please count me in if you guys discuss it on g-talk or anything of that sort.
Cheers,
Prabu
Nos with 1 digit = 9
Nos with two digit = 99 - 9 = 90
Nos with three digit = 999 - 99 = 900
So total digits upto 999 = 2700 + 180 + 9 = 2889
Remaining = 28383 - 2889 = 25494
So the numbers with four digit until the 28383rd term = 25494/4 = 6373 Remainder 2
So 999 + 6373 = 7372
So 2nd term of nest number 7373 = 3
Answer ---- 3
Hope this is correct
Too good!
Is the remainder of any significance here?
Cheers,
Prabu
yes defenetly..... here is how?
Hope you understand how we arrive at the number 7372.
we already have 2 as remainder so the 2 of 7372 is the 28381th term in the series.
but, we have to get the 28383rd term. the differnce is what we got as the remainder.
so to traverse two more terms we get the next number 7373. whose 2nd term is 3.
and that is the answer.
hope you get it clear now....
cheers
vidhyasagar
Number System LOD 2 Ques.No: 62
M is a 2 digit no. which has the property that: the product of factorials of it's digits > sum of factorials of its digits.
how many values of M exist?
a) 56 b) 64 c) 63 d) None of these
Number System LOD 2 Ques.No: 65
If you form a sub-set of integers chosen from between 1 to 3000, such that no two integers add up to a multiple of 9, what is the maximum number of elements in the subset.
a) 1668 b) 1332 c)1333 d)1334
Number System LOD 2 Ques.No: 70
A triangular number is defined as a number which has the property of being expressed as a sum of consecutive natural numbers starting with 1. How many triangular numbers less than 1000, have the property that they are the difference of squares of 2 consecutive natural numbers?
a)20 b)21 c)22 d)23
Number System LOD 2 Ques.No: 71
X and Y are 2 positive integers. Then what will be the sum of coefficients of the expansion of the expression (X+Y)^44?
a) 2^43 b)2^43+1 c)2^44 d)2^44-1
Please explain the answer guys....
thanks in advance
Cheers
vidhyasagar
Number System LOD 2 Ques.No: 62
M is a 2 digit no. which has the property that: the product of factorials of it's digits > sum of factorials of its digits.
how many values of M exist?
a) 56 b) 64 c) 63 d) None of these
Number System LOD 2 Ques.No: 65
If you form a sub-set of integers chosen from between 1 to 3000, such that no two integers add up to a multiple of 9, what is the maximum number of elements in the subset.
a) 1668 b) 1332 c)1333 d)1334
Number System LOD 2 Ques.No: 70
A triangular number is defined as a number which has the property of being expressed as a sum of consecutive natural numbers starting with 1. How many triangular numbers less than 1000, have the property that they are the difference of squares of 2 consecutive natural numbers?
a)20 b)21 c)22 d)23
Number System LOD 2 Ques.No: 71
X and Y are 2 positive integers. Then what will be the sum of coefficients of the expansion of the expression (X+Y)^44?
a) 2^43 b)2^43+1 c)2^44 d)2^44-1
Please explain the answer guys....
thanks in advance
Cheers
vidhyasagar
in a hurry im solved 2 qns:
answer for the first:
only when the two digits have a 1 or 0 as one of the digits then sum>product of factorial of nos. also if both digits r same then sum=product. both cases can be subtracted from 90 (no of 2 digit nos.)
1st case: 17+9=26
2nd case: only 1. so total=27. so total nos: 90-27=63. ans: c
answer for third:
sum of n consecutive nos= n(n+1)
total nos of triangular nos= 44
now the diff between two consecutive squares is always an odd.
here no of even nos is 2*(a multiple of 4) since the multiple can be n or n+1.
so total nos= 44*2/4=22
so the ans= option 'c'.
hope the answrs r correct...
answer for q.71+
ans= 44c0+44c1+44c2+.....44c44
=2^44.
Number System LOD 2 Ques.No: 65
If you form a sub-set of integers chosen from between 1 to 3000, such that no two integers add up to a multiple of 9, what is the maximum number of elements in the subset.
a) 1668 b) 1332 c)1333 d)1334
Please explain the answer guys....
thanks in advance
Cheers
vidhyasagar
the no of ways a number can leave a remainder when divided by 9 =8.
out of these we will choose four:
9x+1, 9x+2, 9x+3, 9x+4 (we can also choose the other four instead of these). now whenever u add ne two of these four they will always leave a remainder with 9.
total no of such nos= 333*4= 1332
we can have one more no of the form 9x (coz it will always leave a remainder when added to others).
so total number of such nos= 1332+1 = 1333
answer option 'c'.
