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Re: If a and b are positive odd integers less [#permalink]
niks18 wrote:
So we don't need the statements actually.

What am I missing here?


Yes we don't need the statements to find out the values and since each statement gives us a unique value of the unit digit not being 6

(D) will be the answer
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Re: If a and b are positive odd integers less [#permalink]
niks18 wrote:

So we don't need the statements actually.

What am I missing here?


I agree with you, the question should have been worded differently imo

knowing a is odd we don't actually need them but we need to go with D i guess what do you think?
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Re: If a and b are positive odd integers less [#permalink]
Hatakekakashi wrote:
niks18 wrote:

So we don't need the statements actually.

What am I missing here?


I agree with you, the question should have been worded differently imo

knowing a is odd we don't actually need them but we need to go with D i guess what do you think?


Hi Hatakekakashi

I think the question is flawed.

Hi Bunuel,

can you provide some clarity.
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Re: If a and b are positive odd integers less [#permalink]
niks18 wrote:

Hi Hatakekakashi

I think the question is flawed.

Hi Bunuel,

can you provide some clarity.


Hi,
yeah Q is flawed


Regards,
HK
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Re: If a and b are positive odd integers less [#permalink]
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EgmatQuantExpert wrote:

Question:



If \(a\) and \(b\) are positive odd integers less than \(30\) and \(Q=7^a+ 5* 3^b\), is the units digit of \(Q\) equal to \(6\)?

1. \(a\) is the square of a prime number.
2. \(a-b=2\)

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) Statement (1) and (2) TOGETHER are NOT sufficient.



Hi..

if positive odd integers is changed to positive even integers, it will be a properly worded question. In present way it is flawed as you don't require any statements..
If it is even, then ans is A
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Re: If a and b are positive odd integers less [#permalink]
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Solution:



We are given:
    \(a\) and \(b\) are positive integers and,
We need to find:
    Units digit of \(7^a+ 5× 3^b\).
Per our conceptual understanding, we know.
when \(5\) is multiplied by an odd number, the units-digit of resultant number becomes \(5\).
Therefore,
    Units digit of \(5× 3^b\) is always equal to \(5\), regardless of the value of \(b\).
Hence, we only need to find the value of \(a\).
Now, for \(7^a+ 5× 3^b\) to end in \(6\), \(7^a\) must end in \(1\) OR a must be of the form \(4k\) for \(7^a\) to give units digit as \(1\).
Let’s now analyse the statements one by one.
Statement 1:
“a is the square of an odd prime number”

The possible values of a less than \(30\) are \(9\) and \(25\).
    \(9= 4*2+1\) \((k=2)\)
    \(25= 4*6+1\) \((k=6)\)

Hence, \(a\) is not of the form \(4k\). Therefore, units digit of \(7^a+ 5× 3^b\) is not \(6\).
Therefore, Statement 1 alone is sufficient to answer the question.
Statement 2:
“a-b=2”
We are given, \(a\) is less than \(30\). Till \(30,\) there are many possible values of \(a\) and \(b\) such that the difference between \(a\) and \(b\) is \(2\).

Thus, from this statement, we cannot be certain that the units digits of \(7^a+ 5× 3^b\) is \(6\) or not.

Therefore, we are getting answer by statement 1 ALONE only but not by statement 2.
Answer: Option A

Originally posted by EgmatQuantExpert on 28 Feb 2018, 12:29.
Last edited by EgmatQuantExpert on 22 Mar 2018, 11:41, edited 1 time in total.
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Re: If a and b are positive odd integers less [#permalink]
EgmatQuantExpert wrote:

Solution:



We are given:
    \(a\) and \(b\) are positive integers and,
We need to find:
    Units digit of \(7^a+ 5× 3^b\).
Per our conceptual understanding, we know.
when \(5\) is multiplied by an odd number, the units-digit of resultant number becomes \(5\).
Therefore,
    Units digit of \(5× 3^b\) is always equal to \(5\), regardless of the value of \(b\).
Hence, we only need to find the value of \(a\).
Now, for \(7^a+ 5× 3^b\) to end in \(6\), \(7^a\) must end in \(1\) OR a must be of the form \(4k\) for \(7^a\) to give units digit as \(1\).
Let’s now analyse the statements one by one.
Statement 1:
“a is the square of a prime number”

The possible values of a less than \(30\) are \(4,9\) and \(25\). However, \(a\) is an odd integer. Thus, the possible values of \(a\) are \(9\) and \(25\).
    \(9= 4*2+1\) \((K=2)\)
    \(25= 4*6+1\) \((k=6)\)

Hence, \(a\) is not of the form \(4k\). Therefore, units digit of \(7^a+ 5× 3^b\) is not \(6\).
Therefore, Statement 1 alone is sufficient to answer the question.
Statement 2:
“a-b=2”
We are given, \(a\) is less than \(30\). Till \(30,\) there are many possible values of \(a\) and \(b\) such that the difference between \(a\) and \(b\) is \(2\).

