If a and b are positive odd integers less

Question

e-GMAT Question:

If  a  and  b  are positive 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 an odd 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.

This is  Question 5 of  The e-GMAT Number Properties Marathon

Go to   Question 6 of the Marathon

Hatakekakashi · Answer

Now we know 5xeven ends in 0 i.e 5 x0 , 5x2 , 5x4 all end in unit digit 0
5xodd ends in 5 i,e 5x1, 5x3, 5x5 all end in unit digit 5

now 7^k (k = 1 to infinity) has a cycle of 4 for its unit digit i.e 7,49,343,2401 etc

thus for  Q=7^a+ 5* 3^b  all we need to find if a is 4/multiple of 4 because for a unit digit of 6 ~ 1+5 is needed according to the given Q value as 5x3^b ~ 5xodd always ends in 5

now we need to know if 7^a ends in 1 / a is a multiple of 4

1) a is the square of prime

a can't be 4 as a is odd positive

sufficient

2) a -b = 2

a and b both are odd 
a can't be 4/multiple of 4

thus

sufficient

(D) imo

niks18 · Answer

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?

Hatakekakashi · Answer

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

Hatakekakashi · Answer

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?

niks18 · Answer

Hi  Hatakekakashi

I think the question is flawed.

Hi  Bunuel ,

can you provide some clarity.

Hatakekakashi · Answer

Hi,
yeah Q is flawed

Regards,
HK

chetan2u · Answer

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

EgmatQuantExpert · Answer

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

niks18 · Answer

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.

Jerry1982 · Answer

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.

amanvermagmat · Answer

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.

varundixitmro2512 · Answer

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

EgmatQuantExpert · Answer

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

If a and b are positive odd integers less

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