If P and Q represent the hundreds and tens digits, respectively, in th

Question

If P and Q represent the hundreds and tens digits, respectively, in the four-digit number x=8PQ2, is x divisible by 8?

(1) P=4
(2) Q=0

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chetan2u · Answer

Divisibility by 8 or  2^3  depends on the last 3 digits, PQ2.

(1) P=4
Nothing about Q.
If Q=0, then No.
If Q=3, then yes.

(2) Q=0
SO the last two digits are 02.
Property of 'Divisibility' by 4 or  2^2  depends on the last 2 digits, Q2 or 02.
Now the number is never divisible by 4, so it will never be divisible by 8 also.
8P02=8P00+2=8P*100+2
8P*100 is divisible by 4, thus 2 is the remainder.
Sufficient as answer is NO.

B

MarmikUpadhyay · Answer

Is 8PQ2 divisible by 8?
Divisibility rule of 8:
If the last three digits of a number are divisible by 8, then the number is completely divisible by 8.
Last 3 digits in 8PQ2 are PQ2.
We will check last three digits only.

Statement 1:
P = 4
=>PQ2 = 4Q2.
If Q = {3, 7}, then PQ2 is divisible by 8.
For all other values of Q, PQ2 is not divisible by 8
Statement 1 is Not Sufficient.

Statement 2:
Q = 0
=>PQ2 = P02
For any value of P, PQ2 is not divisible by 8
Statement 2 is Sufficient.

So, correct answer is option  B .

chetan2u · Answer

You will have to look into your solution.

102, 202, 302 and so on are not divisible by 4, leave aside 8.

TarunKumar1234 · Answer

If P and Q represent the hundreds and tens digits, respectively, in the four-digit number x=8PQ2, is x divisible by 8?

Stat1: P=4
then, x =84Q2, it can be divisible by 8 or not. Not sufficient.

Stat2: Q=0
then, x =8P02, we can have any digit for P, but x won't be divisible by 8.

x =8P02, if last two digits is 02, x won't be divisible by 8.
x =8P02, if last two digits is 22, x won't be divisible by 8.
x =8P02, if last two digits is 42, x won't be divisible by 8.
x =8P02, if last two digits is 62, x won't be divisible by 8.
x =8P02, if last two digits is 82, x won't be divisible by 8.
Sufficient.

So, I think B.

GCMEMBER · Answer

If P and Q represent the hundreds and tens digits, respectively, in the four-digit number x=8PQ2, is x divisible by 8?

(1) P=4

x = 84Q2
For Q=2, 8422 is not divisible
For Q=3, 8432 is divisible
Not sufficient 
 
(2) Q=0

x = 8P02
The last two digits are not divisible by 4. So, 8P02 is not divisible by 8. 
Sufficient

Option B

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Bunuel · Answer

If P and Q represent the hundreds and tens digits, respectively, in the four-digit number x=8PQ2, is x divisible by 8?

For a number to be divisible by 2 the last digit must be divisible by 2 (so the last digit must be even);
For a number to be divisible by 4 the last two digits must be divisible by 4 (04, 08, 12, 16, ..., 96);
For a number to be divisible by 8 the last three digits must be divisible by 8 (008, 012, 016, ..., );
etc.

(1) P=4. If Q=0, then the answer is NO, because 402 is not divisible by 8, but if Q=3, then the answer is YES, because 432 is divisible by 8. Not sufficient.

(2) Q=0. In this case the last tow digits are 02, so x is not divisible by 4, which means that it's not divisible by 8 either. Sufficient.

Answer: B.

Hope it's clear.

bumpbot · Answer

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If P and Q represent the hundreds and tens digits, respectively, in th

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