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P and Q are prime numbers less than 70. What is the units digit of P*Q

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P and Q are prime numbers less than 70. What is the units digit of P*Q [#permalink]

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P and Q are prime numbers less than 70. What is the units digit of P*Q?

(1) Units’ digit of \((P^{4k+2} - Q\)) is equal to 7, where k is a positive integer.

(2) Units digit of the expression \([PQ + Q*(Q+1) - Q^2]\) is a perfect cube



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Last edited by EgmatQuantExpert on 13 Dec 2016, 06:45, edited 5 times in total.

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Re: P and Q are prime numbers less than 70. What is the units digit of P*Q [#permalink]

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St 1- Units’ digit of (P4k+2−Q) is equal to 7, where k is a positive integer.
Diff of P&Q is odd. So either of them should be even. So only prime which is even is 2. So Q-2 and To get the diff. 7, P should be 3 as power of 3 in the form of (4k+2) results the unit digit into 9 and 9-2=7.
So P=3 and Q=2. Ans-Sufficient.

St-2-Units digit of the expression [PQ+Q∗(Q+1)−Q2] is a perfect cube.
Expression modified to (PQ+Q)...
The only unit digit which is perfect cube is 8.
Could not solve hereafter, please help.

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Re: P and Q are prime numbers less than 70. What is the units digit of P*Q [#permalink]

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Hi nilboy

For condition 2, as you already got that the units digit has to be 8.
So, we have Q(P+1)=x8, where x be any number before units digit.
try just 2 -3 prime numbers.
Consider, P=3 Q=2 Q(P+1)=8
PQ=6
Consider, P=3 Q=7 Q(P+1)=28
PQ=21
So, we cannot find the units digit with this condition.
Insufficient

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Re: P and Q are prime numbers less than 70. What is the units digit of P*Q [#permalink]

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EgmatQuantExpert wrote:
P and Q are prime numbers less than 70. What is the units digit of P*Q?

(1) Units’ digit of \((P^{4k+2} - Q\)) is equal to 7, where k is a positive integer.

(2) Units digit of the expression \([PQ + Q*(Q+1) - Q^2]\) is a perfect cube

We will provide the OA in some time. Till then Happy Solving :lol:

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(1) Units’ digit of \((P^{4k+2} - Q\)) is equal to 7, where k is a positive integer.

(2) Units digit of the expression \([PQ + Q*(Q+1) - Q^2]\) is a perfect cube



Perfect squares are integers: 0, 1, 4, 9, 16 and
Similarly, perfect cubes are: 0, 1, 8, 27, 64.


(1) Units’ digit of \((P^{4k+2} - Q\)) is equal to 7, where k is a positive integer.
\((P^{4k+2} - Q\)) = \((P^{4k}*P^{2} - Q\))
as the unit digit is ODD (7) either P or Q is 2 (that's the only even prime) .

IF P is EVEN and Q is ODD PRIME
if P=2 , we know that 2 has a cyclicity of 4 , so , \((P^{4k}*P^{2} )\) will always have a unit digit 4.

how ? \((P^{4k}\) will have a unit digit of 6 and \(P^{2}\) will have a unit digit of 4 and \(6*4 = 24\).

now Q's unit has to be 7.

Q Can be 7, 17, 37 67

Unit digit is 4 in this case .

lets look at other case when Q=2 and P is some ODD prime.

explained below: -

\((P^{4k}*P^{2} - Q\))

\((P^{4k}*P^{2} - 2\))

IF P=3 Q=2 the above expression has a unit digit of 7 . unit digit of P*Q= 6
IF P=7 Q=2 the above expression has unit digit of 7. unit digit of P*Q= 4

This option is not sufficient .



(2) Units digit of the expression \([PQ + Q*(Q+1) - Q^2]\) is a perfect cube

this is one is easy ...
\([PQ + Q*(Q+1) - Q^2]\)
=\([PQ + Q]\)
=\([Q(P + 1)]\)

the above product should result into a unit digit of 0,1,8

P=19, Q=31 unit digit 0
P=2, Q=7 unit digit 1
P=7, Q=11 unit digit 8

Not Sufficient.


