Sum of first N natural numbers=n(n+1)(2n+1)/6 The number of times K is divisible by 4= number of times M is divisible by 8 + number of times M+1 id divisible by 8 =6+7=13 So answer is none of these.
Can anyone help me out with these?
1. Define a number K such that it is the sum of the squares of the first M natural numbers (k = 1^2 +2^2 + 3^2 .....+M^2) where M a) 10 b)11 c)12 d)none of these
2. Find the 28383rd term of series 123456789101112....
a)3 b)4 c)9 d)7
First one I don't have a clue, don't know how to proceed further from K=m(m+1)(2m+1)/6.
And the second one I am getting answer 3 whereas books says 9.
Me too got the same answer for that 2nd digit of 7372 that is 3
Can anyone help me out with these?
1. Define a number K such that it is the sum of the squares of the first M natural numbers (k = 1^2 +2^2 + 3^2 .....+M^2) where M a) 10 b)11 c)12 d)none of these
2. Find the 28383rd term of series 123456789101112....
a)3 b)4 c)9 d)7
First one I don't have a clue, don't know how to proceed further from K=m(m+1)(2m+1)/6.
And the second one I am getting answer 3 whereas books says 9.
Sum of first N natural numbers=n(n+1)(2n+1)/6 The number of times K is divisible by 4= number of times M is divisible by 8 + number of times M+1 id divisible by 8 =6+7=13 So answer is none of these.
Thanks dude but answer in the book says 12. This is sum no 61 of LOD 2 numbers. Although the answer may be wrong.
92) taking out and removing 6! (14*7+14(7*8*9.....13))/(112-21) =14*7(1+8*9*.........13)/91 =14*7(1+8*9*.....13)/13*7 =reminder(14/13)=1
puyz help me out in des problemz... numbers-lod2 89)The computer takes input of a no. N & a X where X is a factor of no. N.In particular case N is equal to 83p796161q and X is equal to 11 where 0
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
If you divide a fraction by a particular number ie downgrade then the remainder must be multiplied by the same number at the end to get the actual remainder ie upgrade. See my soln. The answer in the book is also 7!
92) taking out and removing 6! (14*7+14(7*8*9.....13))/(112-21) =14*7(1+8*9*.........13)/91 =14*7(1+8*9*.....13)/13*7 =reminder(14/13)=1
Answer of 92.
14. 7! + 14 .13! 2.8! -21.6!
= 14(7! +13!) 6!(2.8.7 -21)
= 14( 7!+13!) 91(6!)
=14(7!+13!) 7!(13) = 14.7! + 14.13! 7!.13 7!.13 = rem 7! + rem 0 = rem 7 ! ans
If you divide a fraction by a particular number ie downgrade then the remainder must be multiplied by the same number at the end to get the actual remainder ie upgrade. See my soln. The answer in the book is also 7!
Question no. 1. Find the remainder when 73 * 75 * 78 * 57 * 197 is divided by 34. a) 32 b)30 c)15 d) 28
Question no. 2 Find the least number that when divided by 16,18 and 20 leaves a remainder 4 in each case, but is completely divisible by 7. a) 364 b)2254 c)2964 d)3234 e)2884
Question no. 1. Find the remainder when 73 * 75 * 78 * 57 * 197 is divided by 34. a) 32 b)30 c)15 d) 28
Question no. 2 Find the least number that when divided by 16,18 and 20 leaves a remainder 4 in each case, but is completely divisible by 7. a) 364 b)2254 c)2964 d)3234 e)2884
see the cyclicity of 2 is 4 53=13*4+1 so rightmost digit is 2
22334=22330+4 22334^2 gives a remainder 1 on division by 5
now we see the power its 66.67.133/6=11.67.133 that means an odd no
remainder 1.4 =4 as we can write the term as 22334.(22334^2)^k
yaar implex i coudnt get u.... its asked....20^53 how can u take 2^53..nd solve for rightmost digit preceeding the zeros is it smethin like...u have taken 2^53 and 10^53 seperatly can u b more clear....
