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Grade 12IIT JEE Entrance Exam

The six-digit number 4m61n2 is visible by both 11 and 4. The number of different combinations of m and n that satisfy the above condition are:
A. 4
B. 6
C. 8
D. 10
please explain in detail with answer

Profile image of vindhya
11 Years agoGrade 12
Answers icon

3 Answers

Profile image of Mihit
ApprovedApproved Tutor Answer11 Years ago
ANSWER   IS   4(option A)
if no. 4m61n2 is divisible by both 11 and 4 then n=1,3,5,7,9 
also , if it has to be divisible by 11 then its ;; sum of every second digit – sum of remainig=0/11
here, (4+6+n)-(m+1+2)=0/11
10+n-3-m=0/11
7+n-m=0/11
so  we get that
                                                      n=1  ,      3    ,5      7    9     
m(m will be +ve as it si a digit in a no.)=8   ,  10     1    3      5
so we get 5 combinations
1,8
3,10
5,1
7,3
9,5
 
but combination 3.10 does not satisfy our need as ‘m‘ can only be a 1 digit no. as it is a single digit . SO THE ANSWER IS 4
                                                           - Mihit
                                                             (class 9)
 
Profile image of Mihit
11 Years ago
si-is*

 
Profile image of vindhya
11 Years ago
thanks a lot mihit. good luck!

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