Together:

we know that Either P or Q is Even=2, as per statement A.
If P=2 , Q Can be 7, 17, 37 67

from stmt B
IF P=3 Q=2 the above expression has a unit digit of 7 .
IF P=7 Q=2 the above expression has unit digit of 7.
and \([Q(P + 1)]\) has a unit digit which is perfect cube (0,1,8) .

if P=2 , P+1=3 , Q can be 7,17,37,67 . unit digit of PQ will be 4 .
but if Q=2 , P has a unit digit of 3 or 7 such that the product \([Q(P + 1)]\) will not have a unit digit which is perfect cube (0,1,8) .


answer C
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Last edited by Lucky2783 on 11 Apr 2015, 02:28, edited 2 times in total.

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Re: P and Q are prime numbers less than 70. What is the units digit of P*Q [#permalink]

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Detailed Solution

Step-I: Given Info

We are given two prime numbers \(P, Q < 70\). We are asked to find the units digit of \(P*Q\).

Step-II: Interpreting the Question Statement

The unit’s digit of the product of \(P\) & \(Q\) would depend on the units digit of \(P\) & \(Q\). Hence, our endeavour would be to find the units digit of \(P\) & \(Q\).

Step-III: Statement-I

Statement-I tells us that the difference between two numbers is odd which would imply opposite even/odd natures of the numbers i.e. one is even and the other is odd. Since, there is only on even number (i.e. \(2\)), there are two possible scenarios:

• \(P=2\), so \(P ^{4k+2}\) will have the units digit as \(4\). As a result the units digit of \(Q\) would be \(7\). Hence units digit of product of \(P\) & \(Q\) would be \(4\)

• \(Q=2\), in such a case units digit of \(P^{4k+2}\) will be \(9\), which implies that units digit of \(P\) is either \(7\) or \(3\). Hence, units digit of product of \(P\) & \(Q\) would be either \(4\) or \(6\).

Since we do not have a unique answer, Statement-I is not sufficient to answer the question.

Step-IV: Statement-II

The expression in statement-II can be simplified to \(Q(P+1)\), since its unit digit is a perfect cube, the possible values are \(1\) and \(8\).

If unit digit is \(1\),
• Units digit of \(P\)= \(2\) and units digit of \(Q\) =\(7\), so the units digit of product of \(P\) & \(Q\) would be \(4\)

If unit digit is \(8\),two cases are possible:

• Units digit of \(Q= 2\) and units digit of \(P =3\), so the units digit of product of \(P\) & \(Q\) would be \(6\)

• Units digit of \(Q= 1\) and units digit of \(P= 7\), so the units digit of product of \(P\) & \(Q\) would be \(7\)

Since we do not have a unique answer, Statement-II is not sufficient to answer the question


Step-V: Combining Statements I & II

Statement-I tells us that the units digit of \(P\) & \(Q\) can be \(4\) or \(6\). Statement-II tells us that the units digit of \(P\) & \(Q\) can be \(4\) or \(6\) or \(7\).
By Combining statements- I & II, we still do not have a unique answer.

Thus combination of St-I & II is also insufficient to answer the question.
Hence, the correct answer is Option E

Key Takeaways

1. Know the properties of Even-Odd combinations to save the time spent deriving them in the test
2. In even-odd questions, simplify complex expressions into simpler expressions using the properties of even-odd combinations
3. Know the cyclicity of the numbers to arrive at their units digit


Regards
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Last edited by EgmatQuantExpert on 28 May 2015, 07:16, edited 2 times in total.

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Re: P and Q are prime numbers less than 70. What is the units digit of P*Q [#permalink]

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EgmatQuantExpert wrote:
Detailed Solution

Step-I: Given Info

We are given two prime numbers P, Q < 70. We are asked to find the units digit of P*Q.

Step-II: Interpreting the Question Statement

The unit’s digit of the product of P & Q would depend on the units digit of P & Q. Hence, our endeavour would be to find the units digit of P & Q.

