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
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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 \(6*4 = 2
4\).
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