mani517
Joined: 20 Dec 2005
Posts: 1
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 Posted: Tue Dec 20, 2005 11:46 am Post subject: Question 4 the day: June 21, 2002 - a doubt |
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Hey! This is manikandan from chennai. I'm new to ascent education, I just went through your archives on number system; really helpful for my preparation. 
But I came across a doubt, while checking the solution for the problem named as Question 4 the day: June 21, 2002.
http://www.ascenteducation.com/india-mba/iim/cat/questionbank/Archives/June2002/arith2106.shtml - check this URL for the above mentioned Q.
Problem:
A certain number when successfully divided by 8 and 11 leaves remainders of 3 and 7 respectively. What will be remainder when the number is divided by the product of 8 and 11, viz 88? (1) 3 (2) 21 (3) 59 (4) 68
Correct Answer - (3)
Solution mentioned on your page is:
Solution:
When a number is successfully divided by two divisors d1 and d2 and two remainders r1 and r2 are obtained, the remainder that will be obtained by the product of d1 and d2 is given by the relation
d1r2 + r1.
Where d1 and d2 are in ascending order respectively and r1 and r2 are their respective remainders when they divide the number.
In this case, the d1 = 8 and d2 = 11. And r1 = 3 and r2 = 7
Therefore, d1r2 + r1 = 8*7 + 3 = 59.
 my doubt:
51 is a number that satisfies the problem condition (i.e.) when divided by 8 gives a remainder of 3; and when divided by 11 gives a remainder of 7. This obviously means that the solution for the problem is 51. Since we can derive 'n' number of controversies from: (88n)+51 [where 0<n<infinity]
Is your solution right? if yes, please explain. |
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