deleted the duplicate
Please give the soln for this problem ...
hi Puys,
Please explain how to solve the following ques..
and also explain the approach for all such type of questions
Q12. Number Systems LOD 3.
Find all two-digit numbers such that the sum of the digits constituting the number is not less that 7;
the sum of the squares of the digits is not greater than 30;
the number consistingof the same digits written in the reverse order is not larger than half the given number.
(a) 52 (b) 51 (c) 49 (d) 53 (e) 50 .
Can anyone explain me euler's theorem in detail
I was going through LOD III on Averages and there is bit doubt about quest 22.
" There are 3 classes having 20,25 and 30 students respectively having average marks in an examination as 20,25 and 30 respectively. If the three classes are represented by A,B and C and you have the following information about the three classes, answer the questions that follow :
A -> Highest score 22, Lowest score 18
B -> Highest score 31, Lowest score 23
C -> Highest score 33, Lowest score 26"
If 5 people are transferred from C to B, further 5 more people are transferred from B to A, then 5 are transfered from A to B and finally, 5 more are transferred from B to C
22. The minimum possible average of group B after this set of operation is
a) 21.6 b) 21.4 c)21.8 d) 21.2 e) cannot say
Hi Guys,
find the last two digits of (99)^45329
To get the last two digits we will divide it by 100.
100= 2^2 * 5^2
By EULER THEOREM: a^(p-1)%p=1 where p is prime number.
But here P is not prime,i.e 5 and 2.
So we will go by extension method.
E(5)=25*(1-1/5)=20
E(2)= 4*(1-1/2)=2
99^20%25=1
99^2%4=1
99^LCM(20,2)%100=1
=> 99^20%100=1
45329=20*x+9
99^45329=99^(20x+9)
=> 99^(20x+9)%100
=> 99^9%100 ( since 99^20%100=1)
=> (-1)^9%100
=> -1 %100
=> 99 AnSwEr..!!!
EULER METHOD A.K.A FERMIT LITTLE THEOREM is basically for complicated quant problems.
Binomial method is the best for this problem..!!
I'll vote for binomial only..!! 😃
Scients in NASA were working on breaking a code of mars bound rocket sent by russians.Years of work bore fruit and they realized that the code is a 3 digit no. with different non zero digits A,B and C.They form all the possible 3 digit numbers using all these digits.To save the code they put different numbers so formed into three groups G1,G2 and G3.The sum of the numbers in G1,G2,G3 were found to be 1176,1167 and 321 respectively.Which of the following cannot be any digits of the numbers......a)6 b)1 c)5 d)7...pls do show d entire process.
Scients in NASA were working on breaking a code of mars bound rocket sent by russians.Years of work bore fruit and they realized that the code is a 3 digit no. with different non zero digits A,B and C.They form all the possible 3 digit numbers using all these digits.To save the code they put different numbers so formed into three groups G1,G2 and G3.The sum of the numbers in G1,G2,G3 were found to be 1176,1167 and 321 respectively.Which of the following cannot be any digits of the numbers......a)6 b)1 c)5 d)7...pls do show d entire process.
rahul005 SaysScients in NASA were working on breaking a code of mars bound rocket sent by russians.Years of work bore fruit and they realized that the code is a 3 digit no. with different non zero digits A,B and C.They form all the possible 3 digit numbers using all these digits.To save the code they put different numbers so formed into three groups G1,G2 and G3.The sum of the numbers in G1,G2,G3 were found to be 1176,1167 and 321 respectively.Which of the following cannot be any digits of the numbers......a)6 b)1 c)5 d)7...pls do show d entire process.
The answer is 7.
Sum of all the various numbers=1176+1167+321 => 2664
Let the number be a,b,c
Therefore we can say the six various combinations will be abc,acb,bac,bca,cab,cba.
Now sum of unit digit for all these numbers will be
2a+2b+2c and this will be equal to 4 as unit digit.
By observing we can analyse that the sum is of 2a+2b+2c is 24 which leads to a carry of 2 on the tensdigit and further leading to another carry of 2 on hundred's digit.
