What is the remainder when 51^203 is divided by 7?

Question

What is the remainder when  51^{203}  is divided by 7?

(A) 4
(B) 2
(C) 1
(D) 6
(E) 5

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ScottTargetTestPrep · Accepted Answer

Since 51/7 = 7 R 2, the remainder when 51^203 is divided by 7 is the same as when 2^203 is divided by 7.

Now notice that 2^203 = (2^3)^67 x 2^2 = 8^67 x 4. Since 8/7 = 1 R 1, the remainder when 8^67 is divided by 7 is the same as when 1^67 is divided by 7. Furthermore, the remainder when 8^67 x 4 is divided by 7 is the same as when 1^67 x 4 is divided by 7. Since 1^67 x 4 = 4 and 4/7 = 0 R 4, the remainder is 4.

Answer: A

Archit3110 · Answer

51^{203}  
(7*7+2)^203
last term 2^203 
for terms raised to 2 ; the remainder of 7 is in pattern of 2^1 = 2; 2^2 = 4 ; 2^3 = 1 
the remainder of 2^203 divided by 7 would be the same as 2^7 divided by 7 (203 is a multiple of 7)
2^7 gives remainder 1 when divided by 7 ; 
IMO C

Kinshook  ; could you please check my solution , I am getting answer as 1 ..

nick1816 · Answer

51^{203}  mod 7=  2^{203}  mod 7

2^{203}=2^{3*67+2}= (2^3)^{67} .2^2

2^{203}  mod 7
=  (2^3)^{67} .2^2  mod 7
=  8^{67}*4  mod 7
= 4 mod 7

Kinshook · Answer

Asked: What is the remainder when  51^{203}  is divided by 7?

2^{203} mod 7 = 2^{3*67+2} mod 7 = 8^{67}*2^2 mod 7 = 4 mod 7

IMO A

chrtpmdr · Answer

Imho the correct answer choice is A) 
See the attached file.

CrackverbalGMAT · Answer

CONCEPT : Remainders. -Try to reach a 1 or a –1 by grouping the individual remainders. SOLUTION: Every 51 on dividing by 7 leaves a remainder of (2). Thus,203 51s will leave 203 (2)s as remainders. So, 51^203 / 7 -> (2)^203 / 7 Group three 2s to obtain a 8. On dividing this by 7 you have a 1 as remainder. 67 such groups would be created and two 2s shall be left ungrouped. Thus (2)^203/7 ---> (8^67 ) x (2) x (2) ->(4)/7 = +4 as remainder. Hope this helps. Keep studying and growing.

All the Best!

Devmitra Sen

jerryjz2690 · Answer

51^203 = (49 + 2)^203

51/7 = (49+2)/7 --> R2

51^2/7 = (49+2)^2/7 --> R4

51^3/7 = (49+2)^3/7 --> R1

51^4/7 = (49+2)^4/7 --> R2 (Beginning of New Cycle)

Remainders are in cycles of 3.
203/3 = 67 R2

2nd term on the cycle is R4
Answer A

rsrighosh · Answer

I read everybody's comments here, but I am unable to understand how to proceed. I got so many doubts now

a) how everyone reached a conclusion.... remainder cyclicity of 2 is 3..

i) I realised that 51 mod 7 is 2,  51^2  mod 7 is 4.... bit it is difficult and time consuming to find  51^3  mod 7. Although I still did that to try to understand the concept however i saw  51^3  mod 7 is 1 (which is  2^0 ). Does this mean every time in such divisions ( \frac{a^n}{b} ), "1" as a remainder is common?

ii) After this I had to use calculator to find 51^4 to understand that the value of 51^4 mod 7 is again 2... which is repeating. So is there any easier alternate method to find remainder cyclicity

b) What was the logic behind concluding we need to do this:-  \frac{2^{203}}{7}  --->  8^67   x  (2)  x  (2)  
 
How did everyone reached a conclusion that we have to use 8 here.. is it because 8>7 and remainder cannot be greater than 7?

Is there any online post which can clear the concept of such kind of concepts of remainders?

Abhishek009 · Answer

\frac{51^{203}}{7}

=  \frac{(49 + 2)^{203}}{7}

=  \frac{2^{203}}{7}    { \frac{49}{7}  will have remainder 0 }

\frac{2^3}{7}  Will have remainder as 1

\frac{2^{67*3+2}}{7}

\frac{2^2}{7}  will have remainder as  4 , Hence Answer must be (A) 4

mimishyu · Answer

53^203= (7*7+2)^203, for this polynomial, only 2^203 cannot be divided by 7 since all others will always include
7 in its term
and observe the pattern of the reminder to the exponentiation of 2 divided by 7 :
 2^1 / 7 reminder 2
 2^2 /7 4
 2^3 /7 1
 2^4 /7 2
 2^5 /7 4
 2^6 /7 1
the number of reminder, 2 4 1 recur three at a time
2^203 = 2^(3*67+2) and divided by 7, if we see the sequence above, the reminder must be at the place of 4 
thus ans(A)

rsrighosh · Answer

Today after 4 months I found a post on such remainder question. So here to answering myself

\frac{51^{203}}{7} can be written as \frac{(49+2)^{203}}{7} Now with binomial concept we will be left with \frac{2^{203}}{7} while other terms will all be divisible by 7. Now checking the remainder cyclicity 2 mod 7 = 2 -----> 3k+1 2^2 mod 7 = 4 --> 3k+2 2^3 mod 7 = 1 --> 3k Take consecutive powers of 2 (e.g. 2^1,2^2,2^3 ) until you find the remainder of 1. That will tell you what the cycle is (e.g. it could be q cycle of 3,4,5 etc). Then, once you figure that out, divided the large exponent by that result. So we see that the cyclicity is 2^{3k+p} ( 3 ) So, \frac{2^{203}}{7} can be calculated as follows 203 mod 3 is 2 Hence remainder is 4 Answer: A

ThatDudeKnows · Answer

51^{203} = (49+2)^{203}

49^203 is a multiple of 7 so we can back that out.

2^{203} = 2^{201}*2^2 = (2^3)^{67}*2^2 = 8^{67}*2^2 = (7+1)^{67}*2^2

7^67 is a multiple of 7 so we can back that out.

1^{67}*4 = 4

Answer choice A.

findingmyself · Answer

Here is my answer:

51^{203}= (49+2)^{203} 
Expanding this equation in  (a+b)^2  format
---> (49^{203})+(2*49*2)+(2^{203})

Term 1 and 2 are easily seen they can be divided perfectly by 7, leaving only  2^{203} .
Remainder when  2^{203}  is divided by 7 can be found as follows:

2 divided by 7--> Remainder=2
 2^2  divided by 7--> Remainder=4
 2^3  divided by 7--> Remainder=1
 2^4  divided by 7--> Remainder=2

Cyclicity: (2,4,1). Now divide 203/3. Why divided by 3: Because the remainder pattern is rotating every 3 times. 
203/3-->67 times 3 and remainder 2. Thus 2nd element in the pattern is the remainder which is "4"

What is the remainder when 51^203 is divided by 7?

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What is the remainder when 51^203 is divided by 7?

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