Step-III: Statement-I

Statement-I tells us that the difference between two numbers is odd which would imply opposite even/odd natures of the numbers i.e. one is even and the other is odd. Since, there is only on even number (i.e. 2), there are two possible scenarios:

• P=2, so P ^(4k+2) will have the units digit as 4. As a result the units digit of Q would be 7. Hence units digit of product of P & Q would be 4
• Q=2, in such a case units digit of P^(4k+2) will be 9, which implies that units digit of P is either 7 or 3. Hence, units digit of product of P & Q would be either 4 or 6.

Since, we do not have a unique answer, Statement-I is not sufficient to answer the question.

Step-IV: Statement-II

The expression in statement-II can be simplified to Q(P+1), since its unit digit is a perfect cube, the possible values are 1 and 8.

If unit digit is 1,
• Units digit of P= 2 and units digit of Q =7, so the units digit of product of P & Q would be 4

If unit digit is 8,two cases are possible:

• Units digit of Q= 2 and units digit of P =3, so the units digit of product of P & Q would be 6
• Units digit of Q= 1 and units digit of P= 7, so the units digit of product of P & Q would be 7

Since, we do not have a unique answer, Statement-II is not sufficient to answer the question


Step-V: Combining Statements I & II

Statement-I tells us that the units digit of P & Q can be 4 or 6. Statement-II tells us that the units digit of P & Q can be 4 or 6 or 7.
By Combining statements- I & II, we still do not have a unique answer.

Thus combination of St-I & II is also insufficient to answer the question.
Hence, the correct answer is Option E

Key Takeaways

1. Know the properties of Even-Odd combinations to save the time spent deriving them in the test
2. In even-odd questions, simplify complex expressions into simpler expressions using the properties of even-odd combinations
3. Know the cyclicality of the numbers to arrive at their units digit


Regards
Harsh


hi harsh


is 0 not a perfect cube ?
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Re: P and Q are prime numbers less than 70. What is the units digit of P*Q [#permalink]

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New post 11 Apr 2015, 05:57
on the 2nd stmt, why dont we could use P=2 and Q=3 so that Q(P+1)=9... The unit digit is a perfect cube again..

Also please let me know on how to deploy and find values for finding the unit digits... as we solve these kind..

I mean on your take away "Know the cyclicality of the numbers to arrive at their units digit".. Could you please give a short note on how to work out? It would be helpful..

Thanks..

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Re: P and Q are prime numbers less than 70. What is the units digit of P*Q [#permalink]

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New post 12 Apr 2015, 06:39
Hi Harsh,

One point regarding statement 1; we all know either of P/Q has to be 2.

If Q=2 then P^4k+2=9 which is possible only if k=0 and P=3, but since 0 is not a positive integer so I would assume K to be at least 1, so in this case P^6=9. Well I don't think any prime number raised to the power 6 can be 9 hence this scenario is not possible so the answer must be A.

Please clarify if I am wrong.

Thanks

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Re: P and Q are prime numbers less than 70. What is the units digit of P*Q [#permalink]

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New post 13 Apr 2015, 03:41
Hi Everyone,


Lucky2783- 0 is indeed one of the possibilities :) even in such a case the answer would be 'E' because for units digit of Q (P+1) to be 0, P's units digit has to be 9 and Q can be any other possible prime number < 70. So, it further increases the no. of possibilities of the units digit of product of P and Q.

sheolokesh- 9 is not a perfect cube, it's a perfect square of 3, so it's not a valid case.

atom- we are dealing with the unit's digit of a number, so when we say that \(P^ {4k+2}\) will have the units digit as 9, the number \(P^{4k+2}\) is not equal to 9, but its units digit is.
For example: assuming P as 3, if we take k=1, then \(3^6\) would be 729. We see that the units digit is 9 and not the number itself. So, \(3^{4k+2}\) will always its units digit as 9. Similarly \(7^{4k+2}\) will also have its units digit as 9.

To know more about the cyclicity of the digits, you can register for our free trial where we have a separate chapter (Units Digit) on the concept. You can register here: https://e-gmat.com/registration/

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Hope it helps!

