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Form two numbers. The sum of them must be $100$.

For the first one you must use the digits $4$, $5$ and $6$ taken exactly once in this order. You cannot use other digits.

For the second one you must use the digits $7$, $8$ and $9$ taken exactly once in this order. You cannot use other digits.

The operations you may use for forming each of the numbers are $x + y$, $x - y$, $x \times y$ and $x \div y$. Division is math division, you cannot use computer integer division where 4/5 = 0. Operator precedence is respected. Each operation may be used multiple times

You cannot use other symbols (except $456789+-\times\div)$


Example:

  • 1st number: $4\times5-6$
  • 2nd number: $7+89$

Sum is wrong $14+96=110 \neq 100$, but everything else is ok.

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  • 1
    $\begingroup$ the titles holds no clues. It's just fun (or lack of inspiration, take it as you will). It alludes to the fact that the digits $2$ and $3$ are the only ones not mentioned in the question. $\endgroup$ – bolov Sep 12 '18 at 0:14
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$ 45_{6} = 29 $

$ 78_{9} = 71 $

$ 29 + 71 = 100 $

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    $\begingroup$ Excellent! Probably the expected answer $\endgroup$ – xhienne Sep 13 '18 at 8:32
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    $\begingroup$ Yes, that's my solution. Congratulations! $\endgroup$ – bolov Sep 13 '18 at 10:29
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Implementing lateral thinking

The question does not specify which base, $B$, we are in. Only that we are using the digits $4$, $5$, $6$, $7$, $8$ and $9$ so this means that $B \geq 10$. I will assume $B=12$ so that we additionally have the digits $a$ and $b$.

Then one answer is

$4 \times 5 \times 6 = a0$
$7+8+9 = 20$

$a0 + 20 = 100$

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  • $\begingroup$ not the solution I had in mind but it does respects all the requirements. As far as I am concerned it's correct and it's the best solution yet. $\endgroup$ – bolov Sep 12 '18 at 13:01
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That should work:

Number 1: $4 \times (-5 + 6) = 4$
Number 2: $7 + 89 = 96 $

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    $\begingroup$ Nice answer!! :D $\endgroup$ – El-Guest Sep 12 '18 at 1:50
  • $\begingroup$ Negation isn't in the list of allowed operations though. $\endgroup$ – Bass Sep 12 '18 at 4:04
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    $\begingroup$ From my understanding, negation and "(", ")" are not allowed. So this should be wrong. $\endgroup$ – npkllr Sep 12 '18 at 8:51
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    $\begingroup$ Good try. However, parenthesis aren't allowed because you can't use their symbols ( ). Also, negation even if it uses the allowed symbol $-$ isn't an operation you can use. $\endgroup$ – bolov Sep 12 '18 at 8:58
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    $\begingroup$ Sorry all, I didn't read the question carefully enough. $\endgroup$ – xhienne Sep 12 '18 at 9:08
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With a bit of lateral thinking:

1st number: $ 4 \times 5 - 6 = 14 $
2nd number: $ 7 - 8 - 9 = -10 $

$ 14 + (-10) = 4 $

Decimal to Binary: $ (4)_{10} = (100)_{2} $

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    $\begingroup$ I like your answer. A step in the right direction. And your answer confirms that the lateral thinking tag is appropriate. $\endgroup$ – bolov Sep 12 '18 at 9:05
  • $\begingroup$ Wait, so this is not your expected answer? $\endgroup$ – npkllr Sep 12 '18 at 9:08
  • $\begingroup$ no, it's not the answer I had in mind $\endgroup$ – bolov Sep 12 '18 at 9:08
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    $\begingroup$ Hmmm, ok. How about a small hint: Do we need multiple numeral systems for your solution? $\endgroup$ – npkllr Sep 12 '18 at 9:12
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With some brute force I can confirm what we thought already: we do need some lateral thinking to solve this challenge!

In bash, try all combinations of operators (+, -, * and / and the concatenate character which is an empty string: ' '). No negation, no brackets.

Then print the ones where the outcome is between 95 and 105:

for a in + - \* / ''; do
    for b in + - \* / ''; do
        for c in + - \* / ''; do
            for d in + - \* / ''; do
                echo 4${a}5${b}6 + 7${c}8${d}9 = $(echo "scale=3 ; 4${a}5${b}6 + 7${c}8${d}9" | bc)
            done
        done
    done
done | awk '$NF>95 && $NF<105,1'

Output:

4+5+6 + 78+9 = 102
4+5-6 + 7+89 = 99
4+5*6 + 7*8+9 = 99
4+5*6 + 78-9 = 103
4+5/6 + 7+89 = 100.833
4-5+6 + 7+89 = 101
4-5/6 + 7+89 = 99.167
4*5-6 + 78+9 = 101
4*5/6 + 7+89 = 99.333
4/5+6 + 7+89 = 102.800
4/5*6 + 7+89 = 100.800
4/5/6 + 7+89 = 96.133
4/56 + 7+89 = 96.071
45+6 + 7*8-9 = 98
45-6 + 7*8+9 = 104
45/6 + 7+89 = 103.500

So no luck here yet!

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  • $\begingroup$ Also, $(4\times 5 + 6)+(78-9)=95$ is another close one :D $\endgroup$ – Feeds Sep 12 '18 at 11:48
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Similar to npkllr's answer

$- 4 + 5 - 6 = - 5$

$78 - 9 = 69$

$-5 + 69 = 64$

$DEC(64) = OCT(100)$

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  • $\begingroup$ Negation isn’t allowed $\endgroup$ – DonielF Sep 12 '18 at 13:56
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I doubt this is what you're looking for, and is really stretching the limits of interpreting what you said, but:

1st number: 4×5−6 2nd number: 7+89

Now, only using the allowed digits, the sum of 4 (1 is not an allowed symbol and therefore 14 becomes 4) and 96 is 100

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-1
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4*5/6 = 3.33
7+89 = 96

3.33 + 96 ≈ 100

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  • $\begingroup$ Welcome to puzzling.SE! Unfortunately, the result must be exactly 100 (moreover 99.333 ≈ 99, not 100). $\endgroup$ – xhienne Sep 12 '18 at 13:30
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    $\begingroup$ @xhienne if ≈ stands for "approximately equal to", or "almost equal to" why is 99.333 ≈ 100 wrong? its still almost equal to the 99.333. ≈ does not mean round down. And even so, i could also go with 99 ≈ 100. Regardless of wether the sum is exactly 100. $\endgroup$ – Ano Sep 13 '18 at 6:16

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