@scrabbler -it shall be abc+abd+bcd+acd=abcd.we have to find number of biquadratic equations with real roots satisfying that sum of roots taken three at a time is equal to product of roots.any shortcut for it?
@hexagon said:ok bhai what is the ans
Not certain, but doing a kind of trial and error with checks and balances (long painful approach got 5 sets of distinct positive integers
2, 3, 10, 15 (Edited to add)
2, 3, 9, 18
2, 3, 8, 24
2, 3, 7, 42
2, 4, 6, 12
2, 4, 5, 20
So there should be 6 * 4! = 144?
Re-edit: Ah, I see chill sir has already done it in detail :)
regards
scrabbler
2, 3, 10, 15 (Edited to add)
2, 3, 9, 18
2, 3, 8, 24
2, 3, 7, 42
2, 4, 6, 12
2, 4, 5, 20
So there should be 6 * 4! = 144?
Re-edit: Ah, I see chill sir has already done it in detail :)
regards
scrabbler
@scrabbler said:Why 100?Besides, there could be other answers too. Did little trial and error, found 2, 3, 9, 18 works too.regardsscrabbler
haan yaar galat tha...us samai sahi lag raha tha
In any month harsh deposits m% and withdraws n% of the closing balance of the previous month.if his balance at the end of march (after the withdrawal) is tha same as his balance at the beginning of January (before the deposit),which of the following is true?p. provide sol.
Is it n >m
Is it n >m
@falcao said:@chillfactor -sir when d=3 and c
because of the assumption that c > d (if you will consider other values of c, then you will get repeated solutions)
@rnishant231 said:thoda detail karna bhai..
16^56/58= 2^224/58=2^223/29
applying euler's theorem, we get E(29)=28 Rem[2^28/29]=1
Rem[2^(28*7+27)/29]
Rem[2^27/29]= Rem[(2^2)*((2^5)^5)/29]=Rem[(2^2)*(3^5)/29]=15
therefore, Rem[2^224/58]= 15*2=30
i hope i have explained it clearly. if still doubts, i can try further.
@zuloo said:16^56/58= 2^224/58=2^223/29applying euler's theorem, we get E(29)=28 Rem[2^28/29]=1Rem[2^(28*7+27)/29]Rem[2^27/29]= Rem[(2^2)*((2^5)^5)/29]=Rem[(2^2)*(3^5)/29]=15therefore, Rem[2^224/58]= 15*2=30i hope i have explained it clearly. if still doubts, i can try further.
doubt ka koi sawaal hi nahi hai..
@jain4444 said:x,y and z are positive reals. What is the maximum value of[(x+y+z)^3−x^3−y^3−z^3]^2/[(x^2+y^2+z^2)^3−x^6−y^6−z^6]
24 ?
@Asfakul said:100! when expanded, ends in exactly 24 zeroes in decimal system. In how many other number systems, when converted, 100! still ends in exactly 24 zeroes?(For example, 5! = 120 has one zero when expanded in decimal system and it becomes 71 after conversion in base-7 and has no zeroes at end.)
21 other such base systems
Exponent of 2 in 100! is 97
=> Exponent of 16 in 100! is 24
Exponent of 3 in 100! is 48
=> Exponent of 9 in 100! is 24
Exponent of 5 in 100! is 24
For higher prime factors exponent is less than 24, so they can not be a factor of base N
In base system N, 100! will have 24 zeros at the end, where N = (2^a)(3^b)(5^c), where maximum value of a can be 4, for b its 2 and for c its 1 and atleast one of a, b, c should attain its maximum value
Possible values of a are 0, 1, 2, 3, 4; possible values of b are 0, 1, 2 and that of c are 0 a d 1
=> 5*3*2 - 4*2*1 = 22 (4*2*1 are those cases when none of a, b, c assumes its maximum value) such values of N (but base 10 is already considered)
Hence 21 such other base systems
@Asfakul said:100! when expanded, ends in exactly 24 zeroes in decimal system. In how many other number systems, when converted, 100! still ends in exactly 24 zeroes?(For example, 5! = 120 has one zero when expanded in decimal system and it becomes 71 after conversion in base-7 and has no zeroes at end.)
100! = 2^97*3^48*5^24.....
97/4, 48/2 or 24
so till 5 powers of 2, 3 powers of 3 or 2 power of 5
there would be 21 such combinations
help me with this tough problem
There are N ordered integer quadruples (a,b,c,d) subject to 0≤a,b,c,d≤99990 such that
ad−bc≡1(mod 99991). What are the last three digits of N?
@junefever said:Find remainder when 33333333......86 times is divided by 13?
333333 mod 13 = 0
33 mod 13 = 7 ?
@jain4444 said:x,y and z are positive reals. What is the maximum value of[(x+y+z)^3−x^3−y^3−z^3]^2/[(x^2+y^2+z^2)^3−x^6−y^6−z^6]
24 ?
x = y = z = 1
@hexagon said:i got this quest in this years cat i was not able to solve it1^2 +1^2+2^2+1^2+2^2+3^2+...........1^2+2^2+3^2.......+21^2 find the sum of the series
summation (1 to 21) of n(n+1)(2n+1)/6
@hexagon said:i got this quest in this years cat i was not able to solve it1^2 +1^2+2^2+1^2+2^2+3^2+...........1^2+2^2+3^2.......+21^2 find the sum of the series
19481...
S=summation[(22-n)*n^2]
S=summation[22n^2-n^3]