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`        Highest comman factor of (21m+4) and (14m+3) is? [where m belongs to N]`
one year ago

Arun
23518 Points
```							Dear student please post the optio s tooas these question require options too to check. I am solving this as a question which is similiar to this sokved by me in past m = natural number.   Two numbers:  N1: 21 m + 4,     N2:   14 m + 31)  N2 is always odd, as 14m is even and 3 is odd. So option a is incorrect. N2 is not divisible by 2. So  2 cannot be HCF.2)  N1 = 21 m + 4 ,      N2 = 14 m + 3     Add N1 + N2 =  35 m + 7 = 35 m + 5 + 2    If N1 and N2 are divisible by 5, then N1+N2 must be divisible by 5.     But we see clearly that N1+N2 is not divisible by 5.3)    N1 = 21 m + 4     N2 = 14 m + 3       N1 - N2 = 7 m + 1        3 N2 - 2 N1 = 42 m + 9 - 42 m - 8 = 1      If N1 and N2 are both divisible by 9, then  3 N2 - 2 N1 should be divisible by 9 or it should be 0.  So 9 is not the HCF.
```
one year ago
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