Thus, from this statement, we cannot be certain that the units digits of \(7^a+ 5× 3^b\) is \(6\) or not.

Therefore, we are getting answer by statement 1 ALONE only but not by statement 2.
Answer: Option A


Hi EgmatQuantExpert

based on your solution, the question needs to be corrected. So instead of a & b being positive odd integers, question should mention a & b as positive integers only.
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Re: If a and b are positive odd integers less [#permalink]
Imo it is D.
In statement 2 For a-b = 2, a>b and they will always be consecutive multiple.
7^27 + 5^25 gives Unit digit as 2 always . Therefore using statement 2 we can sufficiently say the answer is NO .
Statement A as well results in an Absolute No.
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Re: If a and b are positive odd integers less [#permalink]
Jerry1982 wrote:
Imo it is D.
In statement 2 For a-b = 2, a>b and they will always be consecutive multiple.
7^27 + 5^25 gives Unit digit as 2 always . Therefore using statement 2 we can sufficiently say the answer is NO .
Statement A as well results in an Absolute No.


Hello

In the present form of question the statements are actually not needed as Niks has explained in his post here:
https://gmatclub.com/forum/if-a-and-b-a ... l#p2022738

So you are correct that as per second statement also the unit's digit will never be 6. But one thing I would like to add is that as per second statement, the unit's digit will NOT always be '2'. It can be '8' also.
Eg., if we take a=3 and b=1, then 7^3 + 5*3^1, will give a unit's digit of '8'. So the unit's digit can be either 2 or 8 but never 6.
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Re: If a and b are positive odd integers less [#permalink]
EgmatQuantExpert wrote:

Solution:



We are given:
    \(a\) and \(b\) are positive integers and,
We need to find:
    Units digit of \(7^a+ 5× 3^b\).
Per our conceptual understanding, we know.
when \(5\) is multiplied by an odd number, the units-digit of resultant number becomes \(5\).
Therefore,
    Units digit of \(5× 3^b\) is always equal to \(5\), regardless of the value of \(b\).
Hence, we only need to find the value of \(a\).
Now, for \(7^a+ 5× 3^b\) to end in \(6\), \(7^a\) must end in \(1\) OR a must be of the form \(4k\) for \(7^a\) to give units digit as \(1\).
Let’s now analyse the statements one by one.
Statement 1:
“a is the square of a prime number”

The possible values of a less than \(30\) are \(4,9\) and \(25\). However, \(a\) is an odd integer. Thus, the possible values of \(a\) are \(9\) and \(25\).
    \(9= 4*2+1\) \((K=2)\)
    \(25= 4*6+1\) \((k=6)\)

Hence, \(a\) is not of the form \(4k\). Therefore, units digit of \(7^a+ 5× 3^b\) is not \(6\).
Therefore, Statement 1 alone is sufficient to answer the question.
Statement 2:
“a-b=2”
We are given, \(a\) is less than \(30\). Till \(30,\) there are many possible values of \(a\) and \(b\) such that the difference between \(a\) and \(b\) is \(2\).

Thus, from this statement, we cannot be certain that the units digits of \(7^a+ 5× 3^b\) is \(6\) or not.

Therefore, we are getting answer by statement 1 ALONE only but not by statement 2.
Answer: Option A





a and b are both "odd positive integers <30"

so if you put b=1, a will be 3. If a will be 3, 7^a will have unit have unit 3.
If you put any value of b( odd value<30), the unit digit value of 7^a will be either 7 or 3.
Also 3^b will always yield an odd value and that odd value when multiplied by 5 will give result in a number of unit digit 5.
so Q can never have an unit digit 6.

IMO answer D
Bunuel plz comment
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Re: If a and b are positive odd integers less [#permalink]
Expert Reply
niks18 wrote:
EgmatQuantExpert wrote:

Question:



If \(a\) and \(b\) are positive odd integers less than \(30\) and \(Q=7^a+ 5* 3^b\), is the units digit of \(Q\) equal to \(6\)?
1. \(a\) is the square of a prime number.
2. \(a-b=2\)

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) Statement (1) and (2) TOGETHER are NOT sufficient.


hi EgmatQuantExpert

\(Q=7^a+ 5* 3^b\), now \(5*3^b\) will have unit's digit of 5 irrespective of value of \(b\) because \(b\) is an odd integer, so if \(a=4k\), then \(7^a\) will have unit's digit of 1 and hence the expression will have a unit's digit of \(1+5=6\).

now it is given that \(a\) is odd integer, so \(a\) will never be of the form 4k. hence the expression will not have a unit's digit of 6.

So we don't need the statements actually.

What am I missing here?


Hey Everyone,

Niks is right, there was a typo in the question because of which none of the statements were required.

We have edited the question.

Apologies for the inconvenience

Thanks,

E-gmat

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Re: If a and b are positive odd integers less [#permalink]
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