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555-605 (Medium)|   Number Properties|                                 
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Bunuel
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AbdurRakib

4^2 - 3^2 = 7
5^2 -4^2 = 9
6^2 -5^2 =11
100^2-99^2 = 199

The difference between the squares of consecutive integers always increases since a^2 -b^2 = (a+b)(a-b)

(a-b) will always be 1 since consecutive integers so as the integers increase a + b will also increase

What you can also figure out from this is that a+b = 25 for this problem

Therefore 2b +1 =25 and b =12, a=13

However you do not need to do this for a DS problem. Its sufficient to know that the difference is unique :)

Hope it is clear!
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Let x and y be 2 consecutive squares such that y>x.

Then root(y) = root (x) + 1

Now let's look at the question.

1) 94 + n and 69 + n are consecutive sqaures

x = 69 + n
y = 94+ n

root(y) = root (x) + 1
Squaring the above equation we get: y = × + 2root (×) +1

2root (×) = y - x - 1 = 94 + n - 69 - n - 1 = 24
Root (x) = 12

x= 144

n = 144 - 69 = 75

y= 94 + 75 = 169

Sufficient

2) 94 + n and 121 + n are consecutive squares.

Sufficient. Can be proven the same way as case 1.
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rishi02
If the positive integer n is added to each of the integers 69, 94, and 121, what is the value of n?

(1) 69 + n and 94 + n are the squares of two consecutive integers

Difference between the two squares is 25 since 94-69=25. This difference is unique.
For example 4^2 - 3^2 = 7
5^2 -4^2 = 9


As can be seen the difference goes on increasing and hence only one unique value is possible. SUFFICIENT

(2) 94 + n and 121 + n are the squares of two consecutive integers

Difference between the squares is 27. Again this difference is unique . SUFFICIENT.

(For those wondering what n is ;
n=75 and the consecutive integeres are 12, 13 & 14)

Interesting application.

Can you elaborate the highlighted Concept ?

Thanks
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Bunuel
If the positive integer n is added to each of the integers 69, 94, and 121, what is the value of n?

(1) 69 + n and 94 + n are the squares of two consecutive integers
(2) 94 + n and 121 + n are the squares of two consecutive integers

Statement 1. Let x and (x+1) be two consecutive integers. Then we have: 69+n=x^2 and 94+n=(x+1)^2. Substitute (69+n) into second equation to get 25+x^2=x^2 + 2x + 1 ==> 2x=24 and x=12 Hence n=75 Sufficient
Statement 2. The same as Statement 1. Sufficient
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Bunuel
If the positive integer n is added to each of the integers 69, 94, and 121, what is the value of n?

(1) 69 + n and 94 + n are the squares of two consecutive integers
(2) 94 + n and 121 + n are the squares of two consecutive integers

Statement 1: 69 + n and 94 + n are the squares of two consecutive integers
The difference between the numbers = 94 - 69 = 25
Let us list down some of the perfect squares.
Since 69 is near to 8^2, I will start from 8^2

64, 81, 100, 121, 144, 169, 196, 225.

Difference between 169 and 144 = 25
Hence 94 + n = 169, and 69 + n = 144

n = 75
SUFFICIENT

Statement 2: 94 + n and 121 + n are the squares of two consecutive integers
Difference between the two = 121 - 94 = 27
Applying the same logic and writing the perfect squares.

100, 121, 144, 169, 196, 225

Hence the numbers are 196 and 169
121 + n = 196 and 94 + n = 169
n = 75
SUFFICIENT

Correct Option: D
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AbdurRakib
rishi02
If the positive integer n is added to each of the integers 69, 94, and 121, what is the value of n?

(1) 69 + n and 94 + n are the squares of two consecutive integers

Difference between the two squares is 25 since 94-69=25. This difference is unique.
For example 4^2 - 3^2 = 7
5^2 -4^2 = 9


As can be seen the difference goes on increasing and hence only one unique value is possible. SUFFICIENT

(2) 94 + n and 121 + n are the squares of two consecutive integers

Difference between the squares is 27. Again this difference is unique . SUFFICIENT.

(For those wondering what n is ;
n=75 and the consecutive integeres are 12, 13 & 14)

Interesting application.

Can you elaborate the highlighted Concept ?

Thanks

the BIG IDEA here:
The difference between squares of two consecutive integers = Sum of the two consecutive integers
eg: \(10^2 - 9^2 = (10+9)(10-9) = 19\) so on and so forth

In Statement 1 we are told that (69+n) & (94+n) are the squares of two consecutive integers,
So use the above idea:
\((94+n)-(69+n) = 25\)
Since we know that the sum of the two consecutive integers is 25 & to find the individual consecutive integers: 25 = 2n+1 (since integers are consecutive)
n = 12 & (n+1) = 13
Now that we have each individual integer:
\(12^2 = (69+n)\)
\(144 = 69 + n\)
\(n = 75\)

Same applies for statement 2
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This is typical DS !

Hope this note helps !
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Bunuel
If the positive integer n is added to each of the integers 69, 94, and 121, what is the value of n?

(1) 69 + n and 94 + n are the squares of two consecutive integers
(2) 94 + n and 121 + n are the squares of two consecutive integers


Answer: Option D

Video solution by GMATinsight

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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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ScottTargetTestPrep
Bunuel
If the positive integer n is added to each of the integers 69, 94, and 121, what is the value of n?

(1) 69 + n and 94 + n are the squares of two consecutive integers
(2) 94 + n and 121 + n are the squares of two consecutive integers

We are given that the positive integer n is added to each of the integers 69, 94, and 121, and need to determine the value of n.

