October 5th, 2010 at 7:55:43 AM
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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.

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!

October 5th, 2010 at 9:22:56 AM
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Just bedween you and me, it seems simple.

1=1

2=2

.

.

.

.

.

50=50

1=1

2=2

.

.

.

.

.

50=50

October 5th, 2010 at 9:28:56 AM
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Where have you used the numbers 51 through 100?Quote:dmJust bedween you and me, it seems simple.

1=1

2=2

.

.

.

.

.

50=50

October 5th, 2010 at 9:34:17 AM
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If you're going to get technical, I withdraw my answer.

October 5th, 2010 at 9:40:34 AM
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Quote:dmIf you're going to get technical, I withdraw my answer.

hehe

When I die, I want everyone who ever worked with me on a group project to lower me into my grave so they can let me down one last time.

October 5th, 2010 at 9:44:12 AM
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Quote:DocWhere 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

October 5th, 2010 at 9:45:59 AM
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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

1=1

2=2

.

.

24=24

26=26

27=27

.

.

50=50

100-99+98-97+96............+52-51=25

October 5th, 2010 at 10:06:29 AM
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Quote:MathExtremistOkay 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

October 5th, 2010 at 10:07:53 AM
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dm - you did not use 25 in any of your equations.

October 5th, 2010 at 10:18:20 AM
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Quote:dmQuote:MathExtremistOkay 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