13 of 100: Operator Search

$10\; \div\; 10\; +\; 10\; \times\; 10=101$

Explanation: See Order of Operations :

1)Since $\times$ and $\div$ come same in precedence we have to judge by associativity which is left to right therefore $\underbrace{10\; \div\; 10\; }_1+\; 10\; \times\; 10$

2) Since precedence of $\times$ is higher over $+$ , next step would be : $1\; +\underbrace{\; 10\; \times\; 10}_{100}$

3) Lastly, simple addition operation will give : $1\; +\; 100\;=\boxed{\color{#D61F06}{101}}$

NOTE:- I just involved precedence and associativity to make solution clearer.

$\begin{array}{c|c|c|c|c|c} \text{Precedence} & \text{Associativity}\\\hline 1) \text{Multiplication, Division} & \text{Left}\to \text{Right}\\\hline 2) \text{Addition, Subtraction} & \text{Left}\to \text{Right}\\\hline \end{array}$

10*10+10/10=101​

Alternatively, $10 \div 10 + 10 \times 10 = 101$ .

Funny how there's an inclination to put the multiplication before the division.

Bonus question

My answer is 4.

Difficult follow-up bonus (assuming, that brackets are possible)

I wrote program on python:

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from itertools import zip_longest
operations = '+-*/'
def exeval(s):
    try:
        res = eval(s)
    except ZeroDivisionError:
        res = 0
    return res
def brackets(s): # braces('(a+a)+(a+a)')('/*-') → '(10/10)*(10-10)'
    return lambda x: ''.join([b+o for (b,o) in zip_longest(s.replace('a','10').split('+'),x,fillvalue='')])
# all combinations of operations of length 3
opses = [op1 + op2 + op3 for op1 in operations for op2 in operations for op3 in operations]
a = list(map(brackets('((a+a)+a)+a'),opses))
b = list(map(brackets('a+(a+(a+a))'),opses))
c = list(map(brackets('(a+a)+(a+a)'),opses))
res = sorted(list(set([exeval(x) for x in a + b + c])))
print(res)
print(len(res))

If brackets are impossible, then use this code

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# …
d = map(brackets('a+a+a+a'),opses)
res = sorted(list(set([exeval(x) for x in d])))

Output was (corrected float values and highlighted if value can be reached without brackets, there are 27 of these):

[ -990 , -900, -190, -100 , -99 , -90, -80 , -20 , -19 , -10, -9.9 , -8, -1 , -0.9, -1/9, 0 , 0.01 , 0.05, 1/11, 0.2, 1/3, 10/11, 1 , 1.1, 10/9, 2 , 3, 5, 9, 9.5, 9.9 , 10, 10.1 , 10.5, 11, 12, 19 , 20 , 21 , 40 , 80 , 90, 99 , 100 , 101 , 110, 120 , 190, 200 , 210, 300, 400, 900, 990 , 1010 , 1100, 2000, 10000 ] 58

Answer: 58 different numbers possible to make with four 10's and three arithmetic operations. (And only 27 possible if we refuse brackets)

Please, if my solution is wrong, say it!

Same solution, just a simple thought process on how to get to the solution: Because 101 is odd, one needs to do a divide to get an odd number (all other operations yield an even number). So try $10 \div 10 = 1$ . Then we only need 100 + 1 to get 101 and since $10 \times 10 = 100$ that does the trick. Therefore the answer is: $10 \div 10 + 10 \times 10 = 101$ or $10 \times 10 + 10 \div 10 = 101$ .

10*10+10/10=101

10 X 10+10/10=101

That’s pretty easy.at first we will make 100 by multiplying (10×10). Then we will make 1 by doing(10/10) .so

    10/10+10×10=1+100=101

Because 10×10 is 100, we know that all we need to do is add 1 to make a sum of 101. Because the only number is 10 in this equation, we know that we need to come up with a reasonable equation to find 1. This equation would be 10÷10 =1. All we need to do now is put an addition symbol in between to add these two equations together.

