Aptitude - Numbers - Discussion
Discussion Forum : Numbers - General Questions (Q.No. 46)
46.
In dividing a number by 585, a student employed the method of short division. He divided the number successively by 5, 9 and 13 (factors 585) and got the remainders 4, 8, 12 respectively. If he had divided the number by 585, the remainder would have been
Answer: Option
Explanation:
5 | x z = 13 x 1 + 12 = 25
--------------
9 | y - 4 y = 9 x z + 8 = 9 x 25 + 8 = 233
--------------
13| z - 8 x = 5 x y + 4 = 5 x 233 + 4 = 1169
--------------
| 1 -12
585) 1169 (1
585
---
584
---
Therefore, on dividing the number by 585, remainder = 584.
Discussion:
39 comments Page 1 of 4.
Rutuja Patil said:
2 years ago
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)
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)
(28)
Tomo Lollen said:
2 years ago
Why we took 13 first and not 5? Please explain?
Sudip said:
4 years ago
Thanks @Manideep.
Naveen kumar v said:
4 years ago
Thanks @Manideep.
(1)
Niru said:
5 years ago
In successive division, last quotient is always 1.
Mohit said:
5 years ago
Thanks @Arati.
(1)
Pabitra said:
6 years ago
Anyone, please explain this for me.
(1)
Pooja said:
6 years ago
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.
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.
(30)
Gowry said:
7 years ago
Thanks @Saurabh.
It is very simple to understand.
It is very simple to understand.
Aliya said:
7 years ago
Thanks @Ankit.
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