hi frnds plz help in solving the question below..
twenty five set problems on data interpretation one each for DI section of 25 catalysts test were prepared by ams research team. The DI section of each catalyst contained 50 questions of which 35 question were unique.what could be the maximum number of question prepared for the DI of all 25 catalyst put together?
a: 1100 b:875 c:1070 D: non of these
My ans for no. of perfect sq. factor of 3600 is 9
is the answer 1070
no dude ans is 12..................like 36 ,144,25 etc are also perfect square factors.....i m looking for reasons behind ans 12......it make has some relation with
pnc
rohitaryan Saysfind the no. of perfect square factors of 3600? also explain.
3600 = 4 x 9 x 100 = 2^2 x 3^2 x 10^2
So, ans is 3
no dude ans is 12..................like 36 ,144,25 etc are also perfect square factors.....i m looking for reasons behind ans 12......it make has some relation with
pnc
no dude ans is 12..................like 36 ,144,25 etc are also perfect square factors.....i m looking for reasons behind ans 12......it make has some relation with
pnc
Ah! I had read the question in a hurry! :banghead:
The prime factors of 3600 are = 2 , 3, 5
3600 = 2^4 x 3^2 x 5^2
Now, so we have the following perfect square factors: 4, 9 and 25
Also, 3600 = 4^2 x 15^2 = 16, 225
3600 = 6^2 x 10^2 = 36, 100
3600 = 12^2 x 5^2 = 144
3600 = 20^2 x 3^2 = 400
3600 = 60^2 x 1 = 3600, 1
3600 = 30^2 x 2^2 = 900
So 12 perfect squares.
Ah! I had read the question in a hurry! :banghead:
The prime factors of 3600 are = 2 , 3, 5
3600 = 2^4 x 3^2 x 5^2
Now, so we have the following perfect square factors: 4, 9 and 25
Also, 3600 = 4^2 x 15^2 = 16, 225
3600 = 6^2 x 10^2 = 36, 100
3600 = 12^2 x 5^2 = 144
3600 = 20^2 x 3^2 = 400
3600 = 60^2 x 1 = 3600, 1
3600 = 30^2 x 2^2 = 900
So 12 perfect squares.
There is a shorter way.
3600 = 2^4*3^2*5^2
Just look for the number of even powers of each prime factor.
For 2, its 0, 2, 4 so 3
For 3, its 0, 2, so 2
For 5, its 0, 2, so 2
=> Number of perfect sq factors = 3*2*2 = 12
There is a shorter way.
3600 = 2^4*3^2*5^2
Just look for the number of even powers of each prime factor.
For 2, its 0, 2, 4 so 3
For 3, its 0, 2, so 2
For 5, its 0, 2, so 2
=> Number of perfect sq factors = 3*2*2 = 12
Another way is arranging the prime factors with the power of 2, i.e. arranging 2^2 ; 2^2 ; 3^2 and 5^2
So, they can be arranged in 4!/2 ways = 24/2 = 12 ways = Total no. of perfect square factors.
hi frnds plz help in solving the question below..
twenty five set problems on data interpretation one each for DI section of 25 catalysts test were prepared by ams research team. The DI section of each catalyst contained 50 questions of which 35 question were unique.what could be the maximum number of question prepared for the DI of all 25 catalyst put together?
a: 1100 b:875 c:1070 D: non of these
The answer is 1070...The explaination goes like this...here there are 25 sets which contains 35 questions each which are unique so total number of questions becomes 25*35=875 questions.....Now since the questions is asking maximum possible number of questions so one question can be repeated atmost in 2 sets .i.e 15*12 and one set will contain all question unique..therefore ..25*35+15*12+15=1070 questions...
when u divide 50 by 11 u can write remainder -5, because it is 5 less than next multiple of 11
similarly for 51 = -4 , for 52 it is = -3
now 50*51*52= -5*-4*-3= -60
on dividing by eleven remainder = -5, that means 11-5= 6
:cheerio:
I also getting remainder 3..guess the answer is given wrong in the book
harsh_hbti Saysthanks the answer is correct..
Hi,
I got the part uptil the equations are derived ... can you please explain how did you get 343 from that ?
