only works with 26+62-126 and not 26+62=129 though. Try finding it.
 

dullard

Elite Member
May 21, 2001
25,983
4,592
126
Originally posted by: Vertimus
Well, you can do that, but there's a better way, one that involves arithmtic and not crossing out.
Doh. I like the crossing out answers. I guess that means you won't accept moving the one onto the equal sign. Then it reads 26 + 62 does not equal 26.
 

Vertimus

Banned
Apr 2, 2004
1,441
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0
Originally posted by: dullard
Originally posted by: Vertimus
Well, you can do that, but there's a better way, one that involves arithmtic and not crossing out.
Doh. I like the crossing out answers. I guess that means you won't accept moving the one onto the equal sign. Then it reads 26 + 62 does not equal 26.

Yep, that's what i mean.