A number when divided by three consequative integer 9,11,13 leaves remainder 8,9and 8 respt
Answers
Answer:
Step-by-step explanation:
Let N be the required number which when divided by the integers a,b and c leaves as reminder p,q and r:
N=ax+p=by+q=cz+r
[x,y and z are the quotients for a,b and c respectively]
The divisors a,b,c and the remainders p,q and r are known to us[and tyhey are o course integers]
x=(cz+r-p)/a
y=(cz+r-q)/b
We try out integral values of z to find simultaneously integral values of x and y.For unfriendly figures or even for modest values of the divisors and the remainders the trial method [for the general case]could be very long and one might be forced to use the computer.
For the problem at hand,trying out z=72 we have x=104 and y=85
We may consider the problem:
What number when divided by 18,24 and 32 leaves remainders 7,19 and 27 respectively?
N=18*x+7=24*y+19=32*z+27
We have z=5 for which x=10,y=7: z is reasonably small in this case for any type of manual enterprise
N=187
There could be other solutions for the sane problem like
z=14 for which x=26, y=19 …….
N=475
z=23 for which x=42,y=31
N=763
N=ax+p=by+q=cz+r
Again,
N’=ax’+p=by’+q=cz’+r
N’-N=a(x’-x)=b(y’-y)=c(z’-z)
x’-x=(N’-N)/a;y’-y=(N’-N)/b;z’-z=(N’-N)/c
N’-N=multiple of the LCM of the divisors a,b and c
Out of the infinite sequence of solutions the appropriate one may be sieved out by using other conditions in the problem
If two of the remainders are the same we can always have an advantage of that to have simplified process [or if the difference divisor-remainder is the same for two or three divisors there are standard short cut methods].
LCM of 18, 24 and 32=288
N’=N+s*288;s: integer
]s=1
N’=187+288=475
s=2
N’=187+2*288=187+576=763
With the knowledge of N’ -N=multiple of the LCM of the divisors we may calculate x’.y’ and z’ from x,y and z using earlier stated the relation
x’-x=(N’-N)/a;y’-y=(N’-N)/b;z’-z=(N’-N)/c
x’=x+16,y’=y+12,z’=z+9
x’>16,y’>12,z’>9
Let’s assume there are no values of x,y and z satisfying:
x<16,y<12,z<9
The first[least valued triplet] does hot satisfy the above for the sated problem
The least values are denoted
x=k1+16;y=k2+12;z=k3+9;k1,k2,k3>0
[for other solution sets the corresponding values are larger]
We kave
18(k1+16))+7=24(k2+12)+19=32(k3+9)+27
or,18*k1+288+7=24*k2+288+19=32*k3+288+27
288=LCM of the divisors and it will appear in the general case for any problem of this type:and the LCM term will it will cancel out. We now have,
8*k1+7=24*k2+19=32*k3+27
Therefore k1,k2 and k3 are themselves solutions to x,y and z
Therefore x,y,z given by
x=k1+16;y=k2+12;z=k3+9
is not the first solution set as claimed by the italicized text. From this contradiction we infer that the first solution set will satisfy
x<16,y<12,z<9
That makes prediction comfortable[when manual enterprise is in our consideration]. Once the first set[smallest set] of (x,y,z) is correctly ascertained we can easily calculate the other sets, the higher ones.
Now we discuss the following properties
x’’=x’+16=x+32;y’’=y’+12=y+24,z’’=z’+9=z+18
x’’>32,y’’>24,z’’>18
Implies 16<x’<32; 12<y’<24;9<z’<18
In view of the above stated intervals we may consider the triplet:(x.y,z)=(26,19,14)
x^(n)=x+n*16 implies x^(n)>n*16
y^(n)=y+n*24 implies y^(n)>n*24
y^(n)=y+n*24 implies y^(n)>*24