Number System LOD 2 Ques.No: 62
M is a 2 digit no. which has the property that: the product of factorials of it's digits > sum of factorials of its digits.
how many values of M exist?
a) 56 b) 64 c) 63 d) None of these
Number System LOD 2 Ques.No: 65
If you form a sub-set of integers chosen from between 1 to 3000, such that no two integers add up to a multiple of 9, what is the maximum number of elements in the subset.
a) 1668 b) 1332 c)1333 d)1334
Number System LOD 2 Ques.No: 70
A triangular number is defined as a number which has the property of being expressed as a sum of consecutive natural numbers starting with 1. How many triangular numbers less than 1000, have the property that they are the difference of squares of 2 consecutive natural numbers?
a)20 b)21 c)22 d)23
Number System LOD 2 Ques.No: 71
X and Y are 2 positive integers. Then what will be the sum of coefficients of the expansion of the expression (X+Y)^44?
a) 2^43 b)2^43+1 c)2^44 d)2^44-1
Please explain the answer guys....
thanks in advance
Cheers
vidhyasagar
q.65
Consider this:
Number System LOD 2 Ques.No: 62
M is a 2 digit no. which has the property that: the product of factorials of it's digits > sum of factorials of its digits.
how many values of M exist?
a) 56 b) 64 c) 63 d) None of these
Number System LOD 2 Ques.No: 65
If you form a sub-set of integers chosen from between 1 to 3000, such that no two integers add up to a multiple of 9, what is the maximum number of elements in the subset.
a) 1668 b) 1332 c)1333 d)1334
Number System LOD 2 Ques.No: 70
A triangular number is defined as a number which has the property of being expressed as a sum of consecutive natural numbers starting with 1. How many triangular numbers less than 1000, have the property that they are the difference of squares of 2 consecutive natural numbers?
a)20 b)21 c)22 d)23
Number System LOD 2 Ques.No: 71
X and Y are 2 positive integers. Then what will be the sum of coefficients of the expansion of the expression (X+Y)^44?
a) 2^43 b)2^43+1 c)2^44 d)2^44-1
Please explain the answer guys....
thanks in advance
Cheers
vidhyasagar
Q.65
consider this :
first write down subset of all integers from 1-45 which do not form an addition of multiple of 9...
the elements under SUBSET are the integers that do not form a multiple of 9 when any 2 nos. are added:
SUBSET NONSUBSET
1 2 3 4 5 6 7 8
9 10 11 12 13 14 15 16 17 18
19 20 21 22 23 24 25 26 27
28 29 30 31 32 33 34 35 36
37 38 39 40 41 42 43 44 45
i have highlighted 9 cuz it is the only multiple of 9 that exists in the subset..so a pattern is being formed here...
the first 4 numbers in a set of 9 consecutive numbers form the members of the subset and the remaining 5 members do not...(excepting the first case where 9 is included in the subset)
hence, amongst first 3000 numbers,
there are 333 sets of 9 numbers each 1-9, 10-18,....and the first 4 numbers from each set form the subset, thus there are 4x333 = 1332 numbers in the subset and an additional 9 tht is included from the first set...hence, total numbers = 1333 which should giv the answer as C...however, this holds true when we consider the numbers from 1 to 2997...so we also have to include 2998, 2999 and 3000 in the list which increases the number to 1336 which is not given in answer...can anyone help from here???
Number System LOD 2 Ques.No: 62
M is a 2 digit no. which has the property that: the product of factorials of it's digits > sum of factorials of its digits.
how many values of M exist?
a) 56 b) 64 c) 63 d) None of these
Number System LOD 2 Ques.No: 65
If you form a sub-set of integers chosen from between 1 to 3000, such that no two integers add up to a multiple of 9, what is the maximum number of elements in the subset.
a) 1668 b) 1332 c)1333 d)1334
Number System LOD 2 Ques.No: 70
A triangular number is defined as a number which has the property of being expressed as a sum of consecutive natural numbers starting with 1. How many triangular numbers less than 1000, have the property that they are the difference of squares of 2 consecutive natural numbers?
a)20 b)21 c)22 d)23
Number System LOD 2 Ques.No: 71
X and Y are 2 positive integers. Then what will be the sum of coefficients of the expansion of the expression (X+Y)^44?
