# Two Nice Problems

#### NewScientist

The first problem is :Use the integers 1, 3, 4 and 6 each once to make the number 24. You may use the operators + - x / (addition, subtraction, multiplicaion and division).

The second problem is to find fractions such that

[a/(10b+c)] + [d/(10e+f)] + [g/(10h+j)] = 1

Where a,b,c,d,e,f,h and j are the digits 1 to 9. You may only use the digits once.

both of these are fun and so I thought I'd share them!

-NewScientist

#### eNathan

idk if thats posible, I would like to see someome do the former problem.

#### NewScientist

Its possible and I've pm'd you the solution.

#### uart

Before I invest any effort in thinking about it, is the first one possible with standard operator precedence and no grouping symbols like brackets and "fraction bar" etc ?

Also must the 1,3,4 and 6 be used as individual integers or can they be combined as digits of a larger integer like 13 or 41 etc, or even 1.3 4.1 etc if decimal points are allowed. Can you rule out all these types of "tricks" for us?

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#### NewScientist

Yes sure, brackets maybe used although anything in brackets can be done as steps in a solution! A Fraction bar may be used as it is a division.

The system used is decimal and the digits given MUST be used as integers.

so - fraction bars and brackets are fine, but squaring etc is banned!! The numeric system is decimal and decimal points are not allowed unless you create them - for example 6/4 = 1.5

-NewScientist

(note that this problem took me about an hour or two and i'm normally quite sharp with these!)

6/(1 - 3/4)

Does that count?

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#### devious_

The second one seems impossible!

#### NewScientist

Yes that solution is correct, and the second one is not impossible - perhaps you could remove your post or edit it so that the text is white and not LaTeX so people will not be forced to see the solution!!

well done!

-NewScientist!

#### wisredz

Nope, he says you have to use the numbers in their original form, you cannot make any other numbers using 1-3-4-6

#### NewScientist

he's fine i've explained why in the PM

#### JonF

i got the first one, the second one is beyond me

#### NewScientist

HINT For the second one is below

Take 3 fractions that sum to 1 and then turn them into equivalent fractions and so on.

Here is one of the fractions:

One of the 3 fractions is 9/12.

-NewScientist

#### rachmaninoff

The 2nd one - the unique answer is

edit: how do you make TeX invisible? the [white] tags don't work...

anyway, the answer is 5/34 + 7/68 + 9/12 = 1

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#### NewScientist

once again don't show it so every one can see it - do it in plain tex tand in white!! stop ruining it!! :P! please!

-NewScientist

#### rachmaninoff

Here's a brain teaser: prove that that answer is unique.

#### NewScientist

Let us theorise that the answer is unique. Therefore we have a bit of fun - find a counterexample and disprove the theory, or prove it algebraically. For the latter i would start by defining every number as 1 + n. Also, I would definitely use (y/10n + p) to express the fractions.

I don't think that a proof is that hard. As you can think about the relationship between[1/(10n+p)] and another number [1/(10a+b)], before adding varying numerators. just a thought.

-NewScientist,

And rachmaninoff - as much as I love your music the way to get rid of the solution is to write it in plain text, eg. [a/(10b+c)] + ..... and then put it into white.

#### fourier jr

has anyone tried bachet's problem? imagine a balance with 2 pans & you have to figure out how to weigh out 1lb-40lbs inclusive using least number of weights as possible. he found that a sequence of weights weighing 1, 2, 4, 8, 16, 32lbs weights worked, & so did weights of 1, 3, 9, 27lbs worked also. the 1st sequence works if only one scale pan is used, and the 2nd sequence works if both scale pans are used (so you can have 9+1 or 9-1, etc)

there was another guy called major macmahon who tried to figure out a harder problem, which would determine all possible sets of weights which would enable us to weigh out any amount of pounds from 1 to n inclusive.

it turns out to be a kind of interesting problem involving polynomials & factoring them in different ways. sorry if i hijacked the thread, i thought this was a kind of interesting related puzzle.

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