What is the remainder when a positive integer P is divided by 20?

Question

Competition Mode Question

(1) P = X^2 where X is a positive integer
(2) X is the square of an even integer

Leadership · Answer

What is the remainder when a positive integer P is divided by 20?

We need to find the definite value of remainder of P/20

(1) P = X^2 where X is a positive integer
 If X (positive integer), then X could be - 1, 2, 3, 4, 5, 6 ....
P= 1, 4, 9, 16, 25, 36,
So, remainder of P/20 will have different values such as 1, 4, 9, 16, 5. Insufficient.

A D  / B C E

(2) X is the square of an even integer
 X could be 2, 4, 6, 8, 10 ... has no relation with P. Clearly Insufficient

A D  /  B  C E

Combining Statements, we get:

X - 2, 4, 6, 8, 10.. 
P - 4, 16, 36, 64 ..

again, remainder of P/20 will have different values such as 4, 16. Combining failed and we get 'E' as the winner.

Archit3110 · Answer

#1
P = X^2 where X is a positive integer
x=2,3,4.. value of P =4,9,16.. 
insufficient
#2
 X is the square of an even integer
X= 4,16,36 ; insufficient
from 1 &2
P =4^2,(16)^2 ... insufficient
IMO E

What is the remainder when a positive integer P is divided by 20?

(1) P = X^2 where X is a positive integer
(2) X is the square of an even integer

freedom128 · Answer

(1) P = X^2 where X is a positive integer
X=2, P=4, remainder is 4
X=4, P=16, remainder is 16
NOT SUFFICIENT

(2) X is the square of an even integer
We have to know P, not X
NOT SUFFICIENT

(1)+(2)
X=4, P=16, remainder is 16
X=16, P=256, remainder is 16
X=36, P=1296, remainder is 16
SUFFICIENT

FINAL ANSWER IS (C)

shameekv1989 · Answer

What is the remainder when a positive integer P is divided by 20?

(1) P = X^2 where X is a positive integer
(2) X is the square of an even integer

i)  P=X^2  => P can be 1,4,16,25 with number of remainders - NS
ii) Nothing about P - NS

Combining i) and ii)
X =  (2m)^2 ; P =  X^2  =>  P=16*m^4  (where m can be any positive integer)

Therefore,  \frac{P}{20}  will give remainder of 16 or (-4) from the first term.

Let's check for  m^4 , 
m = 1 => remainder 1 =>  \frac{P}{20}  will give remainder of -4 = 16
m = 2 => remainder  2^4  = 16 = -4 =>  \frac{P}{20}  will give remainder of -4*-4 = 16
m = 5 => remainder of  \frac{625}{20}  = 5 =>  \frac{P}{20}  will give remainder of -4*5 = -20 = 0

2 different answers - Not Sufficient

Answer E

Jawad001 · Answer

What is the remainder when a positive integer P is divided by 20?

(1) P = X^2 where X is a positive integer
(2) X is the square of an even integer

From statement (1), let x = 1, 2, 3, 4, 5, 6, …………
P = x^2 = 1, 4, 9, 25, 36, …………………
When p/25, then remainder = 1, 4 , 5 , 16 …………….. Not sufficient.

From statement (2), there is no P. Hence, Not Sufficient.

Combining both statements, we get, P = x^2 = 4, 16, 36, 84, 100, 144, ………..
When p/25, Remainder = 4, 16, 16, 4, 0, 4, Not Sufficient.
Answer: E

lacktutor · Answer

What is the remainder when a positive integer P is divided by 20?

( Statement1 ):  P = X^2  where X is a positive integer
-->  P=6^{2}  -->  \frac{36}{20}= 1  (remainder-16)
-->  P=10^{2} -->  \frac{100}{20}=5  (remainder-0)
 Insufficient  
 
( Statement2 ): X is the square of an even integer
the question is about P, not X. Nothing tells us about what P is.
 Insufficient

Taken together 1&2, 
 P= (4^{2})^{2} =256  -->  \frac{256}{20}= 12  (remainder -16)
 P= (10^{2})^{2}= 10000  -->  \frac{10000}{20} =500  (remainder -0)
Still we got more than one value.
Insufficient

The answer is E.

eakabuah · Answer

We are given that P is a positive integer, and we are to determine the remainder when P is divided by 20.

Statement 1: P=X^2, where X is a positive integer.
Possible values of P: 1, 4, 9, 16, 25, 36, etc.
When P=1, the remainder is 1. When P=4, the remainder is 4.
Statement 1 is insufficient since we have more than one possibility as the remainder when P is divided by 20.

Statement 2: X is the square of an even integer.
X=4, 16, 36, 64, 100, etc.
Clearly statement 2 is insufficient because we have no idea the value of P, and as per the question, there are many possibilities that lead to 20 possible remainders when P is divided by 20.

1+2
From 1, we know that P=X^2 and from 2 we know that X can be: 4, 16, 36, 64, 100, etc
Hence P=16, 256, or 10,000, etc.

When P=16, the remainder is 16, however, when P=10,000, the remainder is 0.
Both statements even when taken together are not sufficient.

E is the answer.

freedom128 · Answer

Hi  Bunuel ,

Can you explain why it is E? Statement 2 says X is the  square  of an even integer, so X=2 must not be included. Thus, X=4,16,36,64,... and P=16,256,1296,4096,... Whenever we divide P, the remainder is always 16. Thus, answer should be C.

Thanks for clarification

Bunuel · Answer

P is the fourth power of a positive even integer.

If P = 2^4, then the remainder is 16.
If P = 20^4, then the remainder is 0.

exc4libur · Answer

remainder: p/20

(1) p=x^2, x=pos integer; insufic.
p=1,4,9… remainder=1,4,9…

(2) x=4,16,36,64,100… p=? insufic

(1/2) insufic
p=x^2=(4,16,36,64,100…)^2
p=4^2=16; remainder=16
p=100^2; remainder=0

Ans (E)

Rainman91 · Answer

This approach is wrong because are you going to try every single number from 4 to 8 until you reached 10? multiplying all those numbers would have taken well over 2 mins. You have to realize that 10 to the 2nd power is 100. and that to the next power will be divisible by 20.

bumpbot · Answer

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What is the remainder when a positive integer P is divided by 20?

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