If x, y, and z are positive integers, is xz even?
(1) xy is even.
(2) yz is even.
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is
sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
OA not available with me.
My answer : E
What do you think?
2^495 mod 625
2^495 mod 5 => 3
2^495 mod 125 = 43
125a+43 = 5b+3
Answer = 293
How did you arrive @ the answer ? please explain the last step.
Puys, Please solve this:
2^495 mod 625 = ?
2^495 mod 625
2^495 mod 5 => 3
2^495 mod 125 = 43
125a+43 = 5b+3
Answer = 293
while a=1 and b=33, answer shall be 168, by ur method, how did u choose a=2 and b=58? pls xplain..
cyborg5021a SaysHow did you arrive @ the answer ? please explain the last step.
Sahana Kavitha Sayswhile a=1 and b=33, answer shall be 168, by ur method, how did u choose a=2 and b=58? pls xplain..
Editing My Post..I did it wrong....Ankurb was right...Thanks To both of you.....Editing my post...
If
x, y, and z are positive integers, is xz even?
(1)
xy is even.
(2)
yz is even.
A. Statement (1)
ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2)
ALONE is sufficient, but statement (1) alone is not sufficient.
C.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is
sufficient.
D.
EACH statement ALONE is sufficient.
E. Statements (1) and (2)
TOGETHER are NOT sufficient.
OA not available with me.
My answer : E
What do you think?
x y z
---------------------------
St 1/2 even odd even
St 1/2 odd even odd
St 1/2 even even even
Hence E, is correct as we have a case, where x and z are odd.
x y z
---------------------------
St 1/2 even odd even
St 1/2 odd even odd
St 1/2 even even even
Hence E, is correct as we have a case, where x and z are odd.
Same choice
If x, y, and z are positive integers, is xz even?
(1) xy is even.
(2) yz is even.
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
x______y_____z
Odd___Even__Odd
Even___Odd___Even
Even___Even___Even
So, it can't be determined, that xz is even or odd
Hence ..Option 5
Hi Puys
I'm posting a question from one of the mocks that I gave. Try solving it. I'l post the answer tonight.
Ques> There are 1000 lockers(all closed) and thousand children. The first child unlocks all the doors. The second child locks open doors and opens locked doors that are multiple of 2. Third child locks open doors and opens locked doors that are multiples of 3 and so on till 1000 students have taken their turns.
Find the number of locked and open doors at the end.
Please post your approach. I'l share mine tonight.
Since each door is opened the no of time of its Factors.
so only those door nos having odd factors will be having its Door open
only perfect squares can have odd factors since the are of the form a^2m*b^2m.....
no of factors would be(2n+1)*(2m+1) etc...always odd
total 31 squares from 1 to 961 in 1000
so total open doors is 31
total closed doors is 1000-31
969
Puys, Please solve this:
2^495 mod 625 = ?
2^500mod625 = 1
2^495* 32mod625 = 1
hence 2^495 = 293
P.S->if options were available cud have been very easy
Puys, Please solve this:
2^495 mod 625 = ?
Whats wrong with below method:-
2^495/625
= 1/2
==>1/2
==>Rem=1
2^495 mod 625 = ?
Whats wrong with below method:-
2^495/625
= 1/2
==>1/2
==>Rem=1
Puys, Please solve this:
2^495 mod 625 = ?
Whats wrong with below method:-
2^495/625
= 1/2
==>1/2
==>Rem=1
you cannot divide like this dude..the denominator shud not be split like this...divide by 625
and if u are spliting then use chinese remainder theorem...split like this 625 = 5*125
Help me out with this please...
Q. How many single digit even natural numbers solutions are possible for the equation a+b+c+d = 24, such that a+b > c+d?
I. 10
II. 11
III. 16
IV. 24
V. N.O.T
Help me out with this please...
Q. How many single digit even natural numbers solutions are possible for the equation a+b+c+d = 24, such that a+b > c+d?
I. 10
II. 11
III. 16
IV. 24
V. N.O.T
is it 11???????
a+b > c+d
14 10
16 8
case1- a+b=14 -->(a,b) = (6,8 ),(8,6)
c+d=10-->(c,d) =(2,8 )(4,6)(6,4)(8,2)
so total = 4*2 = 8 values
case 2-a+b=16 (a=8,b=8 )
c+d=8 -->(c,d) = (2,6)(4,4),(6,2)
total 3 values
hence total = 8+3 =11
correct me if i m wrong
2^500mod625 = 1
2^495* 32mod625 = 1
hence 2^495 = 293
P.S->if options were available cud have been very easy
How did u come to the answer in one go? can you explain your last step please?
Btw: original Q was. : Whats the last 4 digit of 2^999?
is it 11???????
a+b > c+d
14 10
16 8
case1- a+b=14 -->(a,b) = (6,8 ),(8,6)
c+d=10-->(c,d) =(2,8 )(4,6)(6,4)(8,2)
so total = 4*2 = 8 values
case 2-a+b=16 (a=8,b=8 )
c+d=8 -->(c,d) = (2,6)(4,4),(6,2)
total 3 values
hence total = 8+3 =11
correct me if i m wrong
Yeah.. my mistake.
I think that should be it... tho I do not have the answer.
How did u come to the answer in one go? can you explain your last step please?
Btw: original Q was. : Whats the last 4 digit of 2^999?
not in one go dude...its acutally helpfull if u had the ioptions and u cud keep each option and check..anyways is it last three digit or four digits?
incrediblylame SaysI do not have the answer atm unfortunately. What's confusing tho, is that the numbers may not be distinct, because the problem doesn't say so. So even 8,8,4,4 could be a case, which changes the answer. There can be more such cases.
i have taken 8 8 4 4 as one of the cases..just chk:)
Q. let n=999......99 be an integer consisting of a string of 2009 nines. Find the sum of digits of n^2.
I. 18072 II. 18081 III. 18090 IV. 18080 V. 18073
Q. let n=999......99 be an integer consisting of a string of 2009 nines. Find the sum of digits of n^2.
I. 18072 II. 18081 III. 18090 IV. 18080 V. 18073
option 2.
2008 * 9 + 8+1 = 18081
Q. let n=999......99 be an integer consisting of a string of 2009 nines. Find the sum of digits of n^2.
I. 18072 II. 18081 III. 18090 IV. 18080 V. 18073
option 2? 18081? just 9*2009:)