Thanks,
Deep
2400 = 3*25*32
7^99 mod 3 = 1
7^99 mod 2^5 = 7^3 mod 32 = 23
7^99 mod 25 = 7.49^49 mod 25 = -7
25k -7 = 32p + 23,==>p = 10
800q + 343 = 3r + 1,==>q = 0
remainder = 343
Ooops.. wrong post quoted in the earlier message...
Can someone please explain how you got 343 aftr the equations are formed.. is it something related to the chinese remainder theorem.. if yes then can you explain what it is and where is it used
Another way is arranging the prime factors with the power of 2, i.e. arranging 2^2 ; 2^2 ; 3^2 and 5^2
So, they can be arranged in 4!/2 ways = 24/2 = 12 ways = Total no. of perfect square factors.
DUDE i know this methiod but just can someone let me know that if atal we dont get a power of 2...means if we get some thing like (2power 2)*(7 power1)..then?
snaval1 SaysDUDE i know this methiod but just can someone let me know that if atal we dont get a power of 2...means if we get some thing like (2power 2)*(7 power1)..then?
Then we ignore 7 power1 because they can't provide a perfect square obviously. So 2 power of 2 can be arranged in 2 ways. .
Answer should be 1.
E(17) = 16, (38^16!)^1777 mod 17 = 38^16k mod 17 = 1
cud sm1 xplain dis.....im nt gettin dis mod thing ryt...
DS: 1 ?
Q. The integers m and p are such that 2
1. the greatest common factor of m and p is 2
2. the least common multiple of m and p is 30
Provide answer with explanation
DS:
Q. The integers m and p are such that 21 ?
1. the greatest common factor of m and p is 2
2. the least common multiple of m and p is 30
Provide answer with explanation
The question can be answered by 1 alone and can't be answered by 2 alone.
Reason:
Using Statement1:
gcd=2.
Therefore, p/m = 2x/2y -----> x/y.
The remainder in x/y can be from 1 to y-1(It can't be 0 since m is not a factor of p).
Since we have cancelled the GCD which is 2, so to find the actual remainder we would have to multiply it back.
So, in any case, the remainder would be greater than 1.
So, statement one alone is sufficient to answer.
Statement 2:
If we consider 2 numbers which have LCM of 30 such as:
5 and 6, we get the remainder as 1.
And, if we consider the numbers to be 15 and 6, the remainder comes to be 3.
So, we can't say that the remainder is definitely greater than 1.
So, statement 2 alone can't answer the question.
@ revival
buddy...cud u plz xplain me the "mod" sol approach to ques below...im nt able to undersatnd the "mod" solution process...
find the remainder when (38^16!)^1777 is divided by 17
DS:
Q. The integers m and p are such that 21 ?
1. the greatest common factor of m and p is 2
2. the least common multiple of m and p is 30
Provide answer with explanation
for this question condition 1 is enough to get the answer..
1>greatest common multiple of m and p is 2...
here we can take examples like..(m,p)or(2m,2p)(m and p are co primes)=(4,10),(8,14),(6,10)......in all these cases we find the remainder will always be greater than 1.and since p>m this condition will always be satisfied
simillarly
2>the least common multiple of m and p is 30
Here if we take examples (m,p)=(15,30),(5,6),(2,15),(3,10)..etc
here the remainder will always be greater than 1 or equal to 1(in case of (5,6))--forgot this case
so only 1 is enough to answer this question
@ revival
buddy...cud u plz xplain me the "mod" sol approach to ques below...im nt able to undersatnd the "mod" solution process...
find the remainder when (38^16!)^1777 is divided by 17
hi if u know the eulers theorem it will be very simple to tackle such question..please search in this thread only u will get the detailed explaination for the same..
now the euler number for 17 is 16 so dividing 1777 by 16 we get remainder as 1..
so the question becomes
38^16!/17...and again the euler number for 17 is 16..so when we divide 16!/16 we get remainder as zero..
therefore it becomes
38^0/17 =1/17...the remainder is one
Ooops.. wrong post quoted in the earlier message...
Can someone please explain how you got 343 aftr the equations are formed.. is it something related to the chinese remainder theorem.. if yes then can you explain what it is and where is it used
we can solve it based on eulers theorem..the question is what is the remainder for 7^99/2400
we can write the same question as (7^99/24)/100
now euler number for 24 is 8
so dividing 99 by 8 we get remainder as 3
so equation becomes=(7^3/24)/100
which is 343/2400
therefore we get 343 as the remainder.