define a number k such that it is the sum of the squares of the first M natural nos(i.e k=1^2+2^2+...+m^2), where m<55. how many valuues of m exist such that k is divisible by 4?

1) 10
2) 11
3) 12
4) 13

• Ans is 12
  
  Detailed solution
  disclaimer: the solution is long but involves small small concepts. As we know while improvising answer for any question we come across many small sub-questions solving which the main question is solved.
  
  K= n(n+1)(2n+1)/6
  a) in what case will k be divisible by 4?
  for k to be divisible by 4, n(n+1)(2n+1) should be divisible by 24.
  
  b) in what case will n(n+1)(2n+1)/6 be divisible by 24?
  24= 2^3 *8
  so it can be represented as a product of two factors in 4 ways.
  they are 24*1
  12*2
  8*3
  4*6
  so in the expression n(n+1)(2n+1), we should either have the above pairs or their multiples.
  
  c) do all the four pairs have the chance of coming in the numerator?
  n(n+1) will be product of two consecutive numbers,
  (2n+1) will be an odd number.
  
  of the 4 ways of representing 24. two are product of two even numbers.(4*6) and 12*2. but our expression will have one even number and two odd numbers. ( cos out of n(n+1) one is even other is odd. and (2n+1) is always odd). so we can rule out cases of 4*6 and 12*2 coming in the numerator. now we're left with 24*1 and 8*3.
  
  
  24 can come twice in the numerator, once in S23 and then in S24.
  similarly 48 in S47 and S48.
  
  now we're left with 8*3. we've to check for multiples of 8 also viz 16,24,32,40,48. as 24 and 48 have been dealt earlier. we have only four multiples. 8, 16, 32, 40. all of which will occur twice but we also need a 3 or multiple of 3.
  
  our 8 multiple can occupy n or (n+1) position. in that case we can look for multiples of 3 in [(n+1) and (2n+1)] and [n and (2n+1)] positions respectively.
  
  d) in what all cases we'll have a multiple of 3 along with 8 in Nr?
  
  lets take the example of our 8 multiple 32. it can come as 31*32 and 32*33. i.e 31,32,33 triplet. in any three consecutive nos we'll have 1 and only 1 multiple of 3. here 32*33 has a three multiple. in case of 16, 15*16 has a 3 multiple. now we're left with the other pair in all the four nos viz 7*8, 16*17, 31*32, 40*41. now we go to the last resort, (2n+1) to look for a 3 multiple.
  
  d) for what all values of n can (2n+1) be a multiple of 3?
  any no can be written in any of these three forms viz. 3m or 3m+1 or 3m+2. only when n is in the form 3m+1, 2n+1 will be divisible by three[ 2(3m+1)+1=6m+2+1=6m+3].
  incidently, all the pairs 7*8, 16*17, 31*32, 40*41 have the first number in the form 3m+1. so the third term will be a multiple of three.
  hence two pairs for each of 8,16,24,32,40,48 will have 24 as one of their multiple. so 12 nos.
  when the smaller concepts are clear the Qn can done in very less time.
  looking forward for smaller approaches.
• 8 years agoHelpfull: Yes(0) No(0)