Is the remainder is 0; when 7^m + 1 is divided by 5. a.

Question

Is the remainder is 0; when 7^m + 1 is divided by 5.
a. m=4a+2
b. m = 3a +1

a. Unit digit is always 0 for - therefore remainder is 0 Sufficient
b. Unit digit could be 2,4,0 - not sufficient

Matador · Answer

please explain your answer...

rdangol · Answer

Plugged in 1,2,3,4 or a and b to derive m.

Is there a quicker way to solve this?

conocieur · Answer

Is the remainder 0; when 7^m + 1 is divided by 5 ? 

In that case we have to start working with 7^m + 1 and a progression of the last digit

7^1 = 7 
7^2 = 49
for 7^3, last digit of 7^2 = 9 then 9*7 = 63 so is 3
for 7^4, last digit of 7^3 = 3 then 3*7 = 21 so is 7
we know for 7^5 last digit would be 7 again.

so we can see that it works for

2, 6, 10, 12, ....

which is actually exactly the sequence 4a+2
from n=0 to infinity

Dilshod · Answer

Hi,

What is OA?

Please, be careful while reading the question. There is not a word regarding the property of m: we don't know whether it is integer or fraction. In fact, it can be smth like 1212/2312.

From st.1 alone we can't get that informtaion, so insuff. Same for st2.
Thus IMO it is C.

Nayan · Answer

hi Dilshod,
 How's C suffient to know that m is not a fraction?

What's the OA?

gmatacer · Answer

i did it this way ...

4a+2 = 3a + 1 => a= = -1. SUbstituting it does not return 0. There is now way remainder can be 0 so E

Dilshod · Answer

Exactly for this reason the ans should be C

Charlie45 · Answer

Ditto

Yurik79 · Answer

4a+2 = 3a + 1 => a= = -1
substituting for m {(7^-2)+1}/5=(1/49+1)/5=(50/49)/5=10/49
C it is!I was wrong with E

Tyr · Answer

You do it like this.
Look at last digits - they will cycle
last_digit(7^0) = 1
last_digit(7^1) = 7
last_digit(7^2) = 9
last_digit(7^3) = 3
last_digit(7^4) = 1
last_digit(7^5) = 7
...
If (m mod 4) = 0 then last_digit(7^m + 1) = 2
If (m mod 4) = 1 then last_digit(7^m + 1) = 8
If (m mod 4) = 2 then last_digit(7^m + 1) = 0
If (m mod 4) = 3 then last_digit(7^m + 1) = 4 

Now

1) m=4a+2 gives us (m mod 4) = 2 => as we noted last_digit(7^m + 1) = 0 => the number WILL be divisable by 5 => the remainder WILL be 0

2) m=3a+1 does not give us any information as far as what (m mod 4) will be => Out
a = 1; m = 4; (m mod 4) = 0
a = 2; m = 7; (m mod 4) = 3
a = 3; m = 10; (m mod 4) = 2
...

The answer is  A

ywilfred · Answer

a) m = 4a+2

If a = 1, m = 7, and 7^6 + 1 = 7^6 + 1 --> Divisible by 5
but if a = 1/8, m = 5/2 and 7^(5/2) + 1 is not divisible by 5.

Not sufficient.

b) m = 3a + 1
If a = 1, m = 4, and 7^4 + 1 is not divisible by 5.
If a = 2, m = 7 and 7^7 + 1 is not divisible by 5
If a = 3, m = 10 and 7^7 + 1 is divisible by 10.

Not sufficient.

Using 1) and 2)
m = 4a + 2
m = 3a + 1

4a + 2 = 3a + 1 -> a = -1. m is only 1 value, -2. So we can figure out the remainder.

Ans C

Angela780 · Answer

Yes Yurik, I have C also. I used the exact same method

Tyr · Answer

Author most likely missed that m is a positive integer.

Is the remainder is 0; when 7^m + 1 is divided by 5. a.

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards