I will try to explain euler's theorem: EULERS'S THEOREM : given: to find remainder when M^a is divided by N NOTE: euler's theorem applicable only if M and N are co-prime to each other(or have no factors in common in oder words) the theorem says that the remainder of M^K wen divided by N is always 1 where k= N (1-(1/x) (1-(1/y) (1-(1/z))...here x,y,z are the primes u obtain wen u write N in standard form
to simplify things i will take ur question: 7^103 wen divided by 100. now 1st step is to check wedr M and N are co-prime to each other. 7 and 100 ar co-prime. So, i can safely apply euler's theorem 2nd step :find "k".. 100=(2^2) x (5^2) so k = 100 (1-(1/2) (1-(1/5)= 40 thus EULER'S THEOREM says that remainder of 7^40 wen divided by 100 is 1, or R((7^40)/100) =1
now question is R((7^103)/100)=R[ (7^40)(7^40)(7^23) /100 ]=R [7^23/100] = 43 (since R(7^40) (7^40) / 100) = 1 X 1=1
hope the explanation helps
#dreamer-on-the-rocks#
do you solve manually or is there a quicker way to solve the last part ie [7^23/100]?
do you solve manually or is there a quicker way to solve the last part ie ?
ok this part is reaallllyy easy......7 is d only 1 digit no. whose last 2 digits repeat in a cycle (which happens in sets of four....ie ( 07,49,43,01) ) .....thus 23%4 = 3 thus 3rd no in the set=43
ok this part is reaallllyy easy......7 is d only 1 digit no. whose last 2 digits repeat in a cycle (which happens in sets of four....ie ( 07,49,43,01) ) .....thus 23%4 = 3 thus 3rd no in the set=43
#dreamer-on-the-rocks#
thanks a lot...my mind didnt jump to the cyclicity :-o
An examination consists of 160 questions. For each correct answer, as student earns 1 mark and for each wrong answer, a student loses (1/4)th of mark. Find the number of distinct net scores that a student could earn in the examination. (1) 790 (2) 797 (3) 795 (4) 801
An examination consists of 160 questions. For each correct answer, as student earns 1 mark and for each wrong answer, a student loses (1/4)th of mark. Find the number of distinct net scores that a student could earn in the examination. (1) 790 (2) 797 (3) 795 (4) 801
c?? 795???
PS: Srihari ji, thode aur bade fonts use karke page hi bhar dete...
6=2*3 the no. 12345678987654321.....999 digits will leave a remainder 2 by 3(since sum of digits =81*58+71,this comes from the chain of 12345678987654321 as 17 digits,n 999 digits wud end as 2345678987656,13 digits wich is d remainder wen 999/17) also will give a remainder of 0 wid 2... so d remainder will b of d form 3m+2=2n so d answer wud be 2... am i correct,,plz correct me if m not 😃
Originally Posted by Srihari123 View Post An examination consists of 160 questions. For each correct answer, as student earns 1 mark and for each wrong answer, a student loses (1/4)th of mark. Find the number of distinct net scores that a student could earn in the examination. (1) 790 (2) 797 (3) 795 (4) 801 i think d answer wud be 795 coz out of -40 to 160 marks,,he cannot get just 159.25,159.5 and 159.75,158.25,158.5,157.25 so 801(-40 to 160=200 nos wid each having 4 multiples of 1/4,including both -40 n 160) - 6 nos=795... correct me if m wrong
Originally Posted by Srihari123 View Post An examination consists of 160 questions. For each correct answer, as student earns 1 mark and for each wrong answer, a student loses (1/4)th of mark. Find the number of distinct net scores that a student could earn in the examination. (1) 790 (2) 797 (3) 795 (4) 801 i think d answer wud be 795 coz out of -40 to 160 marks,,he cannot get just -39.5,-39.75,159.25,159.5 and 159.75,158.25 so 801(-40 to 160=200 nos wid each having 4 multiples of 1/4,including both -40 n 160) - 5 nos=795... correct me if m wrong
ohh Shit....
hey we cant get 157.25 also we cant get 158.25 158.5
12345678987654321..................999digits/6.......what is the remainder./?????plzz help guys?
1234....999 times mod 6
6 = 2*3
if 1234 were repeated 1000 times then it wld hv completed the cycle of 4 digits being repeated 100 times. but since its 999, this number ends in 3. hence rem by 2 is 1.
Quote: Originally Posted by shrikantmishr View Post 12345678987654321..................999digits/6.......what is the remainder./?????plzz help guys?
