The equation 6 × 5 4 ÷ 2 3 + 7)^0 1)^2 = 6 is missing two opening

Question

The equation  6 × 5 – 4 ÷ 2 – 3 + 7)^0– 1)^2 = 6  is missing two opening parentheses. It can be made true if opening parentheses are placed immediately before each of two of the numbers.

Assume that standard order-of-operations principles are followed, whereby multiplication and division are performed before addition and subtraction, unless parentheses indicate otherwise. Select in the table a number before which the first opening parenthesis could be placed and a number before which the second opening parenthesis could be placed, such that the equation would express a true statement. Make only two selections, one in each column.

ID: 700184

KarishmaB · Accepted Answer

Hit and Trial in a question like this will take a long time. We need to look for some logic. On the RHS we have 6 and the LHS ends with a square. We can certainly not have an opening bracket at the beginning since no integer squared will give 6.
Then notice that the first term is 6 and then there is a multiplication sign. So it made sense that 6 is multiplied by 1 and hence the entire thing squared is the square of 1

6 ×  (5 – 4 ÷ 2 – 3 + 7)^0– 1)^2  = 6

The highlighted is likely all 1. This means that 5 – 4 ÷ 2 – 3 + 7)^0 should be 2.

5 – 4 ÷ 2 – 3 + 7)^0 = 2

The first number is 5 so we need some small numbers to be subtracted out of it to give 2. We have a 0 as exponent. It should take care of some of the numbers. 4 ÷ 2 = 2 so we should not subtract out another 3. Hence (3+7) should be made 1 by the exponent.

5 – 4 ÷ 2 – (3 + 7)^0 = 5 - 2 - 1 = 2  Valid

Hence the parentheses are before 5 and before 3. ANSWER

chetan2u · Answer

A new type of question in DI. Purely Logical.

6×5–4÷2–3+7)^0–1)^2=6 
Now, If the entire term is under SQUARE, then the right side has to be a square integer. But 6 is not a square.
Thus,  (   6×5–4÷2–3+7)^0–1)^2=6  is not possible.

Next shift the first parenthesis to one term right to get   6×(5–4÷2–3+7)^0–1)^2=6 
  (5–4÷2–3+7)^0–1)^2=1 
This would be ok as 1 is a square.

We get two solutions
(1)   (5–4÷2–3+7)^0–1)^2=1........5-4÷2–3+7)^0–1=1.........5-4÷2–3+7)^0=2 
Now, power of 0 will make entire thing as 1. Also 4÷2 should be together, otherwise we are likely to get a fraction.
 5-2-3+7)^0=2 
Let us shift '(' towards right.
 5-2-3+7)^0=2  =>  (5-2-3+7)^0=2  =>  1=2 ....No
 5-2-3+7)^0=2  =>  5-(2-3+7)^0=2  =>  5-1=2 ....No
 5-2-3+7)^0=2  =>  5-2-(3+7)^0=2  =>  5-2-1=2 ....Yes

Thus the equation is   6×(5–4÷2–(3+7)^0–1)^2=6

Answer
First parenthesis - Before 5
Second parenthesis - Before 3

Sans8

srishti_6 · Answer

According to general mathematics principles the externalmost bracket is known as the 2nd or 3rd bracket and the internal bracket is called the 1st bracket, I got the answer correct but messed up the order. Can someone please explain the logic behind this?

Sans8 · Answer

chetan2u   KarishmaB   Bunuel  Pls explain the logic behind this?

KarishmaB · Answer

I am not sure what the problem is here. We solve the inner bracket first and the other bracket later.

So first solve highlighted. 
6 × (5 – 4 ÷ 2 – ( 3 + 7 )^0– 1)^2 = 6

I get 
 6 × (5 – 4 ÷ 2 – (10)^0– 1)^2 = 6

6 × (5 – 4 ÷ 2 – 1 – 1)^2 = 6

Next solve highlighted using BODMAS/PEMDAS

6 × ( 5 – 4 ÷ 2 – 1 – 1 )^2 = 6

6 × (5 – 2 – 1 – 1)^2 = 6

6 × (1)^2 = 6

6 = 6

GMATGuruNY · Answer

6×5–4÷2–3+7)^0–1)^2=6

We can PLUG IN THE ANSWERS.

