Given that p is a positive even integer with a positive

Should this be E?

(Unit digit of p^3) - (Unit digit of p^2) = 0 if the units digit is 0, 1, 5, 6.

The answer in these cases would be 3, 4, 8, 9.

Even if 0 is neither posive or negative, we still have 3 options left.

However, only 9 is mentioned from among these choices, so could be D as well... any suggestions?

E It cannot be determined from the given information.

As p is even,
p = 10x+y

y = {0,2,4,6,8}

Unit's digit of p^3 = {0, 4, 6}
Unit's digit of p^2 = {0,2,4, 6, 8}

As unit's digit p^3 -unit's digit p^2 =0

Common units digit between p^3 and p^2 = {0, 4,6}

We can get {0,4,6} in p^2 or p^3 from {0,2,4,6,8}

So cannot determine.
Answer: E

What the heck I was thinking

Answer should be D because last digit of p must be +ve and even.

unit digits could be 0, 1, 5 and 6 but only 6 satisfies the conditions given in the question. so D make sense..

From the condition,

p is even and not divisible by 10 (because units digits is positive)

so p=2k, and k can not be multiple of 5 or 10

p^3-p^2 = p^2(p-1) = (2k)^2*(2k-1)

Since we know that units digit of p^3-p^2 is 0 and k con not be 5 or 10,
the only way to get 0 at the end of the number is if 2k-1=5

=> 2k=6 => p=6 => p+3=9

(D)

Hey U2...
Can you tell me where I missed this question?

My solution is below... goes without saying got it wrong...

If units digit = 6 then units(216)-units(36) = 0
p+3 = 9

Hence D

Heman

Since Q says unit digit has to be +ve 0 is out
if 4 is units digit and say # is 4--> 4^3 = 64 & 4^2=16 . Subrtacting units digit does not result in 0.

Try the rest. Only when 6 is in the units digit will 6^3= 216 , 6^2=36. Subtracting units digit will result in zero.

Hence reqd answer = 6+3 = 9 Choice D

Heman

yes... zero is neither positive nor negative so you must leave it alone

I just picked p=2, 4, 6 and got p^3-p^2= 4, 8, 0

p=6 and p+3=9

P is a even number with a positive number in the unit digit
Hence unit digit can be 2,4,6 or 8

2^2 = 4 2^3 = 8
4^2 = 6 4^3 = 4
6^2 = 6 6^3 = 6
8^2 = 4 8^3 = 2

Hence only 6 in the unit digit satisfies the condition given.

hence unit digit of p+3 = 9

Argh, stung again by not reading the question correctly.

Thought it could have been 5 or 6, so picked e.

Until I reviewed the answer and re-read them q stem.

As stated above; numbers ending with 0, 1 , 5 , 6 have squares, cubes or any high power returning the same units digit at the end

Example:

5 ^ 10000 would also have 5 at the end

Given: P is an even positive number (i)
and units digit of p^3 - units digit of p^2= 0 (ii)

Required: Units digit of p+3

For (i), The units digit can be 2,4,6 or 8

For (ii), the numbers for which the units digit of cube and square is same are {0, 1, 5, 6}
But we know that p is positive and even.
Hence p = 6

p+3 = 9 Option D

Units digit of cube and square should be equal.
3 units digit satisfy the above condition- 1, 5, 6
p+3 for above units digit are 4,8,9.
Among the options 9 is there.
Hence answer D) 9

I understand that units digits must end in 1 , 5 , and 6 .
Therefore, P + 3 will be 4 , 8 , and 9
But as there are three options , we will not be able to determine a unique answer .
So the answer must be E even though one of the options (that is, 9) is mentioned .
If E would have be any other digit than 4 or 8 then answer would have been D .

Your comments .

Hi sb0541 and obliraj

There is only one possible answer out of 1,5, and 6..
Do not forget the info given.. p is a positive even integer so 6 is the only possibility
And thus answer is 6+3=9..

\(p^3-p^2\) has units digit as 0... So p^3 and p^2 should have same digit..
Possible values 1, 5, 0, and 6...
we are looking for POSITIVE even integers
0 and 6 are even and only 6 is positive and even..
So p=6, then p+3=6+3=9

D

Official answer from Manhattan Prep.

The variable p is a positive even integer and the units digit of this number is also positive. Therefore, the units digit has four possible values: 2, 4, 6, or 8. (O is not a possibility, since 0 is not positive.)

This is useful information because the units digit of a number is the only value that determines the new units digit when that number is squared or cubed. For instance \(1,394^{2} =\) some very large number whose units digit is 6, because \(4^{2} = 16\).

In other words, only the units digit of \(p\) will determine the units digit of \(p^{3}\) and \(p^{2}\) . Further, since the units digit of \(p^{3} – p^{2} = 0\), you’re looking for a circumstance in which the square of the units digit is equal to the cube of the units digit.

When does this occur? Test the possible units digits 2, 4, 6, and 8 to see.

Only when \(p = 6\) are the units digits of \(p^{3}\) and \(p^{2}\) equal—and therefore the units digit of  \(p^{3} – p^{2} = 0\). As a result, the units digit of p must equal 6, and the units digit of \(p + 3\) must equal 9.

The correct answer is (D).