Hey everyone! help me in these..... 1.What is the remainder when (2222)^5555 + (5555)^2222 is divided by 7 ? 2.How many sets of three distinct factors of the no n=26*34*52 can be made such that the factors in each set has a common factor 1 with respect to every other factor in that set? i) 236 ii) 360 iii) 104 iv) 380 v) None of these 3. last three digits of 777^777 thanks in advance.....
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Answer to Q2.
(6 + 1)(4 + 1)( 2 + 1) - 1 = 104
for explanation chk http://www.cat4mba.com/math-e-book/combinations
Number of ways of selecting one or more things from p identical things of one type qidentical things of another type, r identical things of the third type and n different things is given by: - (p+1) (q+1) (r+1)2n - 1
i gues ques 1 was asked once before : answer 0
ques 3. find remainder when 777^777 is divided by 1000, why do people ask such questions which one in his
minds would not even attempt in the CAT. this is not JEE. next someone would ask the first 3 digits. !!!
let me try for one digit,
777 = 7 mod 10
7^777 = x mod 10
cyclicity = 4
777 = 1 mod 4
hence the last digit is 7.
who ever has enough free time could try three digits.
I do agree . . .
but if i am not wrong the above question was asked in last CF mock . . .
i never thought real CAT questions were as tough as mock questions. eg. asking last 3 digits bring the problem into a realm where only maths background guys could solve. even if you knew how to solve that prob. no guarantee that if the question appears with a little twist in the real cat, one could nail it in 3 minutes using a pen and paper, unless he has a real high IQ.
remainder is 0
use euler th:
euler no of 7 is 6
so 2222^5555/6 will reduce to 2222^5
nd
5555^2222 will reduce to 5555^2
now divide 2222/7 and find out the remainder= 3
so 2222^5 will be having remainder 3^5
furthet 3^5 will leave remainder of 5
similarly find remainder of 2nd term , it will leave remainder of 2
now sum up both the remaindersi,e 7 n divide it by 7...so remainder z 0
try to find out using cyclicity...as raj xplained....i guess ansr z 001...(all flukee...)
2222 = 3 mod 7
5555 = 4 mod 7
problem reduces to 3 5555 + 4 2222 mod 7
3 has cyclicity 6 and 4 has cyvlicity 3 wrt 7
5555 = 5 mod 6
2222 = 2 mod 3
total remainder = 243 + 16 = 259 = 0 mod 7
hence remaindr = 0
Any body has enough time to explain it in details so that ppl like me can understand it.
Q. What is the remainder when (2222)5555 + (5555)2222 is divided by 7 ?
I am trying to explain the way rajorshi solved it.
2222 = 3 mod 7 means 2222 divided by 7 gives a reminder 3
so reminder of (2222)5555 divided by 7= Reminder of 35555 divided by 7
Similarly 5555 = 4 mod 7
and the question reduces to 3 5555 + 4 2222
3 has cyclicity 6 and 4 has cyvlicity 3 wrt 7
what does this means ?? the reminder of 3 , 37 , 313 . . . . . . divided by 7 is same
3 when divided by 7 gives reminder 3
32 when divided by 7 gives reminder 2
33 when divided by 7 gives reminder 6 ( 2 x 3)
34 when divided by 7 gives reminder 4 ( 6 x 3)
35 when divided by 7 gives reminder 5 ( 4 x 3)
36 when divided by 7 gives reminder 1 ( 5 x 3)
37 when divided by 7 gives reminder 3 ( 1 x 3)
So reminder of 35555 = reminder of 35 5555 = 5 mod 6 243
similarly we get 16
and thus the reminder = reminder of 243 + 16 = 259
hence remainder = 0
Anita remainder cnt b greater thn 6....!!!!!
y dnt u try eulers method...just give it a try..its really helpfull..but in da end use dat 1 in vich u r comfrtble...
regards..
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