Latest L&T Aptitude Question SOLUTION: What is the greatest number of 4 digits that when divided by any of the numbers 6, 9, 12, 17 leaves a remainder of 2? 1) 9792 2) 9982 3) 9794 4) 9906 5) None of these

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(#M40156517) L&T Numerical Ability question Keep an EYE

What is the greatest number of 4 digits that when divided by any of the numbers 6, 9, 12, 17 leaves a remainder of 2?
1) 9792
2) 9982
3) 9794
4) 9906
5) None of these

Asked In L&T ABHINAV PUSHPAM (7 years ago)

Unsolved Read Solution (4)

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Read Solution (4) :

Aditya  (7 years ago)

take L.C.M of all numbers i.e 6,9,12,17
the L.C.M is 612 therefore 612 is divisible by all numbers
so multiple of 612 will also be divisible by all the above numbers
so 612*17=9792 which is divisible by all
add 2 to 9792 to get a remainder of 2 as required in the question..
so ans is 9792+2=9794

Is this solution Helpfull? Yes (6) | No (2)  

Roger Anton  (7 years ago)

Soln to find n-digit greatest number divided by a, b, c leaving same remainder K

https://books.google.co.in/books?id=CIzCCwAAQBAJ&lpg=SA2-PA4&ots=2CwuyFvMc1&dq=n-digit%20greatest%20number%20divided%20by%20a%2C%20b%2C%20c&pg=SA2-PA4#v=onepage&q&f=false

9999/612
Q=16 R=207
ans = 9999-207+2
=9794

Is this solution Helpfull? Yes (4) | No (3)  

Gaurav Singh  (7 years ago)

Apply hit and trial method at each option
9792/9= 0 remainder (failed)
9982/9= 1 remainder (failed)
9794/6= 2
9794/9= 2
9794/12=2
9793/17= 2
condition is satisfy so answer is option c

Is this solution Helpfull? Yes (3) | No   

BHARAT KUMAR  (7 years ago)

Take LCM of all the numbers 6,9,12,17 =612
and devide each option
(3) 9794

Is this solution Helpfull? Yes (1) | No   

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