a) 2^43 b)2^43+1 c)2^44 d)2^44-1
Please explain the answer guys....
thanks in advance
Cheers
vidhyasagar
Q.65
consider this :
first write down subset of all integers from 1-45 which do not form an addition of multiple of 9...
the elements under SUBSET are the integers that do not form a multiple of 9 when any 2 nos. are added:
SUBSET
1 2 3 4
9 10 11 12 13
19 20 21 22
28 29 30 31
37 38 39 40
NONSUBSET
5 6 7 8
14 15 16 17 18
23 24 25 26 27
32 33 34 35 36
41 42 43 44 45
i have highlighted 9 cuz it is the only multiple of 9 that exists in the subset..so a pattern is being formed here...
the first 4 numbers in a set of 9 consecutive numbers form the members of the subset and the remaining 5 members do not...(excepting the first case where 9 is included in the subset)
hence, amongst first 3000 numbers,
there are 333 sets of 9 numbers each 1-9, 10-18,....and the first 4 numbers from each set form the subset, thus there are 4x333 = 1332 numbers in the subset and an additional 9 tht is included from the first set...hence, total numbers = 1333 which should giv the answer as C...however, this holds true when we consider the numbers from 1 to 2997...so we also have to include 2998, 2999 and 3000 in the list which increases the number to 1336 which is not given in answer...can anyone help from here???
Q.65
consider this :
first write down subset of all integers from 1-45 which do not form an addition of multiple of 9...
hence, amongst first 3000 numbers,
there are 333 sets of 9 numbers each 1-9, 10-18,....and the first 4 numbers from each set form the subset, thus there are 4x333 = 1332 numbers in the subset and an additional 9 tht is included from the first set...hence, total numbers = 1333 which should giv the answer as C...however, this holds true when we consider the numbers from 1 to 2997...so we also have to include 2998, 2999 and 3000 in the list which increases the number to 1336 which is not given in answer...can anyone help from here???
dude if we consider general equations of the form 9x+r (wher r=remainder)
x can assume 333 values n 2998 2999 3000 will also be included
yes defenetly..... here is how?
Hope you understand how we arrive at the number 7372.
we already have 2 as remainder so the 2 of 7372 is the 28381th term in the series.
but, we have to get the 28383rd term. the differnce is what we got as the remainder.
so to traverse two more terms we get the next number 7373. whose 2nd term is 3.
and that is the answer.
hope you get it clear now....
cheers
vidhyasagar
I Got you! Wonderful!
Thanks a ton!
Cheers,
Prabu
the no of ways a number can leave a remainder when divided by 9 =8.
out of these we will choose four:
9x+1, 9x+2, 9x+3, 9x+4 (we can also choose the other four instead of these). now whenever u add ne two of these four they will always leave a remainder with 9.
total no of such nos= 333*4= 1332
we can have one more no of the form 9x (coz it will always leave a remainder when added to others).
so total number of such nos= 1332+1 = 1333
answer option 'c'.
Please make the following doubts clear....
1. 9x+1, 9x+2, 9x+3, 9x+4 are the four terms which are multiplied with 333 to get 1332. i am clear with this. but 9x which also occurs 333(same number of times as 9x+1, 9x+2, 9x+3, 9x+4) is not multiplied with 333. Instead 1 is added to 1332 to get 1333. please help me out here....
2.9x+1, 9x+2, 9x+3, 9x+4 in these terms you have substituted x=1,2,3... and get the total number of terms as 333 . But x can be substituted values from 0..(i.e x=0,1,2,3... ), because 1,2,3,4 will also be a part of the sub-set which is of the form 9*(0)+1,9*(0)+2,9*(0)+3,9*(0)+4. I am rephrasing the doubt of ajayreddy here .... Please help me out here also....
Note: The answer given in arun sharma is 1333.
Please clear my doubts guys.... i am breaking my head for the past 2-3 days to get the answer.....
Thanks
Vidhyasagar
Number System LOD 2 Ques.No: 71
X and Y are 2 positive integers. Then what will be the sum of coefficients of the expansion of the expression (X+Y)^44?
a) 2^43 b)2^43+1 c)2^44 d)2^44-1
Cheers
vidhyasagar
The sum would be
2^44
i.e. 44Co +44C1 ............. + 44C44
Please make the following doubts clear....
