Math Problems That Equal 100

Question 1

$$ iii 3 \cdot 3 + 3 - 3 + 3 / iii = 100. $$

$$iii\cdot3\cdot3\cdot3 + iii\cdot3\cdot3 - 3\cdot3 + 3/three = 100.$$

Oleg567Oleg567 16.8k ii gold badges 36 silverish badges 74 bronze badges · Answer · 2016-10-04 20:23:48Z

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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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Question 2

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.

Question 3

For another variation non using "adjoined" 3s:

$$iii \times three \times 3 \times 3 + 3 \times 3 + 3 \times 3 + 3 - 3 \div 3 - 3 \div iii = 81 + ix + 9 + 3 - ane - 1 = 100$$

dxivdxiv 70.8k 6 aureate badges lx silvery badges 115 bronze badges · Answer · 2016-10-04 05:35:07Z

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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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Question 4

Is there a limit on the number of times you lot can use the performance?

Otherwise it seems y'all could do

$$iii\times 33 + 3\div 3 +3\div 3 - 3\div 3$$

Question 5

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.

Answer · 2016-10-04 05:53:01Z

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.

Question 6

$$3333/33-33/three+3\cdot 3+iii/3=\\101-11+9+1=100$$

If one is immune to skip any of the 4 operations we tin do

$$33\cdot 3 + 3/3=99+one=100$$

mathreadlermathreadler 24.4k 9 gold badges 33 argent badges 83 bronze badges · Answer · 2016-10-04 20:39:15Z

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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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Answer 2 · 2016-10-04 05:15:18Z

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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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0

Answer 3 · 2016-10-04 05:16:09Z

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.

SquirtleSquirtle half-dozen,380 28 silver badges 57 statuary badges · Answer 5 · 2016-10-04 05:19:12Z

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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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one

$\begingroup$ 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