Now since 2a+2b+2c will give 4 as unit digit.
We can place options.
a) 6 => 2*6 + 2b + 2c= 4(considering unit digit)
2b + 2c= 2
b can be 5 and c can be 1.
Similary for 1 and 5.
d)7 => 14+ 2b + 2c =4
2b + 2c = 0 ( 4 and 1 )
All the four options are correct..!!!
SOMEBODY HELP !!
Hi every1!
This question is from Arun Sharma Ch - 1 (Number Systems)...
LOD 3 - Ques 44.
N = 202 X 20002 X 200000002 X 20000000000000002 X 200000000....(31 zeroes). The sum of digits in this multiplication will be:
(a) 112
(b) 160
(c) 144
(d) Cannot be determined
Any help would me much appreciated...
Thanks in advance. 
Hi every1!
This question is from Arun Sharma Ch - 1 (Number Systems)...
LOD 3 - Ques 44.
N = 202 X 20002 X 200000002 X 20000000000000002 X 200000000....(31 zeroes). The sum of digits in this multiplication will be:
(a) 112
(b) 160
(c) 144
(d) Cannot be determined
Any help would me much appreciated...
Thanks in advance.
Hey what a coincidence !!!! I just racked my brains over this one on the morning of 7th aug 😃 The answer to this one can be gauged my oserving the pattern of the numbers that appear ....
lets break this into the multiplication byy pairs ..
1) 202 X 20002 --> 202 X 2 is 404 so we the soln for this is 4040404 look there are 4 4's here :)
2) 4040404 X 200000002 --> 4040404 X 2 is 8080808 so the answer for this becomes 808080808080808 ..voila !! what have we got 8 8's :)
3)808080808080808 X 20000000000000002 --> 808080808080808 X 2 is
1616161616161616 so the answer for the above multiplication is 16 16's ....carrying this forward the next multiplication yields 32 32's whic gives the answer as 160 (3+2 = 5 is occuring 32 times = 160 )
Hope this answers your problem ...
Can anybody tell me the solutions to problem 47 to 49 from the number systems set....this one relates to the wickets by 7 bowlers problem ...TIA
2ndApr2011 SaysCan anybody tell me the solutions to problem 47 to 49 from the number systems set....this one relates to the wickets by 7 bowlers problem ...TIA
I thought it would be better if i mention the problem here :
At a particular time in the 21st cebtury there were seven bowlers in the indian cricket team's list of 16 players shortlisted to play the next wc . Statisticians discovered that if u looked at the no of wickets taken by any of the 7 bowlers in the current indian cricket team , the number of wickets taken by them had a strange property . The numbers were such that for any team selection of 11 players (having 1 to 7 bowlers) by using the number of wickets taken by each bowler and attaching coefficients 0f +1 , 0 , -1 to eachvalue available and adding the resultant values , any number from 1 to 1093 both included could be formed . If we denote W1,W2,W3,W4,W5,W6,W7 as the 7 values in the ascending order what would be the answer to the following questions :
47) Find the value of W1+2W2+3W3+4W4+5W5+6W6
a) 2005 b) 1995
c)1985 d)NOT
4
Find the index of the largest power of 3 contained in the product W1 W2 W3 W4 W5 W6 W7a) 15 b) 10
c) 21 d) 6
49) If the sum of the seven coefficients is 0 , find the smallest number that can be obtained
a) -1067 b)-729
c) -1040 d) -1053
Please help !!!!
q. what is the total number of factors of 16!.
a.2016 b.1024 c.3780 d. 4032 e. none of these
my answer is coming as 5376 , so option e. none of these.
but the answer is option d.
please solve this and let me know whether i am wrong or the amswer.
q. what is the total number of factors of 16!.
a.2016 b.1024 c.3780 d. 4032 e. none of these
my answer is coming as 5376 , so option e. none of these.
but the answer is option d.
please solve this and let me know whether i am wrong or the amswer.
16! = (2^15).(3^6).(5^3).(7).(11^1).(13^1)
So, no. of factors = 16*7*4*3*2*2=5376 is what I am getting.
