Question

A number when divided by a divisor leaves a remainder of 5 and when divided by twice the divisor leaves a remainder of 45. Find the divisor.

Answer 1

Let d is divisor , Q1 and Q2 are quotient of 1st and 2nd cases respectively.

first case ;
number = d × Q1 + 5

2nd case;
number = 2d × Q2 + 45

from 1st and 2nd cases ;
d × Q1 + 5 = 2d × Q2 + 45
d × Q1 - 2d × Q2 = 45 - 5
d( Q1 - 2Q2) = 40
d = 40/( Q1 - 2Q2)

hence, divisor may be possible 40 or factor of 40 { e.g 2 , 5 , 8, 10, 20 } .

now , according to Euclid division lemma
we know,
a = bq + r
where 0 ≤ r < b
hence, 0 ≤ 5 < d { in first case
and 0 ≤ 45 < 2d
or 0 ≤ 22.5 < d { In second case

hence, d > 22.5
so, it's possible only when ,
we take d = 40

hence, divisor = 40

Answer 2

Answer:

40

Step-by-step explanation:

Let the number is N , divisor is d and a is the quotient

Then N=ad+5

N-5=ad……………….(1)

Let N=2d*b+45

N-45=2bd………………….(2)

Subtracting(2) from(1)

40 =(a-2b)*d

Thus d is a divisor of 40

so d=2,4,5,8,10 ,20 or 40………….(3)

Now divisor is always greater than remainder and remainder is 45

when N is divided by 2d

so 2d > 45

d > 22.5………………(4)

Thus from (3) and (4) we conclude that d=40

Thus divisor d=40