Statement One Alone:

69 + n and 94 + n are the squares of two consecutive integers.

From statement one, we can say that for some positive integer x, 69 + n = x^2 and 94 + n = (x + 1)^2. Let’s subtract the first equation from the second equation:

(94 + n) - (69 + n) = (x + 1)^2 - x^2
25 = x^2 + 2x + 1 - x^2
25 = 2x + 1
24 = 2x
12 = x

Since we know x = 12, we can substitute this into the first equation to determine the value of n:

69 + n = 12^2
69 + n = 144
n = 75

Statement one alone is sufficient to answer the question. Eliminate answer choices B, C and E.

Statement Two Alone:

94 + n and 121 + n are the squares of two consecutive integers.

We can use the same method that we used in statement one to solve for n. Therefore, without performing the actual calculations, we can conclude that we can find a unique value for n. Statement two alone is also sufficient to answer the question.

Answer: D

Quote:
Let’s subtract the first equation from the second equation:

ScottTargetTestPrep please could you help me understand what prompted you to subtract the first and second equations? I want to understand your thought process while attempting this question
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Hoozan
please could you help me understand what prompted you to subtract the first and second equations? I want to understand your thought process while attempting this question

We are trying to solve a system of equations consisting of 69 + n = x^2 and 94 + n = (x + 1)^2. We have two equations and two unknowns, so the natural thing to is to eliminate one of the variables. Subtracting the equations is one way of doing it, you could also write n = x^2 - 69 using the first equation and substitute this for n in the second equation:

94 + n = (x + 1)^2

94 + (x^2 - 69) = x^2 + 2x + 1

25 = 2x + 1

24 = 2x

12 = x

As you can see, we obtain the same result. So pretty much the only reason I subtracted the equations is so that one of the variables will be eliminated.
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OFFICIAL SOLUTION






-KT

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An excellent strategy on a problem like this that tests an unusual number property is to start with simple numbers and look for the pattern.

Here, let's square a few small consecutive integers and see what happens:

1^2 = 1
2^2 = 4 --- increase of 3
3^2 = 9 --- increase of 5
4^2 = 16 --- increase of 7

Every time you square the next integer, the distance from the last squared integer goes up by 2. This is the pattern.

Statement 1:

You can see that: (94 + n) - (69 + n) = 25

In other words, the second integer squared is an increase of 25 from the first squared integer. Since the pattern tells you that the increase goes up by 2 each time, you know that there is exactly one pair of consecutive integers whose squares have an increase of 25 from the first to the second.

You could find those integers if you wanted to (no need to do so though), so you'd be able to calculate n. Sufficient.

In fact, you didn't even need to calculate the 25. Whatever the increase is, because of the pattern you know that there can be only one answer.

Statement 2:

By the same logic, there can be only one pair of consecutive integers whose squares have an increase of: (121 + n) - (94 + n) = 27

Since you could find those integers, you'd be able to calculate n. Sufficient. The answer is D.

Takeaway: When number properties questions involve big numbers, start small and use patterns (and remember not to overcalculate on Data Sufficiency). Tough-looking problems like this one can be solved efficiently in this manner.
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wait why can we just subtract these equations...?
ScottTargetTestPrep


We are given that the positive integer n is added to each of the integers 69, 94, and 121, and need to determine the value of n.

Statement One Alone:

69 + n and 94 + n are the squares of two consecutive integers.

From statement one, we can say that for some positive integer x, 69 + n = x^2 and 94 + n = (x + 1)^2. Let’s subtract the first equation from the second equation:

(94 + n) - (69 + n) = (x + 1)^2 - x^2
25 = x^2 + 2x + 1 - x^2
25 = 2x + 1
24 = 2x
12 = x

Since we know x = 12, we can substitute this into the first equation to determine the value of n:

69 + n = 12^2
69 + n = 144
n = 75

Statement one alone is sufficient to answer the question. Eliminate answer choices B, C and E.

Statement Two Alone:

94 + n and 121 + n are the squares of two consecutive integers.

We can use the same method that we used in statement one to solve for n. Therefore, without performing the actual calculations, we can conclude that we can find a unique value for n. Statement two alone is also sufficient to answer the question.

Answer: D
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Sa800
wait why can we just subtract these equations...?


This is a standard algebraic step used to eliminate a variable.

Since both equations contain the same n, subtracting one equation from the other cancels it out, leaving only one variable, x, which we can then solve for.

This was also explained HERE.
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Hi Sa800,

You're looking at Scott's step where he wrote (94 + n) - (69 + n) = (x + 1)^2 - x^2, and you're asking why that subtraction is even allowed. Good instinct to pause on it.

Here's the key idea: an equation is a statement that two things are truly equal. Statement (1) hands you two of them that are both true at the same time:

- 69 + n = x^2
- 94 + n = (x + 1)^2

So 69 + n and x^2 are the same number. Likewise 94 + n and (x + 1)^2 are the same number. When you "subtract the equations," all you're doing is subtracting equal amounts from equal amounts - take equals away from equals and you're still left with equals. That's why the new line is also a valid equation.

The payoff is that the n cancels: (94 + n) - (69 + n) = 25, and the n is gone. You've eliminated a variable in one clean move, which is exactly why subtracting is the smart choice here.

Notice Scott himself shows the alternative right after: instead of subtracting, he solved one equation for n and substituted into the other, and got the same x = 12. Subtracting and substituting are just two doors into the same room - both are legal because each equation is a true statement of equality.

Answer: D

Sa800
wait why can we just subtract these equations...?

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