16! = (2^15).(3^6).(5^3).(7).(11^1).(13^1)
So, no. of factors = 16*7*4*3*2*2=5376 is what I am getting.
You are rite brother..same thing i tried..none of these should be the answer..:D
Hi every1!
This question is from Arun Sharma Ch - 1 (Number Systems)...
LOD 3 - Ques 44.
N = 202 X 20002 X 200000002 X 20000000000000002 X 200000000....(31 zeroes). The sum of digits in this multiplication will be:
(a) 112
(b) 160
(c) 144
(d) Cannot be determined
Any help would me much appreciated...
Thanks in advance.
Check the series. It has a special sequence.
In the 1st term - 1 zero is present i.e 2 - 1 i.e (2^1) - 1.
In the 2nd term - 3 zeros are present i.e 4 - 1 i.e (2^2) - 1.
In the 3rd term - 7 zeros are present i.e 8 - 1 i.e (2^3) - 1.
In the 4th term - 15 zeros are present i.e 16 - 1 i.e (2^4) - 1.
In the 5th term - 31 zeros are present i.e 32 - 1 i.e (2^5) - 1.
Starting with first two terms -
20002 * 202 -
For this firstly do -
20002 * 2 ( Unit's digit of 202 ) = 40004
And 40004
+ 4000400 ( Start from the unit's digit and put the no. of zeros = (2^1) i.e 2 ( Underlined ) according to the no. considered and the no. adjacent to these zeros, is the replica of the no. obtained i.e 40004).
On adding, = 4040404 -> It has 4 4's.
Now consider -
4040404 * 200000002 -
Again taking unit's digit of 200000002 i.e 2 and perform multiplication. It will be -
4040404 * 2 = 8080808
And 8080808
+ 808080800000000 ( Start from the unit's digit and put no. of zeros = (2^3) i.e 8 ( Underlined ) according to the no. considered and the no. adjacent to these zeros, is the replica of the no. obtained i.e 808080
.On adding, = 808080808080808 -> It has 8 8's.
Similarly do for 808080808080808 * 20000000000000002 -
It will be - 16161616161616161616161616161616 -> It has 16 16's. ( This time put no. of zeros = (2^4) i.e 16 according to the no. considered and the no. adjacent to these zeros, is the replica of the no. obtained i.e 1616161616161616 ).
Lastly do for 16161616161616161616161616161616 * 200000000000000000000000000000002 -
It will be - 3232323232323232323232323232323232323232323232323232323232323232-> It has 32 32's. ( This time put no. of zeros = (2^5) i.e 32 according to the no. considered and the no. adjacent to these zeros, is the replica of the no. obtained i.e 32323232323232323232323232323232 ).
Finally,
Frequency of 3 is 32. So 32 * 3 = 96
And frequency of 2 is also 32. So 32 * 2 = 64
Therefore, Sum of digits = 96 + 64 = 160 i.e option (b).
This is what i manipulated..Hope u got it too:D
Can any1 please tell me,
How to solve questions of the form -
3x4^2 + 4x5^2 + 5x6^2 + ... (15 terms)
I couldn't devise a formula for this. Is it a GP or an AP?
Please tell if there's a formula for these kind of questions.
Can any1 please tell me,
How to solve questions of the form -
3x4^2 + 4x5^2 + 5x6^2 + ... (15 terms)
I couldn't devise a formula for this. Is it a GP or an AP?
Please tell if there's a formula for these kind of questions.
what is expected in this ?
Plz post the full question
if the Question is to get the sum to 15 terms .
then this is my approach...
the series is of the form : n(n+1)^2
we need to find it sum from n = 3 to n = 15.
now what we can do is find its sum form n = 1 to n = 15 and then subtract the sum of the first two terms form the original sum.
soln: Tn = n(n^2 + 2n + 1)
= n^3 + 2n^2 + n
now, Sn = sum of n terms .
Sn = Sum to n terms of Tn .
