# Remainder problem

• Jan 20th 2009, 02:31 AM
thereddevils
Remainder problem
(1) A number when divided by a divisor leaves a remainder of 24 . When twice the original number isdivided by the same divisor , the remainder is 11. What is the value of the divisor ?

(2) When 242 is divided by a certain divisor the remainder obtained is 8 . When 698 is divided by the same divisor the remainder obtained is 9 . However , when the sum of the two numbers 242 and 698 is divided by the divisor , the remainder obtained is 4 . What is the value of the divisor ?
• Jan 20th 2009, 05:31 AM
sumit2009
Quote:

Originally Posted by thereddevils
(1) A number when divided by a divisor leaves a remainder of 24 . When twice the original number isdivided by the same divisor , the remainder is 11. What is the value of the divisor ?

Ans-divisor is 37
let the number be X and divisor be D
then X=D*q+24 --------------------(1)Euclid division lemma

2X=2[D*q+24] ------------------muliplying by 2

=D*2q + 48

=D*2q + 37 +11

if we take D=37 then, 37 can be taken common from first 2 terms
2X =37[2q+1] + 11

=37[Q] + 11 2q+1=Q (say)

which is in accordance with 2nd condition.
• Jan 20th 2009, 05:46 AM
AlvinCY
Since the remainder is 24 and 11 respectively, we know that the divisor must be greater that 24 (otherwise a remainder of 24 is not possible) I used modular arithmetic to find the divisor to be 37, but simple guess and trial would have gotten you there!

By trial and error:

$61\div37=1r24$

$122\div37=3r11$

and...

By similar methods, the divisor must be greater than 4, 8 and 9... so start testing divisors from 10...

I know $242\div d=$ Answer remainder 8...

divisor of 10 gives remainder of 2
divisor of 11 gives remainder of 0
divisor of 12 gives remainder of 2
divisor of 13 gives remainder of 8

$242\div13=18r8$

$698\div13=53r9$

$242+698=940$

$940\div13=72r4$

Voila!