Find the remainder when 65203 is divided by 7.
A. 4
B. 2
C. 1
D. 6
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & \frac{{{{65}^{203}}}}{7} \cr & Or,\,\frac{{{{\left( {63 + 2} \right)}^{203}}}}{7} \cr & 63\,{\text{is}}\,{\text{divisible}}\,{\text{by}}\,7, \cr & {\text{so}}\,{\text{remainder}}\,{\text{will}}\,{\text{depend}}\,{\text{on}}\,{\text{the}}\,{\text{powers}}\,{\text{of}}\,2 \cr & Or,\,\frac{{{2^{203}}}}{7} \cr & {\text{Its}}\,{\text{remainder}}\,{\text{would}}\,{\text{be}}\,{\text{same}}\,{\text{as}} \cr & \frac{{{2^3}}}{7} \cr & {\text{Now,}}\,{\text{Required}}\,{\text{Remainder}}\,{\text{wold}}\,{\text{be}}, \cr & \frac{8}{7} = 1\cr & {\text{Required}}\,{\text{remainder}} = 1 \cr} $$Note: We have manipulated the powers in the form of (4x + n). It means 203 is taken as,
203 = 4x + n = 4 × 50 + 3.
We neglect power which is in the multiple of 4.
This Question Belongs to Arithmetic Ability >> Number System
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Comments ( 9 )
Related Questions on Number System
Three numbers are in ratio 1 : 2 : 3 and HCF is 12. The numbers are:
A. 12, 24, 36
B. 11, 22, 33
C. 12, 24, 32
D. 5, 10, 15
Remainder is 4
Why giving wrong ans
65^203/7=?
Procedure:
65^1/7=2(remainder)
65^2/7=2*2(remainder of previous)/7=4(remainder)
65^3/7=4(previous remainder)*2(first remainder)/7=1 stop the process because it's repeating....
No of terms in cycle=4
Power of number=203
R[203/4]=R[2] =2th term from the cycle is 4
So, remainder is 4
So option 1 is correct 💯💯
marked wrong... remainder will be 4
the answer should be 4.Option A
4 is the remainder
after (2^203)/7
((2^3)^67*2^2)/7= ((7+1)^67*4)/7
1^67*4/7=4/7
remainder =4
It should be 4
65^203=(63+2)^203
Since 63 is a multiple of 7, we need to find the remainder when 2^203 is divided by 7
2^n for n = natural numbers is 2,4,8,16,32,64,128,...
When the above is divide by 7, we get a remainder of 2,4,1,2,4,1,2,...
Notice that the remainder follows a cycle which change at every 4th power of 2
Now, 2^203=2^(3∗67+2)
Hence the remainder is 4
Since 7 is prime, we shall use Fermat’s Little Theorem.
65^6 == 1 mod 7; hence, (〖65〗^6 )^33=6^198==1 mod 7.
But, 6 == -1 mod 7; hence, 6^5==-1 mod 7.
Hence, 65^203 == -1 mod 7 = 6 mod 7
Answer: Option C
By Sanjay Mohan Bhatnagar (www.mathsacad99.com)
65^203/7
Or,(63+2)203/7 => 2^203/7
And 203/6 = 5
So, 2^5/7 = 32/7 ==> 4 as remainder.