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`        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. 4B. 6C. 8D. 10please explain in detail with answer `
2 years ago

Mihit
32 Points
```							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=0/11here, (4+6+n)-(m+1+2)=0/1110+n-3-m=0/117+n-m=0/11so  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      5so we get 5 combinations1,83,105,17,39,5 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)
```
2 years ago
Mihit
32 Points
```							si-is*
```
2 years ago
vindhya
50 Points
```							thanks a lot mihit. good luck!
```
2 years ago
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### Course Features

• 731 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions