Aptitude - Numbers - Discussion

From L.C.M and H.C.F methods
We have two formulas
i.e
1-> To find the greatest number that will divide by x,y&z leaving remainders a,b,c,respectively.

Required no.= H.C.F of(x-a),(y-b)&(z-c).

2->To find the least number which when divided by x,y,z leaves the remainder a,b,c, respectively.
It is always observed that (x-a)=(y-b)=(z-c)=k
Required no.=(L.C.M of x,y&z)-k

But in question, we have not asked for find out the least or greatest number, So by ignoring that least and greatest number,we have to notice on the given dividends and remainders.

Here,
we have 3 dividends and 3 remainder i.e,
Dividends =5,9,13.
Remainders =4,8,12.

Now,we have to substract the remainders of the respective dividends to check if the differences are same then here we apply the L.CM rule but if the differences are different then we will apply the H.C F rule.

But here the differences between every dividend and remainder is same.i.e,
(5-4)=(9-8)=(13-12)=(k)1.

As the differences are same here we apply the formula 1 to find the remainder.
Required number = (L.C.M of x,y&z)-k.
= (L.C.M of 5,9&13)-1,
= 585-1.
= 584 (remainder) answer.

Hope, my solution will be logical:

Suppose, the number is G.
According to the question:
G = 5a + 4 [where, a is an integer].
G = 9b + 8 [where, b is an integer].
G = 13c + 12 [where, c is an integer].

Now,
5a + 4 = G.
=> 5a + 5 - 1 = G.
=> 5(a + 1) = G + 1.

9b + 8 = G.
=> 9(b+) = G + 1.

13c + 12 = G.
=> 13(c + 1) = G+1.

Here, [5(a + 1)], [9(b + 1)], and [13(c + 1)] have same value [i.e: (G+1)].
So, the LCM of [5(a + 1)], [9(b + 1)], and [13(c + 1)] will also be (G + 1).

In other words,
LCM of [5(a + 1)], [9(b + 1)], and [13(c + 1)] = 585(a + 1)(b + 1)(c + 1).
So, 585(a + 1)(b + 1)(c + 1) = (G + 1).
Or, G = 585(a + 1)(b + 1)(c + 1) - 1.

Now, The number divided by 585.

=> G/585 = [585(a + 1)(b + 1)(c + 1) - 1]/585.
= [585(a + 1)(b + 1)(c + 1) - 585 + 584]/585.
= [{585(a + 1)(b + 1)(c + 1) - 1} + 584] = {(a + 1)(b + 1)(c + 1) - 1} + 584/585.

Surely {(a + 1)(b + 1)(c + 1) - 1} is an integer. So, the remainder is 584.
Then, the answer is 584.

Dividend=divisor*quotient+remainder ------> (A)
Here, n=585*Q+R -----> (1)

Given,
If divided by 5, then remainder = 4, let quotient = x.
Therefore, n=5x+4 -----> (2)

Now if x is divided by 9, then remainder = 8, let quotient=y;
Therefore, x=9y+8 ----->(3)

& if y is divided by 13, then remainder = 12, let quotient=z
Therefore, y=13z+12 -----> (3).

From eq. (2) x=9y+8.
Therefore x = 9 (13z+12)+8 -----> (putting value of z from eq.(3))
x = 117z + 108 + 8
= 117z +116 -------> (5)

put (5) in (2)
we get;
n = 5(117z+116)+4.
= 585 * z +580+4.
n = 585 * z +584.

In above eq;
(refer to formula (A))
divisor= 585,
quotient= z,
& remainder = 584 -->(Answer)

@All.

Successive division means, if a number X is successively divided by say p and q, then first X is divided by p and then the quotient obtained when X is divided by p is then divided by q.

Here let X be the number, since it is successively divided by 5, 9 and 13. We obtain.

X=5q1+4------- (1) (4 is remainder) and then q1 is divided by 9 then.
Q1=9q2+8------- (2) (given 8 is remainder) then.
Q2=13q3+12------ (3) (given 12 is remainder).

Substituting q2 in eq2, we get q1=117q3+116.
Then substituting q1 in equation 1 we get.
X=585q3+584.
Hence 584 is the remainder.

Hope it helps everyone.

There is a shortcut to solve this problem.

If a number (say X), when divided by d1, d2, d3 (either successively or independently) leaves remainders r1, r2, r3 and if (d1 - r1) = (d2 - r2) = (d3 - r3) = k.
then,

X = L.C.M (d1, d2, d3) - k.

From the question.
(d1,d2,d3) = (5,9,13),
(r1,r2,r3) = (4,8,12).
k = 1.
X = L.C.M(5,9,13)-1 = 585-1 = 584.

(584%585) = 584 (i.e. the remainder).

First thing to keep in mind is, the number is divided SUCCESSIVELY by 5, 9, 13.

Thus N = 5a + 4 or 5a - 1.

a = 9b + 8 or 9b - 1.

b = 13c + 12 or 13 - 1.

So the remainder in each case is 1. LCM of 5, 9, 13 is 585 so the remainder will be same (its a theorem) if divide by the LCM.

Therefore N = 585d - 1 rem= 584.

@Maindeep.. How come assumed x=5a+4,a=9b+8,b=13c+12.
Because it should have been like this x=5a+4=9b+8=13c+12=585d+r.

Where r is the required remainder ...So there will be 3 equations to solve for a, b, c and x but unknown are for.(I am only confused about your assumption..please through some light on that).

Hello everyone.

Find reminder this type problem use formula ( 1st reminder + 2nd reminder * 1st factor + 3rd reminder * 1st factor * 2nd factor ) then divided by given number.

Here, ( 4+8*5+12*5*9) /585 = 584/585 = 584 Ans because we know if ( a/a+1 ) then reminder always = a.

@ Arti

Let the number be x.

Now given x=5a+4, a=9b+8, b=13c+12...

By reverse substitution you get

a = 9(13c + 12) + 8 = (13*9)c + 116

x = 5[(13 * 9)c + 116] + 4 = 585c + 584

So this when divided by 585 you get 584 as reminder.

Thank you :)

On dividing a number by a,b,c if we get a-k, b-k nd c-k as remainder respectively then that number will be N*LCM of[a,b,c]-k.
therefore:
=N*[lcm of 5,9,13]-1
=N*[585]-1
=N*584
=584 here N=1