Dween
Joined: Jan 24, 2010
• Posts: 339
October 5th, 2010 at 7:55:43 AM permalink
Last night, I had a dream about a puzzle. The rules to the puzzle were vivid enough that I remembered it the next morning. I haven't tried to solve it, and frankly don't even know if it has a solution, or if it would be "fun" to do. I just wanted to post it before it left my brain.

Using all of the integers 1 through 100 inclusive, can you construct 50 mathematical equations to give answers 1 through 50?
The numbers 1-100 should each be used only once. You may not repeat a number within an equation, or in the entire puzzle, save for the answers of the 50 equations.

In the dream, there was an optional rule: Equations could be a single number if you'd like. (ie 53 = 53)
If this rule was not in effect, that would mean each equation would use exactly two numbers.

Basic math functions are allowed. Add, subtract, multiply, divide. Use square roots and exponents if necessary.

Can this be solved?
Can you solve it with just plus, minus, times and divide?
Can it be solved with pairs, ie no single number equations?

I'll see if I can come up with a solution as well, but wanted to give some time for the forum members to tackle it.
-Dween!
dm
Joined: Apr 29, 2010
• Posts: 699
October 5th, 2010 at 9:22:56 AM permalink
Just bedween you and me, it seems simple.

1=1
2=2
.
.
.
.
.
50=50
Doc
Joined: Feb 27, 2010
• Posts: 6873
October 5th, 2010 at 9:28:56 AM permalink
Quote: dm

Just bedween you and me, it seems simple.

1=1
2=2
.
.
.
.
.
50=50

Where have you used the numbers 51 through 100?
dm
Joined: Apr 29, 2010
• Posts: 699
October 5th, 2010 at 9:34:17 AM permalink
If you're going to get technical, I withdraw my answer.
Ayecarumba
Joined: Nov 17, 2009
• Posts: 6497
October 5th, 2010 at 9:40:34 AM permalink
Quote: dm

If you're going to get technical, I withdraw my answer.

hehe
Simplicity is the ultimate sophistication - Leonardo da Vinci
MathExtremist
Joined: Aug 31, 2010
• Posts: 6526
October 5th, 2010 at 9:44:12 AM permalink
Quote: Doc

Where have you used the numbers 51 through 100?

Okay then - use the answer above, except the last one:

50 + SUM[53..98]*((100+51)-(99+52)) = 50
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
dm
Joined: Apr 29, 2010
• Posts: 699
October 5th, 2010 at 9:45:59 AM permalink
BUT, I resubmit a NEW one.

1=1
2=2
.
.
24=24
26=26
27=27
.
.
50=50
100-99+98-97+96............+52-51=25
dm
Joined: Apr 29, 2010
• Posts: 699
October 5th, 2010 at 10:06:29 AM permalink
Quote: MathExtremist

Okay then - use the answer above, except the last one:

50 + SUM[53..98]*((100+51)-(99+52)) = 50

I don't think that is allowable-used 50 twice(X+Y=X) and not X=X
WestSider
Joined: Jun 18, 2010
• Posts: 4
October 5th, 2010 at 10:07:53 AM permalink
dm - you did not use 25 in any of your equations.
MathExtremist
Joined: Aug 31, 2010
• Posts: 6526
October 5th, 2010 at 10:18:20 AM permalink
Quote: dm

Quote: MathExtremist

Okay then - use the answer above, except the last one:

50 + SUM[53..98]*((100+51)-(99+52)) = 50

I don't think that is allowable-used 50 twice(X+Y=X) and not X=X

You make the rules, so you get to say what's allowable. The equation reduces to 50 + 0, which does indeed equal 50.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563