question_answer

Let, x be the greatest number of 4 digits, which when divided by 15, 20 and 28 leaves in each case the remainder 2. The sum of digits of x is

A) 19                    

B) 21     

C) 23        

D)        25

Correct Answer: C

Solution :

Sol.

LCM of 15, 20 and 28 is

LCM = \[2\times 2\times 5\times 3\times 7=420\]

Largest number of 4-digits = 9999

Now, \[\frac{9999}{420}\,=\,23\frac{339}{420}\]

\[\therefore \] Remainder = 339

\[\therefore \] Four digit number divisible by 15, 20 and 28

9999 - 339 = 9660

So number when divided by 15, 20 and 28 leaves Remainder

= 9660 + 2 = 9662

\[\therefore \] Sum of digits = 9+6 + 6+2=23