= sum to n terms (n^3 ) + 2 * sum to n terms (n^2) + sum to n terms (n)
= {n(n+1)/2}^2 + 2 * n(n+1)(2n+1)/6 + n(n+1)/2
here n = 15.
so the expression becomes ,
S15 = {15(16)/2}^2 + 15*16*31/3 + 15*16/2
= (120)^2 + 5 * 16 *31 + 120
= 14400 + 2480 + 120
= 17000.
now the Sum of first two terms : 1*2^2 + 2*3^2 = 4+ 18 = 22.
subtract 22 from S15 .
therefor answer is 16978.
Please correct me if I am wrong.
what is expected in this ?
Plz post the full question
if the Question is to get the sum to 15 terms .
then this is my approach...
the series is of the form : n(n+1)^2
we need to find it sum from n = 3 to n = 15.
now what we can do is find its sum form n = 1 to n = 15 and then subtract the sum of the first two terms form the original sum.
soln: Tn = n(n^2 + 2n + 1)
= n^3 + 2n^2 + n
now, Sn = sum of n terms .
Sn = Sum to n terms of Tn .
= sum to n terms (n^3 ) + 2 * sum to n terms (n^2) + sum to n terms (n)
= {n(n+1)/2}^2 + 2 * n(n+1)(2n+1)/6 + n(n+1)/2
here n = 15.
so the expression becomes ,
S15 = {15(16)/2}^2 + 15*16*31/3 + 15*16/2
= (120)^2 + 5 * 16 *31 + 120
= 14400 + 2480 + 120
= 17000.
now the Sum of first two terms : 1*2^2 + 2*3^2 = 4+ 18 = 22.
subtract 22 from S15 .
therefor answer is 16978.
Please correct me if I am wrong.
Firstly, yes we had to find the sum till 15 terms.
Secondly, what i can decipher from your post is that the solution is absolutely correct but i think there's a problem at the very beginning.
If you're calling Sn as a sum of n terms and got it's general form as n(n+1)^2
it should give us the sum of 3 terms if we put n=3( Keeping in mind that we will subtract S2 from S15 later).
instead we get the third term i.e 3.4^2 . Hence, the general terms doesn't serve our purpose. Hence, I think n(n+1)^2 isn't the sum of n terms.
Correct me if I'm wrong. I couldn't find a flaw in the solution.
P.S - the correct answer is 27110
what is expected in this ?
Plz post the full question
if the Question is to get the sum to 15 terms .
then this is my approach...
the series is of the form : n(n+1)^2
we need to find it sum from n = 3 to n = 15.
now what we can do is find its sum form n = 1 to n = 15 and then subtract the sum of the first two terms form the original sum.
soln: Tn = n(n^2 + 2n + 1)
= n^3 + 2n^2 + n
now, Sn = sum of n terms .
Sn = Sum to n terms of Tn .
= sum to n terms (n^3 ) + 2 * sum to n terms (n^2) + sum to n terms (n)
= {n(n+1)/2}^2 + 2 * n(n+1)(2n+1)/6 + n(n+1)/2
here n = 15.
so the expression becomes ,
S15 = {15(16)/2}^2 + 15*16*31/3 + 15*16/2
= (120)^2 + 5 * 16 *31 + 120
= 14400 + 2480 + 120
= 17000.
now the Sum of first two terms : 1*2^2 + 2*3^2 = 4+ 18 = 22.
subtract 22 from S15 .
therefor answer is 16978.
Please correct me if I am wrong.
Firstly, yes we had to find the sum till 15 terms.
Secondly, what i could decipher from your post is that the solution is absolutely correct but i think there's a problem at the very beginning.
If you're calling Sn as a sum of n terms and got it's general form as n(n+1)^2
it should give us the sum of 3 terms if we put n=3( Keeping in mind that we will subtract S2 from S15 later).
instead we get the third term i.e 3.4^2 . Hence, the general terms doesn't serve our purpose. Hence, I think n(n+1)^2 isn't the sum of n terms.
Correct me if I'm wrong. I couldn't find a flaw in the solution.
P.S - the correct answer is 27110