 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Divisible by 28 (Posted on 2014-12-21) Each of a, b and c is a positive integer with a ≤ b ≤ c ≤ 70.

Find total number of triplets (a, b, c) such that a2 + b2 + c2 is divisible by 28.

 No Solution Yet Submitted by K Sengupta No Rating Comments: ( Back to comment list | You must be logged in to post comments.) possible solution | Comment 1 of 4

There are only 8 values for N^2 mod28:  0, 1 , 4, 8, 9, 16, 21, 25.

To make the amount a^2+b^2+c^2 equal to 28*k (k a non-negative integer) we will choose  a triplet  (a ,b, c)   fulfilling   a+b+c =0 mod 28.

It is obtainable by numbers whose mod28 representation is   (a,b,c) = (0, 0, 0)  or (4, 8, 16).

5 numbers correspond to   0:  14, 28, 42, 56, 70      (each of those squared is divisible by 28)           (i)

10 numbers correspond to   4:  2,12,16,26,30,40,44,54,58,68      (each of those squared is 28k+4)    (ii)

10 numbers correspond to  8:   6,8,20,22,34,36,48,50,62,64        (each of those squared is 28k+8)   (iii)

10 numbers correspond to 16:  4,10,18,24,32,38,46,52,60,66      (each of those squared is 28k+16)   (iv)

(0, 0, 0)   combination yields choice of 3 numbers from the   (i) set i.e.    5+10+20=35 choices

(4, 8, 16)   combination -   choice of 3 numbers,   one each of the (ii), (iii), (iv) sets
i.e.    10^3=1000 choices

Answer:   there are 1035 possible solutions.

Hope I did not err.

Apparently I did:  overcounted - I considered triplets 111,112,113,...555 without noticing that 112,121,211 and similar count as one , not three. I got 5^3=125.- wrong thinking

The correct number is 35 (5 by using the same number e.g. 14,14,14 ;
20 -one repeated twice and the third one of the remaining four  e.g. 14,14,28;

10-three distinct numbers e.g. 14,42,56).

So the correct answer is 1035.

Thanks due to Dej Mar for pointing out my error.

BTW, the creation of list like 4,10,18,24,32,38,46,52,60,66 is extremely easy :

1.     First you evaluate n^2 MOD 28 for n=0
to 7,  since (14-n)^2=n^2   taken MOD 28 .

2.  Find the n (within the range 0-7)that  corresponds to 16(for OUR example), get 4.

3, Clearly 14-4=10 is the next in the list, so we have a couple (4,10) to begin with .

4. Add to this couple (14,14), get (4,10,18,24,)

5. The extension of this set is made by adding 14 to each of the two last  members, ...and  again  not go over 70.

- piece of cake.

Edited on December 22, 2014, 12:51 am
 Posted by Ady TZIDON on 2014-12-21 12:13:32 Please log in:
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