Data Sufficiency

If M and N are integers,is (10^M+N)/3 an integer?

1.N=5

2.MN is even

Any integer power of 10 when divided by 3 the remainder will be 1. so there will be always a case of fraction. How can we see the options as we have a definite answer before going to options?

Hence IMO D Not sure

shovan85 wrote:Any integer power of 10 when divided by 3 the remainder will be 1. so there will be always a case of fraction. How can we see the options as we have a definite answer before going to options?

Hence IMO D Not sure

OA is E
I marked A

Because i think any power of 10 will be like 10, 100, 1000, 10000 ......

now if we add 5 to it even then it will be divisible by 3 like 105, 1005, 10005..

If MN is even than we can't figure out much because it can be even * even or even * odd.
so thought its not enough.

I am not able to understand why answer is not A

goyalsau wrote:

shovan85 wrote:Any integer power of 10 when divided by 3 the remainder will be 1. so there will be always a case of fraction. How can we see the options as we have a definite answer before going to options?

Hence IMO D Not sure

OA is E
I marked A

Because i think any power of 10 will be like 10, 100, 1000, 10000 ......

now if we add 5 to it even then it will be divisible by 3 like 105, 1005, 10005..

If MN is even than we can't figure out much because it can be even * even or even * odd.
so thought its not enough.

I am not able to understand why answer is not A

Oh My God!! I have taken it as 10^(M+N) not (10^M)+N..... LOL thats why it was not making sense

goyalsau wrote:

shovan85 wrote:Any integer power of 10 when divided by 3 the remainder will be 1. so there will be always a case of fraction. How can we see the options as we have a definite answer before going to options?

Hence IMO D Not sure

OA is E
I marked A

Because i think any power of 10 will be like 10, 100, 1000, 10000 ......

now if we add 5 to it even then it will be divisible by 3 like 105, 1005, 10005..

If MN is even than we can't figure out much because it can be even * even or even * odd.
so thought its not enough.

I am not able to understand why answer is not A

U r right!! Be sure u do it from an official source.

question must be written properly.

It is actually ((10^M)+N)/3 is integer?

IMO answer should be A.

goyalsau wrote:If M and N are integers,is (10^M+N)/3 an integer?

1.N=5

2.MN is even

goyalsau wrote:If M and N are integers,is (10^M+N)/3 an integer?

1.N=5

2.MN is even

I think it is a pretty good question with all its traps. I DO NOT think it should be written with any more or any less sighs, brackets etc....why? Well one should know that 10^m has to be solved before adding ........

1 - (10^m+n)/3, always try to break the fraction and see the individual components. 10^m/3 + n/3 will it be an integer? we know that 10^m will always give us a remainder of 1. if n = 5 we are done, the remainder for 10^m+n = 0.

2 - mn even does not tell anything. we could have m = 2, n = 2 remainder is 0. m = 3 n = 4, remainder not zero........

Hence A solves it.

It has to be option E:

It says just integers may be positive or negative.
Option 1 works fine for positive integers and for the other scenario it doesn't

Option 2 can be eliminated as discussed in the above posts.

narik11 wrote:It has to be option E:

It says just integers may be positive or negative.
Option 1 works fine for positive integers and for the other scenario it doesn't

Option 2 can be eliminated as discussed in the above posts.

Excellent point!

Agreed it is E.

goyalsau wrote:If M and N are integers,is (10^M+N)/3 an integer?

1.N=5

2.MN is even

i hate it when you guy dont post precise question
is it (10^M)+N or 10^(M+N)??????

diebeatsthegmat wrote:

goyalsau wrote:If M and N are integers,is (10^M+N)/3 an integer?

1.N=5

2.MN is even

i hate it when you guy dont post precise question
is it (10^M)+N or 10^(M+N)??????

It is neither!

If M and N are integers,is (10^M+N)/3 an integer?

hence solve 10^m first add n to it and then finally divide the whole by 3 OR break the fraction apart which will read as 10^m/3 + n/3 again by basic math rules we will solve the exponent first then divide by 3 and solve n/3 first and then add the two parts.

The question is clear!

Cheers

The issue is that the question was not written correctly.

(10^M)+N and 10^(M+N) are VERY different.

uwhusky wrote:The issue is that the question was not written correctly.

(10^M)+N and 10^(M+N) are VERY different.

As shown in posts above the questions is -

(10^M+N)/3

I do not see any confusion with that.

The question is neither of (10^M)+N nor 10^(M+N)

We can simply apply PEMDAS - Parenthesis > Exponents > Multiplication > Division > Addition > Subtraction

to solve this clinically......

Anyways I am thick or something, but that is how I have been doing maths all my life, and I have been pretty good so far

uwhusky wrote:Since GMAT isn't written in symbols, whether this question was written correctly is besides the point. Obviously there's some confusion whether it's 10 ^(M+N) or (10^M) + N, and I guess you knew all along. Whoopee for you!

I am not trying to get in an argument here, but was perplexed as to how we (many people) in the group could not understand the question as is.

As for the part about GMAT not being in symbols - there are so many questions that use symbols on this forum and I do not remember any of them being an issue earlier. I still do not know how to superscript / subscript numbers, variables on the forum.

Cheers

narik11 wrote:It has to be option E:

It says just integers may be positive or negative.
Option 1 works fine for positive integers and for the other scenario it doesn't

Option 2 can be eliminated as discussed in the above posts.

sharp thinking...
It is really important to recall the basics,
You did it pretty well.

Thanks..