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Re: units digit [#permalink]
I think its C. Whats OA?
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Re: units digit [#permalink]
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sm021984 wrote:
What is the units digit of z if (385a)(225b) = z, where a and b are positive integers?

1) a = 3
2) a + b = 11


The unit's digit of z will be decided by the units digit of the two numbers.

Statement 1: a = 3
(3853)(225b) = z
Depending on what b is, the unit's digit of z can take different values.
If b = 1, unit's digit of z is 3.
If b = 2, unit's digit of z is 6.
etc
Not sufficient

Statement 2: a + b = 11
Again, there are many possible cases:
a = 2, b = 9, unit's digit of z is 8.
a = 3, b = 8, unit's digit of z is 4.
etc

Taking both together, we know a = 3 and b = 8. So unit's digit of z must be 4.

An interesting variation of this question could be:

What is the units digit of z if (385*a)(225*b) = z, where a and b are positive integers?

1) a = 3
2) a + b = 11

Try this one.
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Re: units digit [#permalink]
answer would still be C.
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Re: units digit [#permalink]
Thanks Karishma ......
Similar to the variation suggested by you , i came across a question which is

If z = 25x125y, what is the value of z ?

1) y = 2

2) x=-3y/2

This confuses me ....
Can x or y have a negative value in this context?
If so, kindly explain the concept.
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Re: units digit [#permalink]
VeritasPrepKarishma wrote:
sm021984 wrote:
What is the units digit of z if (385a)(225b) = z, where a and b are positive integers?

1) a = 3
2) a + b = 11


The unit's digit of z will be decided by the units digit of the two numbers.

Statement 1: a = 3
(3853)(225b) = z
Depending on what b is, the unit's digit of z can take different values.
If b = 1, unit's digit of z is 3.
If b = 2, unit's digit of z is 6.
etc
Not sufficient

Statement 2: a + b = 11
Again, there are many possible cases:
a = 2, b = 9, unit's digit of z is 8.
a = 3, b = 8, unit's digit of z is 4.
etc

Taking both together, we know a = 3 and b = 8. So unit's digit of z must be 4.

An interesting variation of this question could be:

What is the units digit of z if (385*a)(225*b) = z, where a and b are positive integers?

1) a = 3
2) a + b = 11

Try this one.


What is the units digit of z if (385*a)(225*b) = z, where a and b are positive integers?

1) a = 3
2) a + b = 11

-->
This should be B right.

My Explanation(ME not OE ;))
385*a = last digit is 5 if a = odd, 0 if a = even
225*b = last digit is 5 if b = odd, 0 if b = even
1) Tells me a = odd, but nothing about b
So b =odd/even ==> Last digit = 0/5
Not sufficient
2) a+b=11 ==> one will be odd and the other even because 11 is odd
sum of an (odd and even)/(odd and odd) only gives an odd number
So answer will be 0 in all cases
So Sufficient

My Answer(MA should be OA 8-) ):
B
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Re: units digit [#permalink]
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tt2011 wrote:
answer would still be C.


Actually in this case answer would not be C.

Look carefully...
What is the units digit of z if (385*a)(225*b) = z, where a and b are positive integers?

1) a = 3
2) a + b = 11

The given equation is (385*a)(225*b) = z
The unit's digit of z is the unit's digit of 385*a*225*b i.e. the unit's digit of 385*225*a*b. Both 385 and 225 end with a 5 and their multiplication gives us the unit's digit of 5. Now the unit's digit of z depends on whether a*b is even on odd. If a*b is even, the unit's digit of z will be 0. If a*b is odd, the unit's digit will be 5. Hence all we need to know is whether one of a and b is even.

Statement 1 tells us that a is odd. We still don't know about b so not sufficient.

Statement 2 tells us that a+b = odd. We know that Even + Odd = Odd. We cannot get an odd by combining two odds or two evens. Hence one of a and b is even. Therefore, the unit's digit of z must be 0. Sufficient.

Answer B. (That is why I said it is a more interesting variation!)

Similarly, if the question instead read: "What is the units digit of z if (385*a)(226*b) = z, where a and b are positive integers?", you don't need either statement. you can directly say that the unit's digit will be 0. Of course this isn't allowed in the GMAT format but generally speaking...
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Re: units digit [#permalink]
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sm021984 wrote:
Thanks Karishma ......
Similar to the variation suggested by you , i came across a question which is

If z = 25x125y, what is the value of z ?

1) y = 2

2) x=-3y/2

This confuses me ....
Can x or y have a negative value in this context?
If so, kindly explain the concept.


Given z = 25x125y, I would assume here that x and y are digits from 0 to 9. The negative sign doesn't make sense except if y = 0 giving us x = 0 too in which case statement 2 is sufficient. But statement 1 contradicts statement 2 which is not acceptable. Hence, either there is a typo somewhere or it is a bad question.

and krishp, you got the correct solution. I wouldn't have posted mine had I seen yours before posting because your solution covers the logic sufficiently. Good work!
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Re: units digit [#permalink]
Thanks Karishma for the variation question.
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Re: units digit [#permalink]
VeritasPrepKarishma wrote:
sm021984 wrote:
Thanks Karishma ......
Similar to the variation suggested by you , i came across a question which is

If z = 25x125y, what is the value of z ?

1) y = 2

2) x=-3y/2

This confuses me ....
Can x or y have a negative value in this context?
If so, kindly explain the concept.


Given z = 25x125y, I would assume here that x and y are digits from 0 to 9. The negative sign doesn't make sense except if y = 0 giving us x = 0 too in which case statement 2 is sufficient. But statement 1 contradicts statement 2 which is not acceptable. Hence, either there is a typo somewhere or it is a bad question.

and krishp, you got the correct solution. I wouldn't have posted mine had I seen yours before posting because your solution covers the logic sufficiently. Good work!


Karishma - That is fine. It is good that you posted your solution because after posting my solution, I remembered that it would be easy to multiply the 2 digits ending with 5s because then we will have to deal only with one 5 and then even/odd. But was very sleepy then.

Also I wrongly mentioned that every odd is sum of (odd and odd). This is wrong (My bad - difficult to write down everything :( when posting, though I knew this)
Not sure if anyone even noticed these 2 flaws : 1 in my approach, 2 in my explanation
So good that you posted your solution because this was exactly what I was going to post today.



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