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Special equation....
by Pooja Dutta - Wednesday, 28 May 2008, 12:17 PM
  Hey please help me to understand this.
Sachin asked Anil for his birthday Anil replied "Take the date and month of my birthday. Multiply the date with 12 and month with 31 and add up the two product. The sum is 452. Find Anils's bithday.

Using remainder only, the solution given is

12D+31M=452 --------(1)
12D+31M/12 =452/12

D+2M+7m/12 =37+8/12

When L.H.S is divide by 12 the remainder is equal to the remainder obtain when 7M is divided by 12

Rem(7M/12)=Rem(452/12)=8

(I have understood till this, please help me understand the following steps)

=> 7M=12a+8 => rem(5a/7)=6

=> 5a=7b+6 => rem(2b/5)=4

=> 2b=5c+4 =>rem(c/2)=0

when c=0 b=2 a=4 M=8

Substituting this in 1 we get D=17

The other value of D are obtain by adding(or subtracting) the coefficent of M(i.e 31) from this value D.
Similarly for M by add or sub the coeffienct of D,

(D,M) are [17,8] (48,-4),(79,-16),(-14,20),(-45,32) etc.

Can you make me understand this steps. Help Help
Re: Special equation....
by TG Team - Thursday, 29 May 2008, 05:07 PM
  Hi Pooja smile
12x + 31y = 452                                           (1)
=> 12x = 452 - 31y = 12(37 - 2y) + 8 - 7y        (2)
As LHS of (2) is a multiple of 12 so must be RHS
=> 8 - 7y = 12k                                            (3)
=> 7y = 8 - 12k = 7(1 - k) + 1 - 5k                  (4)
As LHS of (4) is a multiple of 7 so must be RHS
for k = 3, 
1 - 5k = 1 - 5*3 = -14 is a multiple of 7 hence the RHS.
putting k = 3 in (3)
we get, y = (8 - 36)/7 = -4. But y need to be positive.
So next suitable values for (k,y) after (3,-4) will be
(-4,8), (-11,20), (-18,32) and so on.
But here y cannot be greater than 12 so y = 8
using (1), x = (452 - 31*8)/12 = 17.
Hence his date of birth is 17th August.smile

Re: Special equation....
by King Anand - Thursday, 29 May 2008, 06:04 PM
  17th august......mental trial ....smile
Re: Special equation....
by rahul gupta - Thursday, 29 May 2008, 07:17 PM
 

hi pooo...smile

here task is to find out the value of M such that 7M gives the remainder 8 on dividing by 12......

one way is hit 'n' trial......

another is the given one.....

ie....

if 7M is divided by 12 giving remainder 8, using remainder theorem,

let qoutient be 'a'.....

7M = 12a + 8.............................................(1)

=>7M = 5a +1 +7( a + 1 )

=>7M -7( a + 1 ) = 5a + 1

ie. 5a when divided by 7 gives remainder -1 or 6

again to find a way is hit 'n' trial or we know that 5a when divided by 7 gives rem 6

let quotient be 'b'......

ie. 5a = 7b + 6............................................(2)

=>5a - 5b - 5 = 2b + 1

ie. 2b when divided by 5 gives remainder -1 or 4

again to find a way is hit 'n' trial or we know that 2b when divided by 5 gives rem 4

let quotient be 'c'........

ie. 2b = 5c + 4.............................................(3)

=> 2b - 4c - 4 = c

=>2(b - 2c - 2) = c

i.e c is divisible by 2................

for least c = 0;

=> b = 2;....................from (3)

=>a = 4;.....................from(2)

=>M = 8;......................from (1)

therefore, D=17 using (1) that u have given..........

ie. 17th august.....

Re: Special equation....
by The Contender - Friday, 30 May 2008, 03:13 AM
  12x + 31y = 452

12 x=452-31y
y should be even so y can be 12,10,8,6,4,2

only y=8 satisfies the equation so

y=8, x=17
Re: Special equation....
by amiable gal - Sunday, 1 June 2008, 06:33 PM
 

hello everyone!!!

The approach i adopted was...

eqn->

12d + 31m = 452

now, we have to subtract a multiple of 31 from 452 such that the remaining part of 452 is a multiple of 12.

only even multiples of 31 less than 452 can only help...

and they are...62,124,186 and 248(2nd,4th,6th and 8th multiple of 31 respectively)

only 248 when subtracted from 452 gives a multiple of 12(i.e. 204).

so...

m=8 (248/31)

and d= 17 (204/12) !!!!!

Take Caresmile

Re: Special equation....
by Pooja Dutta - Thursday, 5 June 2008, 02:13 PM
 
              Thank You all for your replies.


Re: Special equation....
by SaiKiran Allenki - Monday, 18 August 2014, 10:23 PM
  Concept: When a division is split in to a sum of 2 division's( with the same divisor as the original divisor), the original remainder will be equal to sum of the remainders of 2 individual divisions.
A1 = A2+A3
Rem(A1/x) = Rem(A2/x)+Rem(A3/x)

Over here,
7M = 12a+8 => 7(a+1) + 5a + 1
Divide on both sides with 7
Rem(7M/7) = Rem(7(a+1)/7) + Rem (5a/7) + Rem (1/7).
Rem(7M/7) = 0
Rem(7(a+1)/7)= 0
Rem(1/7) = 1
Therefore from the above concept mentioned to neutralise the 1
Rem(5a/7) = -1(i.e. 6).

This way the problem is solved.
Hope this helps you.