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Re: P and Q are prime numbers less than 70. What is the units digit of P*Q [#permalink]

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Hi,

Wanted to know are these questions expected in GMAT? Because these are not really doable in 2 minutes :(

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Re: P and Q are prime numbers less than 70. What is the units digit of P*Q [#permalink]

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New post 19 Apr 2015, 22:39
natashakumar91 wrote:
Hi,

Wanted to know are these questions expected in GMAT? Because these are not really doable in 2 minutes :(



Hi natashakumar91- the questions are slightly on the difficult side but with clarity of concepts and process, they are indeed doable in the given time frame. The first priority should be learning the right approach to the problems without considering the time taken. You would realize that once you have the right approach, time will take care of itself :)

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Re: P and Q are prime numbers less than 70. What is the units digit of P*Q [#permalink]

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New post 29 Dec 2016, 08:54
Hi Harsh,


after simplification of statement II we reach at Q(P+1)= perfect cube

Prime numbers P & Q are smaller than 70 so P+1 can't be more than 68 as well.

I see it like Q*(Q*Q) = Perfect cube so (P+1) = (Q*Q) and there is only one case (P = 3) below 70 where (P + 1) is a perfect square.

Q (3+1) = Q*Q*Q = 8 = 2*2*2

Hence Q = 2 and P = 3 and we have Unit Digit as 6.

answer is B

I am eager to know if I have missed some basic concept.



EgmatQuantExpert wrote:
Detailed Solution

Step-I: Given Info

We are given two prime numbers \(P, Q < 70\). We are asked to find the units digit of \(P*Q\).

Step-II: Interpreting the Question Statement

The unit’s digit of the product of \(P\) & \(Q\) would depend on the units digit of \(P\) & \(Q\). Hence, our endeavour would be to find the units digit of \(P\) & \(Q\).

Step-III: Statement-I

Statement-I tells us that the difference between two numbers is odd which would imply opposite even/odd natures of the numbers i.e. one is even and the other is odd. Since, there is only on even number (i.e. \(2\)), there are two possible scenarios:

• \(P=2\), so \(P ^{4k+2}\) will have the units digit as \(4\). As a result the units digit of \(Q\) would be \(7\). Hence units digit of product of \(P\) & \(Q\) would be \(4\)

• \(Q=2\), in such a case units digit of \(P^{4k+2}\) will be \(9\), which implies that units digit of \(P\) is either \(7\) or \(3\). Hence, units digit of product of \(P\) & \(Q\) would be either \(4\) or \(6\).

Since we do not have a unique answer, Statement-I is not sufficient to answer the question.

Step-IV: Statement-II

The expression in statement-II can be simplified to \(Q(P+1)\), since its unit digit is a perfect cube, the possible values are \(1\) and \(8\).

If unit digit is \(1\),
• Units digit of \(P\)= \(2\) and units digit of \(Q\) =\(7\), so the units digit of product of \(P\) & \(Q\) would be \(4\)

If unit digit is \(8\),two cases are possible:

• Units digit of \(Q= 2\) and units digit of \(P =3\), so the units digit of product of \(P\) & \(Q\) would be \(6\)

• Units digit of \(Q= 1\) and units digit of \(P= 7\), so the units digit of product of \(P\) & \(Q\) would be \(7\)

Since we do not have a unique answer, Statement-II is not sufficient to answer the question


Step-V: Combining Statements I & II

Statement-I tells us that the units digit of \(P\) & \(Q\) can be \(4\) or \(6\). Statement-II tells us that the units digit of \(P\) & \(Q\) can be \(4\) or \(6\) or \(7\).
By Combining statements- I & II, we still do not have a unique answer.

Thus combination of St-I & II is also insufficient to answer the question.
Hence, the correct answer is Option E

Key Takeaways

1. Know the properties of Even-Odd combinations to save the time spent deriving them in the test
2. In even-odd questions, simplify complex expressions into simpler expressions using the properties of even-odd combinations
3. Know the cyclicity of the numbers to arrive at their units digit


Regards
Harsh

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P and Q are prime numbers less than 70. What is the units digit of P*Q [#permalink]

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New post 30 Dec 2016, 13:05
annusngh wrote:
Hi Harsh,


after simplification of statement II we reach at Q(P+1)= perfect cube

Prime numbers P & Q are smaller than 70 so P+1 can't be more than 68 as well.

