Can you make this equation true?

$\Large 99 + 9 \div 9 = 99 + 1 = 100$

99- (9/9) = 100 is an answer, but following PEMDAS, the parenthesis is not needed, so we can rewrite the problem as 99- 9/9 = 100. Thus, the least amount of operators needed would be two ( the subtraction and division ).

$\huge 99+\dfrac99$

I know I'm going to get yelled at for this, but I think it's funny I got the correct answer incorrectly.

I understand you are able to CONCATENATE, but I didn't realize you are not able to UNCONCATENATE (i.e. Split the 100 into 1 00.

My solution was:

99 - 99 = 10 * 0

I know it's incorrect. I just think it's interesting.

$99+\frac{9}{9}$ HAHAAAHH!!!

9÷9+99=100

99+(9-9)! =100 ergo god

99+9/9=99+1=100

[99 is combined as written that you can combine]

9/9+99=100. 9/9=1,1+99=100. Easy Peeasy Lemon Squezy. im 11 years old by the way...

99+9/9=100

(999 - 99 ) : 9 = 100

99 + 9 ÷ 9 = 100

99+(9-9)! = 100

99+9/9=100 (2 operations) It cannot be one as none of the possible solutions are 100;
999+9=1008
999-9=990
999/9=111
999x9=8991
99+99=198
99-99=0
99/99=1
99x99=9801
9-999=-990
9/999=1/111

9 9 - 9 9 = 1 x 0 0? That'd be two then? :)

( 99 - 9 ) / .9

The answer is 2. => 99+9/9

99+(9/9) - 2 operators

99+1=100

an easier solution is 999 ÷9= 100 so it would be 1

99 + 9/9 = 99 +1 = 100 using 2 signs.

((9/9) +9)*((9/9) +9)= 10 *10 =100

9 divided by 9 = 1 + 99 = 100

9/9+9 9=1+99=100 by PEMDAS

99+9/9 =100 that is two operations.

9/9=1+99=100 so the operator is 2

9/9 + 99 = 100

99 + 9/9 =100

99 + (9 / 9) = 99 +1 = 100

This would be simpler if we just address the problem as if it were code in Java. Ex: 9 9 9 9 = 100 we could solve this quite easily if we concatenate the digits into 9999 and used only one operator. 9999 != 100. Simple.

9/9 = 1

1+99 = 100

2 operations.

99+9/9 = 100

First, we could do 9/9 on the two left-most nines to get 1. Then we add with the remaining concatenated two nines (99). 1 + 99 = 100. In other words, 9/9 + 9 9 =100

9/9+99=100

9÷9 +99=100. Easier expression as the division is first in the equation and there was no need to change to a fraction.

9/9 + 99 = 100

two operators used

99+9/9=100

99+9/9 = 99+1 = 100

99+(9/9)=99+1=100

Written in the question: With your answer, define N as the least amount of mathematical operators used. Find N.

I think, the value of N is not 'least', it is 'only'.

PEMDAS = Please excuse my dear aunt Sally. BODMAS = Billy only does math after supper ??? Or is there already a better memory aid?

$99+$ $\frac{9}{9}$ $=100$

99+9/9= 100

=99+9/9 =99+1 =100