1. 9x+1, 9x+2, 9x+3, 9x+4 are the four terms which are multiplied with 333 to get 1332. i am clear with this. but 9x which also occurs 333(same number of times as 9x+1, 9x+2, 9x+3, 9x+4) is not multiplied with 333. Instead 1 is added to 1332 to get 1333. please help me out here....
2.9x+1, 9x+2, 9x+3, 9x+4 in these terms you have substituted x=1,2,3... and get the total number of terms as 333 . But x can be substituted values from 0..(i.e x=0,1,2,3... ), because 1,2,3,4 will also be a part of the sub-set which is of the form 9*(0)+1,9*(0)+2,9*(0)+3,9*(0)+4. I am rephrasing the doubt of ajayreddy here .... Please help me out here also....
Note: The answer given in arun sharma is 1333.
Please clear my doubts guys.... i am breaking my head for the past 2-3 days to get the answer.....
Thanks
Vidhyasagar
as for ur first doubt: if we take 9x more than once then 9a+9b will be divisible by 9(for x=a,b).
ur second doubt is very valid. infact i too noticed it only after u posted. but remember that when x=333, 9x+4 doesnt hold. so we will definitely have 1336 nos.
Arun Sharma used to put me in an ambivalence too. then i brought this up with an ims math faculty. she too reassured me that many answers given r confusing and unconvincing.
but the questions are really thought provoking. no doubts abt that
Number System LOD 2 Ques.No: 70
A triangular number is defined as a number which has the property of being expressed as a sum of consecutive natural numbers starting with 1. How many triangular numbers less than 1000, have the property that they are the difference of squares of 2 consecutive natural numbers?
a)20 b)21 c)22 d)23
answer for third:
sum of n consecutive nos= n(n+1)
total nos of triangular nos= 44
now the diff between two consecutive squares is alsways an odd.
here no of odd nos is 2*(a multiple of 4) since the multiple can be n or n+1.
so total nos= 44*2/4=22
so the ans= option 'c'.
hope the answrs r correct...
here no of odd nos is 2*(a multiple of 4) since the multiple can be n or n+1.
so total nos= 44*2/4=22
Somebody please explain these highlighted steps.
I dont understand.
here no of odd nos is 2*(a multiple of 4) since the multiple can be n or n+1.
so total nos= 44*2/4=22
Somebody please explain these highlighted steps.
I dont understand.
no of even nos=2*(multiple of 4). last post edited. sorry for the mistakes..
ne ways the logic is as follows: whenever u have 4 or a multiple of 4 when u reduce by 2 u will get an even.but in this case either n or (n+1) can be a multiple of 4. for eg: n, n+1 can be (3,4) or (4,5). so we will need to multipy 44/4 with 2 that gives 22 even nos.
So total no of odd nos. will be 44-22=22.
odd nos r always the difference between consecutive primes. so 22 is the answer... hope am clear.
Number System LOD 2 Ques.No: 74
what is the highest power of 3 available in the expresion 58!-38!?
a)17 b)18 c)19 d)none of these
Ans is given as a)17
this is how i did the sum....
58! contains 27 3's.
38! contains 17 3's.
so 58!-38! will have 27-17 = 10 3's. But that is not the correct answer according to the book....
help me out....
Thanks in advance....
cheerz
vidhyasagar
Number System LOD 2 Ques.No: 74
what is the highest power of 3 available in the expresion 58!-38!?
a)17 b)18 c)19 d)none of these
Ans is given as a)17
this is how i did the sum....
58! contains 27 3's.
38! contains 17 3's.
so 58!-38! will have 27-17 = 10 3's. But that is not the correct answer according to the book....
help me out....
Thanks in advance....
cheerz
vidhyasagar
Assuming
58! contains 27 3's.
38! contains 17 3's.
So 58! - 38! has a common of 17 3s
So answer shld be 17
Assuming
58! contains 27 3's.
38! contains 17 3's.
So 58! - 38! has a common of 17 3s
So answer shld be 17
it can be explained as follows:
58!-38!=38!(58!/38!-1)
now definitely (58!/38!-1) is not divisible by 3 (since it will leave a remainder of -1). so the highest power will be that in 38! i.e.17...
it can be explained as follows:
58!-38!=38!(58!/38!-1)
now definitely (58!/38!-1) is not divisible by 3 (since it will leave a remainder of -1). so the highest power will be that in 38! i.e.17...
Please explain the highlighted line in detail.....
Thanks
Vidhyasagar