I see it like Q*(Q*Q) = Perfect cube so (P+1) = (Q*Q) and there is only one case (P = 3) below 70 where (P + 1) is a perfect square.

Q (3+1) = Q*Q*Q = 8 = 2*2*2

Hence Q = 2 and P = 3 and we have Unit Digit as 6.

answer is B

I am eager to know if I have missed some basic concept.



Hey Annu,


I think you did not read the Second statement correctly.

It clearly states that the Units digit of the expression \([PQ+Q∗(Q+1)−Q^2]\) is a perfect cube"

After simplification once you got \(Q*(P+1)\), you assumed this expression to be a perfect cube, which is incorrect!

What you needed to do was assume the units digit of expression to be a perfect cube and the possible values would have been 1 and 8.

I would request you to try the question again and even go through the solution once. In case you still have any doubts, do post it here. :)


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Re: P and Q are prime numbers less than 70. What is the units digit of P*Q [#permalink]

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New post 14 Feb 2017, 20:17
is there any way to solve any high-level problems in less than 3 minutes?

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Re: P and Q are prime numbers less than 70. What is the units digit of P*Q [#permalink]

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Excuse me, but I don't think this question is properly designed.
Statement 1, is
"1) Units’ digit of (P^4k+2)−Q is equal to 7, where k is a positive integer."
If P and Q are prime numbers less than 70, they both have to be positive, and one of them has to be equal to 2. You can derive from Statement 1, if the result is odd having a 7 as a units digit, either P or Q is even. And, if P and Q are prime, either of them can be number 2.
If you plug in the 2 in that expression, :
a. If Q is 2, Then P must be 3 to the power of 2., and yields k equal to zero, which CAN NOT be, since k is positive according to St1. So Q can never be number 2.
b. Hence, P is equal to 2. If you plug in the 2 as P in the expression, you can subtract a prime number Q, from any number ending in 4 in the units digit place (64,1024, etc), which leaves Q as being possibly 7,17,37, 47 or 67. In either case, u are able to answer the main question, "What is the units digit of P*Q?", so it's sufficient A.
The units digit of P*Q is 4., since P=2, and Q ends in 7, no matter the other digits in it.

If you analyze separately St2, "Units digit of the expression [Q(P+1)] is a perfect cube", it's insufficient. At this moment, you this should be over, but If you analyze it in light of St1, you arrive to: Q(P+1)=Q*3=XXX8, which means Q can be either 6,16, 26, but you know it ends with 6, and since P is 2, you know The units digit of P*Q is 2, which contradicts info gathered from St1.
Both Statements should always share TRUE information, no matter if it's sufficient or not.
This is just mind blowing for me. I 've read that the GMAT statements ARE ALWAYS TRUE, and in this question they do not complement each other.

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P and Q are prime numbers less than 70. What is the units digit of P*Q [#permalink]

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New post 03 Aug 2017, 04:14
Hi Harsh!

In statment two you offer Q=1 P=7 as a solution yielding a units number 8 since 1(7+1)=8. The information given in the prompt states that both P and Q are prime and 1 is not a prime number, thus 8 expressed in the form 8*1 or 1*8 is not possible altogether in this question.

On a second note, I think there is also the case where 0 is a perfect cube, even though considering would only increase the number of possibilities.

Regards!

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Re: P and Q are prime numbers less than 70. What is the units digit of P*Q [#permalink]

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New post 03 Aug 2017, 05:13
Statement 1 : Let k=1 then P^6-Q unit digit to be 7 we have two values of P and applying Q same in both the cases
So P=13 and Q=2 gives the unit digit 7 and P=23 and Q=2 will also give 7 as a unit digit
Not sufficient

Statement 2 : P*Q-Q unit digit is a perfect cube so the possible value for the perfect cube are 1,8
Insufficient as there will be two possible values of P and Q

Combining 1 and 2
P=23 and Q=2 will satisfy both the statements so it will be a possible value hence P*Q=46

Ans: C

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Re: P and Q are prime numbers less than 70. What is the units digit of P*Q   [#permalink] 03 Aug 2017, 05:13
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P and Q are prime numbers less than 70. What is the units digit of P*Q

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