x is a positive integer such as 0

Question

x is a positive integer such as 0<x<22.
Is x a perfect square?

1) x^2+1 is multiple of the seventh prime number
2) x is even

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prateekchugh · Answer

x is a positive integer such as 00; 0 x= 4,9,16 1) x^2 +1 is multiple of the seventh prime number Prime Number: 2,3,5,7,11,13,17 17 is the seventh prime number We can Write: x^2 +1= 17K (k=0,1,2,3,4,.....) K=1; x^2 +1= 17; x^2 =16 YES K=2; x^2 +1= 34; x^2 =33 NO This statement is insufficient 2) x is even x=2; Is x a perfect square? NO x=4; Is x a perfect square? YES This statement is insufficient Combine statement 1) and 2) x^2 +1= 17K x^2 = 17K -1 from 2nd statement----> x=even; x^2= even x^2 = 17K -1---> K need to be odd k=1; x^2+1 = 17k; x^2 = 16; x=4---> Even k=3; x^2+1 = 17k; x^2 = 50; X will not be even and not a perfect square k=5; x^2+1 = 17k; x^2 = 84; X will not be even and not a perfect square k=7; x^2+1 = 17k; x^2 = 118; X will not be even and not a perfect square Answer is C

IanStewart · Answer

I haven't read the rest of your solution, but you should be careful when combining information in a question stem with information in a statement. Here we know in advance that x is a positive integer between 1 and 21 inclusive. It's not possible that x^2 = 33, because there is no positive integer that x could equal in that case. So that example doesn't agree with the information in the question, and it's thus not a relevant example to consider. It doesn't establish anything about whether Statement 1 is sufficient here.

It's true that Statement 1 is not sufficient. It's possible that x = 4, in which case x^2 + 1 = 17. But it's possible that x = 13, in which case x^2 + 1 = 170. So we have two possible values of x for which x^2 + 1 is a multiple of 17, one of which is a perfect square, the other of which is not.

Statement 2 is clearly insufficient alone, but using both Statements, the only even value of x that works is 4, though I don't see any particularly clean way to demonstrate that, using GMAT-level math, besides testing the relevant values.

Kinshook · Answer

Given: x is a positive integer such as 0<x<22.
Asked: Is x a perfect square?

1) x^2+1 is multiple of the seventh prime number
Prime numbers = {2,3,5,7,11,13, 17 ,...}
x^2 + 1 = 17k; where k is an integer
x^2 = 17k - 1 = {16,169}
x = {4,13}
4 is a perfect square but 13 is not.
NOT SUFFICIENT

2) x is even
x = {2,4,6,8,10,12,14,16,18,20}
4 & 16 are perfect squares but other are not
NOT SUFFICIENT

(1) + (2)
1) x^2+1 is multiple of the seventh prime number
2) x is even
x = 4 is a perfect square
SUFFICIENT

IMO C

Archit3110 · Answer

Target find value of X +be integr which is b/w 0 to 22
#1
X^2+1 is multiple of 7th prime ie 17
X can be 4, 13, insufficient
#2
X is even insufficient
From 1&2 
X has to be 4 
Sufficient option c

TestPrepUnlimited · Answer

I'd like to focus on statement 1 here, the rest of the question is straightforward once we figure out which values of x are viable for statement 1.

Statement 1 translates to:  x^2 + 1 = 17*i , where  (i) x is a positive integer between   , and  (ii)  i  is a positive integer.

In order to find possible pairings of x and i, we have two ways of testing values (and since we cannot break down  x^2 + 1  we'll have to test values...).

Method 1: Plug in x from 1 to 21, and check if the right side is a multiple of 17.

Method 2:  x^2 = 17*i - 1  Plug in  i  from 1 up until the value of  17*i - 1  exceeds  21^2 = 441  and check if  x^2  is a square.

Now typically we would choose method 1 because there are fewer values to plug in for method 1. However, checking for divisibility for 17 is tedious, so method 2 would actually be faster here. Memorizing the squares up until  20^2  is also standard preparation for GMAT.

Thus we go down the values by adding 17 at a time, the list we would be looking at is:

16, 33, 50, 67, 84, 101, 118, 135, 152, 169, 186, 203, 220, 237, 254, 271, 288, 305, 322, 339, 356, 373, 390, 407, 424, 441 .

16 and 169 should jump out immediately, and 441 is also  21^2  so we have to include that. Thus x = {4, 13, 21} are all viable x's. Among those x = 4 is a perfect square but x = {13, 21} are not, so statement 1 is insufficient.

Combining with statement 2 we can only choose x = 4 therefore the answer is C.

x is a positive integer such as 0<x<22. Is x a perfect squar

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