If opening parentheses are placed before 6, we get:
 (6×5–4÷2–3+7)^0–1)^2=6 
Taking the square root of both sides, we get:
 6×5–4÷2–3+7)^0–1=\sqrt{6} 
Not viable.

If opening parentheses are placed before 5, we get:
 6×(5–4÷2–3+7)^0–1)^2=6 
Dividing both sides by 6, we get:
 (5–4÷2–3+7)^0–1)^2=1 
Taking the square root of both sides, we get:
 5–4÷2–3+7)^0–1=1 
 5–4÷2–3+7)^0=2 
This seems viable.

Testing the remaining answer choices:
Before 4 -->  5–(4÷2–3+7)^0=2  --> 5-1=2 --> Nope
Before 2 -->  5-4÷(2–3+7)^0=2  --> 5-4=2 --> Nope
Before 3 -->  5–4÷2–(3+7)^0=2  --> 5-2-1 = 2 --> Success!

Answer: Before 5 and 3

srishti_6 · Answer

KarishmaB  Thank you for your reply. My sincere apologies I believe my question could have been framed better. My problem is not with which bracket to solve first. In the question stem they are asking that the "first" parenthesis should be placed immediately before which number, and as you mentioned innermost bracket should be solved first, so I inferred that the innermost bracket (which, as per me is the first bracket) which is before 3 should be the answer of the first column. I thus messed up the order of the answers and I am unable to understand the logic.

chetan2u · Answer

Sans8  and   srishti_6 ,

We are talking of two  opening  parenthesis, so what would they be like   (  ( .
Next we look for first  opening  parenthesis. It would surely be red colored one,   (,   between   (   and  (.

I hope it settles the query.

Dbrunik · Answer

how do you even begin a problem like this?

Shivang29 · Answer

This was my thought process,

knowing that we don't have a way to bring sqrt(6) into the picture, we know that the first paratheses was not going in front of 6.

Next I looked at the expression right before ^0, knowing that anything to the power of 0 we were going to get 1 and seeing that right after we are subtracting from 1 the only way we were going get 6 was if the expression right before ^0 was going to be 2. From here you can trial and error to find that the 2nd bracket should go before the 3 but the way I thought about it was that 5-2-3+7 would equal 1 and we were trying to get 2. so I had to find a way to go from 5-2 =3 -3 + 7 to 3-1 which could only happen if (3+7) were to somehow become 1, from there if we put the bracket in front of 3 we would have the whole expression 5 -2-3 +7 equal 2 and we could put a bracket in front of 5 to get the whole expression to equal 1.

Hope this made some sense.

Adit_ · Answer

As soon as one sees the squared term, it needs to be of the form likely 6*1 when u also notice 6 in the far left. This is likely going to be the case as the factors of 6 are 2,3,6 and here we also notice none of 2,3,6 arent square number but 1 is! Therefore we have to reduce the right side number to 1 somehow and now moving to inner bracket: We have to do a bit of trial and error atleast thats what I did added to the fact that 4/2 gives 2 and we have a 5 subtracing another 1 more and i get to 2! So 3+7 raised to 0 pushed this value to 1 and when u subtract the entire right portion you indeed land up with 1. Hence 6*1=6

Dbrunik

how do you even begin a problem like this?

Maverick6 · Answer

Could we put both opening parenthesis before 5? The equation still seems to work but I am not sure if this restricted by the clause "before  each   of two of the numbers "?

chetan2u · Answer

If you place both opening parenthesis before 5, the equation will not be correct.
6×((5...)^0-1)=6×(1-1)=0 and not 6..

The equation 6 × 5 4 ÷ 2 3 + 7)^0 1)^2 = 6 is missing two opening

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First parenthesis immediately before	Second parenthesis immediately before
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GMAT Two-part Analysis (TPA) Questions

The equation 6 × 5 4 ÷ 2 3 + 7)^0 1)^2 = 6 is missing two opening

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My Rewards