1234....999 times mod 6
6 = 2*3
if 1234 were repeated 1000 times then it wld hv completed the cycle of 4 digits being repeated 100 times. but since its 999, this number ends in 3. hence rem by 2 is 1.
for 3. (1+2+3+4)*100 - 4 = 996 mod 3 = 0.
hence them rem will be of the form 2k+1=3k.
apply this to the options and get the ans.
dude i hv a doubt... 6=2*3 the no. 12345678987654321.....999 digits will leave a remainder 2 by 3(since sum of digits =81*52+71,this comes from the chain of 12345678987654321 as 17 digits,n 999 digits wud end as 1234567898765,13 digits wich is d remainder wen 999/17) also will give a remainder of 1 wid 2... so d remainder will b of d form 3m+2=2n+1 so d answer wud be 5... am i correct,,plz correct me if m not
Originally Posted by testdjjaydev View Post Originally Posted by Srihari123 (Number System-II)" />"] " />An examination consists of 160 questions. For each correct answer, as student earns 1 mark and for each wrong answer, a student loses (1/4)th of mark. Find the number of distinct net scores that a student could earn in the examination. (1) 790 (2) 797 (3) 795 (4) 801 i think d answer wud be 795 coz out of -40 to 160 marks,,he cannot get just -39.5,-39.75,159.25,159.5 and 159.75,158.25 so 801(-40 to 160=200 nos wid each having 4 multiples of 1/4,including both -40 n 160) - 5 nos=795... correct me if m wrong
ohh Shit....
hey we cant get 157.25 also we cant get 158.25 158.5
similarly -39.25... -38.5 no dude u r wrong again,,d answer wud b 795,,i hv solved it above(edited)..plz check n let me know if m wrong 😃
Quote: Originally Posted by shrikantmishr View Post 12345678987654321..................999digits/6.......what is the remainder./?????plzz help guys?
1234....999 times mod 6
6 = 2*3
if 1234 were repeated 1000 times then it wld hv completed the cycle of 4 digits being repeated 100 times. but since its 999, this number ends in 3. hence rem by 2 is 1.
for 3. (1+2+3+4)*100 - 4 = 996 mod 3 = 0.
hence them rem will be of the form 2k+1=3k.
apply this to the options and get the ans.
dude i hv a doubt... 6=2*3 the no. 12345678987654321.....999 digits will leave a remainder 2 by 3(since sum of digits =81*52+71,this comes from the chain of 12345678987654321 as 17 digits,n 999 digits wud end as 1234567898765,13 digits wich is d remainder wen 999/17) also will give a remainder of 1 wid 2... so d remainder will b of d form 3m+2=2n+1 so d answer wud be 5... am i correct,,plz correct me if m not
oh damn. i read the question as 1234....999 times. srry yaar.
here lets see. the number is something like. 12345678987654321.....999 times
forming pairs of 10 each -> 1234567898 7654321987 now both these pairs will complete their cycles of 50 each. but since there are 999 digits hence the last digit,ie, 7 wont be there. hence the number ends in 8. so completely divisible by 2. by 3; adding these pairs we get 105. 105*50 = 5250 - 7 = 5243 mod 3 = 2.
guyz pls help me in solving this ques.. A test has 80 questions. there is 1 mark for correct answer, while there is -ve penalty of -0.5 for a wrong answer and -0.25 for an unattempted question. what is the nos. of questions answered correctly if the student has a score of 34.5 marks and left 10 questions unanswered.pls explain with steps
guyz pls help me in solving this ques.. A test has 80 questions. there is 1 mark for correct answer, while there is -ve penalty of -0.5 for a wrong answer and -0.25 for an unattempted question. what is the nos. of questions answered correctly if the student has a score of 34.5 marks and left 10 questions unanswered.pls explain with steps
thanks in advance
My take is 48. Multiply by 4 to each score to make calculation easy. So, we have +4, -2 and -1 marks and student has scored 138 marks. Now, student gets -10 for not attempting 10 questions. So, he has to score 148 with other questions.
Now, say he got 'x' questions correct and 'y' questions incorrect.
So, 4x-2y = 148 And x+y = 70 => x = 48 => Student answered 48 questions correctly.
In a famous Bel Air Apartments in Ranchi, there are three watchmen meant to protect the precious fruits in the campus. However, one day a theif got in without being noticed and stole some precious mangoes. On the way out however, he was confronted by the three watchmen, the first two of whom asked him to part with 1/3rd of the fruits and one more. The last asked him to part with 1/5th of the mangoes and 4 more. As a result he had no mangoes left. What was the number of mangoes he has stolen? the given answer is 15 pls explain in details with step.:-(
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