Math Problems That Equal 100
vii Answers 7
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$$ iii 3 \cdot 3 + 3 - 3 + 3 / iii = 100. $$
$$iii\cdot3\cdot3\cdot3 + iii\cdot3\cdot3 - 3\cdot3 + 3/three = 100.$$
answered October 4, 2016 at xx:23
Oleg567Oleg567
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Well, conspicuously $(333-33)\div iii = 100$.
Now, you simply have to write something equivalent that does non use parenthesis. Then effigy a mode to use add-on.
answered Oct iv, 2016 at v:xv
Graham KempGraham Kemp
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You can notation that $\frac 33=1$, and so you tin can add or subtract any number you want by adding or subtracting $\frac 33$. We probably want to make things bigger, and then choose improver. At present we just need to use up multiplication and subtraction. And then $3 \times 3 -3=6$ and add $94$ terms of $\frac 33$ to be done.
answered Oct 4, 2016 at 5:16
Ross MillikanRoss Millikan
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For another variation non using "adjoined" 3s:
$$iii \times three \times 3 \times 3 + 3 \times 3 + 3 \times 3 + 3 - 3 \div 3 - 3 \div iii = 81 + ix + 9 + 3 - ane - 1 = 100$$
answered October 4, 2016 at 5:35
dxivdxiv
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Is there a limit on the number of times you lot can use the performance?
Otherwise it seems y'all could do
$$iii\times 33 + 3\div 3 +3\div 3 - 3\div 3$$
answered Oct 4, 2016 at 5:nineteen
SquirtleSquirtle
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If we tin employ addition and segmentation more ane times then it can be solved easily.
By using division iii times and addition two times we tin can write expression,
$$3÷three+iii÷3+33×three-3÷3$$ At present using order of functioning(PEMDAS) solve expression.
$$one+1+99-1$$ $$=100$$ It can besides solved past various types, like
$$33×three+\dfrac{3+3}{3}-three÷3$$ $$99+\dfrac{3+3}{three}-one$$ $$99+2-1$$ $$=100$$ I hope information technology will aid you.
answered Oct iv, 2016 at v:53
Deepak SuwalkaDeepak Suwalka
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$$3333/33-33/three+3\cdot 3+iii/3=\\101-11+9+1=100$$
If one is immune to skip any of the 4 operations we tin do
$$33\cdot 3 + 3/3=99+one=100$$
answered Oct 4, 2016 at 20:39
mathreadlermathreadler
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Math Problems That Equal 100,
Source: https://math.stackexchange.com/questions/1953035/to-write-100-from-a-bunch-of-threes
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$\begingroup$ This is the same every bit $(iii\times 33) + (3\div iii) + (three\div 3) - (3\div 3) = 99+1+1-i=100$ $\endgroup$
Oct four, 2016 at v:20