Aptitude - Numbers - Discussion
Discussion Forum : Numbers - General Questions (Q.No. 131)
131.
A number when divided successively by 4 and 5 leaves remainders 1 and 4 respectively. When it is successively divided by 5 and 4, then the respective remainders will be
Answer: Option
Explanation:
4 | x y = (5 x 1 + 4) = 9 -------- 5 | y -1 x = (4 x y + 1) = (4 x 9 + 1) = 37 -------- | 1 -4 Now, 37 when divided successively by 5 and 4, we get 5 | 37 --------- 4 | 7 - 2 --------- | 1 - 3 Respective remainders are 2 and 3.
Discussion:
22 comments Page 2 of 3.
Gaurav said:
10 years ago
Here 29 is fit for this by hit and trail method when 29/4 give reminder 1 and when 29/5 gives reminder 4 so why 4, 1 is not answer.
Alam milan said:
9 years ago
Can you please tell the answer for this?
A number x when divided by 9 leaves a remainder 8. What is the remainder when we divide x^24 by 9?
A. 3
B. 2
C. 7
D. 1
A number x when divided by 9 leaves a remainder 8. What is the remainder when we divide x^24 by 9?
A. 3
B. 2
C. 7
D. 1
(1)
Kunal madankar said:
9 years ago
When a number is successively divided by 7, 5 and 4 it leaves remainders of 4, 2 and 3 respectively. What will be the respective remainders when the smallest such number is successively divided by 8, 5 and 6?
(1)
Ripa Roy said:
8 years ago
37/4=remainder 1.
Then how 1 - 3? I don't understand it, please explain it.
Then how 1 - 3? I don't understand it, please explain it.
Sasikala said:
8 years ago
@Ripa Roy.
x=4a+1 and a=5b+4.
when successively divide a number, quotient will be taken as dividend.
to find the number:
a=5(1)+4=9,
a=9.
Substitute a value to x then x=4(9)+1=36+1=37.
x=37.
when 37 is successfully divided by 5 and 4,
37/5=quotient 7 and remainder 2,
then 7/4=quotient 1 and remainder 3,
so answer is 2 and 3.
x=4a+1 and a=5b+4.
when successively divide a number, quotient will be taken as dividend.
to find the number:
a=5(1)+4=9,
a=9.
Substitute a value to x then x=4(9)+1=36+1=37.
x=37.
when 37 is successfully divided by 5 and 4,
37/5=quotient 7 and remainder 2,
then 7/4=quotient 1 and remainder 3,
so answer is 2 and 3.
(6)
Mr. MP said:
8 years ago
I took number as 37. I calculated by trial and error.
When a number is divided by 4 remainder is 1 and successive division (this is nothing but QUOTIENT OF FIRST division). Initially, I used 5 as the number and divided by 4 5 5/4 I got remainder 1 and quotient also 1. Now a successive division of 1/5 (here 1 is the quotient of previous division) won't give rem 1. So I used 9 so that 9/5 remainder is 4. Now 9 quotient of step 1 division 4*9 =36 so if I take 37 as the number remainder is 1.
See now 37/4 = 9 quo and 1 rem matches the rem in question.
Successive division 9/5 1 quo and 4 rem, matches rem in question.
Now as asked in the question first div by 5 and then 4 find rem.
37/5 = 2 rem quo is7 and 7/4 1 quo 3 rem.
So the answer is 2, 3.
When a number is divided by 4 remainder is 1 and successive division (this is nothing but QUOTIENT OF FIRST division). Initially, I used 5 as the number and divided by 4 5 5/4 I got remainder 1 and quotient also 1. Now a successive division of 1/5 (here 1 is the quotient of previous division) won't give rem 1. So I used 9 so that 9/5 remainder is 4. Now 9 quotient of step 1 division 4*9 =36 so if I take 37 as the number remainder is 1.
See now 37/4 = 9 quo and 1 rem matches the rem in question.
Successive division 9/5 1 quo and 4 rem, matches rem in question.
Now as asked in the question first div by 5 and then 4 find rem.
37/5 = 2 rem quo is7 and 7/4 1 quo 3 rem.
So the answer is 2, 3.
(1)
Neeraj said:
6 years ago
4Q+1=n.
5q+4=Q.
n = 4(5q+4)+1 = 20q+17= 5 * 4q + 5 * 3 + 2.
Now n is devided by 5 so the remainder is 2.
And now devide 4q+3 by 4 so remainder is 3.
5q+4=Q.
n = 4(5q+4)+1 = 20q+17= 5 * 4q + 5 * 3 + 2.
Now n is devided by 5 so the remainder is 2.
And now devide 4q+3 by 4 so remainder is 3.
(7)
Ziyaul Haque said:
5 years ago
@Sasikala.
Why b=1?
Why b=1?
Raihan said:
5 years ago
Thanks @Jay.
Sahil said:
3 years ago
Original number is assumed as x.
The quotient after dividing by 5 is assumed in the given solution as 1. This could be assumed as any number and has no effect on the remainder. This is because there are many numbers such as 17, 37, 57 etc. that satisfy the criteria of leaving 1 and 4 as remainders after successively dividing by 4 and 5.
If we assume 1 as the quotient after division by 5, we get two equations:
4 | x
5 | y (remainder: 1)
|1 <== assume any number here (remainder: 4)
1*5 + 4 =y ===> y=9;
y*4+1=x ====> x=37;
Once we get the original number as 37, divide by 5, get the remainder 2 and quotient 7. Divide 7 by 4, get the remainder 3. Hence the answer 2,3.
The quotient after dividing by 5 is assumed in the given solution as 1. This could be assumed as any number and has no effect on the remainder. This is because there are many numbers such as 17, 37, 57 etc. that satisfy the criteria of leaving 1 and 4 as remainders after successively dividing by 4 and 5.
If we assume 1 as the quotient after division by 5, we get two equations:
4 | x
5 | y (remainder: 1)
|1 <== assume any number here (remainder: 4)
1*5 + 4 =y ===> y=9;
y*4+1=x ====> x=37;
Once we get the original number as 37, divide by 5, get the remainder 2 and quotient 7. Divide 7 by 4, get the remainder 3. Hence the answer 2,3.
(10)
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