While shifting his departmental store, Mr. Trump found the

Question

While shifting his departmental store, Mr. Trump found the number of articles in his store to be 7^10. He had 8 rooms each of equal capacity to store these articles. If at the end he was left with n articles for which he had no space, which of the following could the minimum possible value of n?
A) 0
B) 1
C) 2
D) 3
E) 4

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MahmoudFawzy · Answer

the question is simply asking about the remainder of  \frac{7^{10}}{8}

7^{10}  can be rewritten as  (8-1)^{10} , which can be simplified to  (-1)^{10}  , which is equal to 1 -->  B

(as I know: source of question is Jamboree)

globaldesi · Answer

last digit of 7^10 will be 9
and as per question
8*x=n = 7^10
the nearest values of n can be 1
hence B

Abhi077 · Answer

This question tests our knowledge about recognizing pattern. 7^1/8 gives us a remainder of 7, 7^2/8 gives us a remainder of 1, and so on we get a pattern of 7 1 7 1. for every even digit of the exponent we get a 1, hence the minimum value should be 1

prachilulla · Answer

You can think of this logically rather than mathematically:
if you have 7^10 articles and they need to be divided equally in 8 rooms.... you can think as 7^8 articles were equally divided in 8 rooms. The remaining 7^2 i.e. 49 still remain... with a remainder of 1, 48 (6*8= 48) articles can still be divided equally among the 8 rooms . Hence B

chrtpmdr · Answer

Solved it with divisibility, cycle of 7 is:

7
49
343
2401

Possible Remainders when divided by 8 are either 7 or 1, as 1 is the only answer choice available -> 1

CAMANISHPARMAR · Answer

The question is what is the remainder when 7^10 is divided by 8.

7^1 is 7 and 7/8 leaves a remainder of 7.
7^2 is 49 and 49/8 leaves a remainder of 1
7^3 is 343 and 343/8 leaves a remainder of 7

continuing so on.... when 7 is raised to an even number and divided by 8; the remainder is 1 and when 7 is raised to an odd number and divided by 8 the remainder is 7. Hence the correct answer is the remainder of 1 as 7 is raised to power 10.

The correct answer is option B

msrivas2 · Answer

use the cyclicity of 7 :
(7,9,3,1)

7^1 = 7
7^2 = 49
7^3 = 693 
7^1 = 4851
i.e. take the last digits until the cycle repeats again with starting number.

Now we have 8 rooms so start counting from the cyclicity in (7,9,3,1) till 8 (startover the counting from begining) and we will have the answer 1 which is the remainder.

HowdyPartner · Answer

Use the units digit rule: Units digits are as follows: 7^1 = 7, 7^2 = 9, 7^3 = 3 7^4 = 1 and then repeats --> From this we, need to count up to 10 to see where we are, and there is a remainder 2 -- so we are at 9 units digit. When 9 is divided by 8, there is 1 digit left over. So B is our answer here for the minimum possible value of n.

Abhishek009 · Answer

\frac{7}{8} = 7 
 \frac{7^2}{8} = 1 
 \frac{7^3}{8} = 7

Thus, a pattern is repeated,  \frac{7^{Even}}{8}  = Remainer 1

Thus,  \frac{7^{10}}{8} = 1 , Answer will be (B)

IanStewart · Answer

Just in case some readers might find some of the solutions misleading: the several solutions here that rely on units digits are not correct. We're dividing by 8, and from a units digit alone, you cannot predict what remainder you will get when you divide by 8. It's true that the units digit of 7^10 is 9, but if you think of what remainders you get when you divide 9, 19, 29, and 39 by 8, you can see you can get any odd remainder at all (from 1 through 7). Units digits would be relevant if we were dividing by 10, but we aren't doing that here. The solutions which investigate the remainders pattern that you get when you divide successive powers of 7 by 8 are all good solutions, as are the ones that rely on modular arithmetic (which you don't really need on the GMAT, but which is used in some of the fastest solutions above).

I'd add that the question itself doesn't make any sense. Assuming we're dividing 7^10 by 8 (something the question doesn't even actually say), there is no "minimum possible value" of the remainder n; n has a single well-defined value. We aren't minimizing anything here.

I'd add that the question itself doesn't make any sense. Assuming we're dividing 7^10 by 8 (something the question doesn't even actually say), there is no "minimum possible value" of the remainder n; n has a single well-defined value. We aren't minimizing anything here.

luisdicampo · Answer

Deconstructing the Question  
We are given  7^{10}  total articles and  8  rooms of equal capacity.

If each room holds  k  articles, then the total stored is  8k  and the leftover is  n , so
 7^{10}=8k+n  with  0\le n<8 .

Therefore  n  is the remainder when  7^{10}  is divided by  8 .

Step-by-step  
Compute modulo  8 :
 7\equiv -1 \pmod{8} .

Raise to the 10th power:
 7^{10}\equiv (-1)^{10}\pmod{8} .

Since the exponent is even,
 (-1)^{10}=1 ,
so
 7^{10}\equiv 1\pmod{8} .

Thus the leftover is
 n=1 .

Answer: B

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