                                                     Summary:
                                                                                    10×10+10÷10=101

Answer- 10 ÷ 10 + 10 × 10 = 101. This is because: 10 ÷ 10 = 1; 10 × 10 = 100 [BODMAS - bracket, of, division, multiplication, addition, subtraction (order of operations)] ; 1 + 100 = 101

101 is 1+100 so you want one operation that will give you 1 and another that will give you 100. Then, you need a plus in between to add the 100 and 1.

10divided by10 plus10 times10= 1+100=101

10*10+10/10=101

10*10+10÷10=101

10×10+10÷10=101. 1)10×10=100; 2)10÷10=1; 3)100+1=101. Remember kids, multiplication and division come before addition and subtraction.

(10 x 10) + (10/10) = 101,

USE LOGIC!

10*10+10/10

Rishabh Cool, I totally don't understand your interpretation of the Order of Operations. The Third Directive clearly states, "multiplication and division, in order from left to right". To me, that says "multiplication on the left, division on the right". What does it say to you? I came up with the same answer but I wrote it the other way 'round: 10 \times 10 + 10 \frac 10 = 101.

What did I do wrong here? Why didn't LaTeX print out my operational symbols?

I was right only because the bonus question gave away the answer: if I'm replacing 101 with something that makes the question impossible, it follows that 101 has to be possible, even without actually doing anything. :/

As for the bonus and difficult bonus themselves, I cheated a bit and just brute forced it with Lua because the search space is tiny: there are only $\boxed{27}$ possible results made from using the 64 combinations of 3 basic operations: my screenshots are huge.

To actually answer the bonus question: $\boxed{3}$

10 / 10 + 10 * 10 = 1 + 100 = 101

By the use MDAS, you can actually take the answer "101" =10 x 10 + 10 / 10 =100+1 =101

10 divided by 10 = 1,then 10 times 10 =101

This equation needs brackets (10/10)+(10x10)

There is a loop hole :

Bonus question: Replace 101 with another number between 0 and 200 that makes this problem impossible.

Which means that there is an answer :-D

10*10 + 10/10 This simplifies into 100 + 1 = 101

You've got 10*10+10/10=101

10 × 10 + 10 ÷ 10 = 101

First step multiple:

(10 × 10) + 10 ÷ 10 = 101

100 + 10 ÷ 10 = 101

Second step divide:

100 + (10 ÷ 10) = 101

100 + 1 = 101

Third and Final step Add them:

100 + 1= 101

101 =101

10*10+10/10

can we repeat the arithmetic operations in bonus question?

$10 \div 10+10\times 10=101$ or $10\times 10+10 \div 10=101$ .

Do 10/10+10x10, which is 1+100=101

This is super simple!! Just look at the first to tens, and divide them. Answer is one. With the next two, 10*10=100. 100+1=101

10/10+10*10

10x10 + 10/10

You have to think about how to get the one, since that would appear to be the most difficult. Once you realize that 10/10 = 1, you can easily get 100 by multiplying 10 by 10, and you can add the two together to get 101. Hence, 10 x 10 + 10/10 is the solution.

10 x 10 + 10 / 10

10*10+10/10

(10x10)+(10/10)=101

10÷10+10×10=101

10/10+10*10=101

1st BRACKET, 2nd OF, 3rd ÷, 4th ×, 5th +, 6th -
10×10+10÷10=101

10 / 10 + 10 x 10 = 101

10*10+10/10

BODMAS rule state that first of all brackets will be solved, then division , then multiplication , the addition and at last subtraction. (However solving subtraction before addition doesn't lead to any change in the answer.) So now we have 101 which we know can be written as (ten times ten) + 1

Therefore using logical reasoning we got the answer to be 10 × 10 + 10 ÷ 10.

You can better understand it if I write (10×10)+(10÷10).. But both are same things..

10x10+10÷10=101

Its 10×10+10÷10 which is equal to 101.

10 x 10 + 10 / 10 = 101

$10\quad \times \quad 10\quad +\quad 10\quad \div \quad 10\quad =\quad 101$

It is recommended to revise BODMAS (operator precedence) rules if you think this solution is wrong ;)