1MN ÷ 3 = MN, Replace the letters with appropriate digits.
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I think the digit will be 1 and 0 at the place of digit 1 there can be another digit
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Given 1MN = MN
If 1MN is divisible by 3 then sum of
all the digits is divisible by 3
=> 1 + M + N = 3 × { 1 , 2 , 3 }
Let 1 + M + N = 3 × 2 = 6
M + N = 5 ------( 1 )
1MN/3 = MN
=> 1MN = 3( MN )
=> 1×100+M×10+N×1 = 3[M×10+N×1]
=> 100+10M+N= 3[10M+N ]
=> 100 + 10M + N = 30M + 3N
=> 100 = 30M + 3N - 10M - N
=> 100 = 20M + 2N
Divide each term with 2 , we get
=> 50 = 10M + N
=> 10M + N = 50 ----( 2 )
Subtract ( 1 ) from ( 2 ), we get
9M = 45
=> M = 45/9 = 5
Substitute M = 5 in equation ( 1 ),
we get
M + N = 5
=> 5 + N = 5
=> N = 5 - 5 = 0
Therefore ,
M = 5 , N =0
1MN/3 = MN
150/3 = 50
••••
If 1MN is divisible by 3 then sum of
all the digits is divisible by 3
=> 1 + M + N = 3 × { 1 , 2 , 3 }
Let 1 + M + N = 3 × 2 = 6
M + N = 5 ------( 1 )
1MN/3 = MN
=> 1MN = 3( MN )
=> 1×100+M×10+N×1 = 3[M×10+N×1]
=> 100+10M+N= 3[10M+N ]
=> 100 + 10M + N = 30M + 3N
=> 100 = 30M + 3N - 10M - N
=> 100 = 20M + 2N
Divide each term with 2 , we get
=> 50 = 10M + N
=> 10M + N = 50 ----( 2 )
Subtract ( 1 ) from ( 2 ), we get
9M = 45
=> M = 45/9 = 5
Substitute M = 5 in equation ( 1 ),
we get
M + N = 5
=> 5 + N = 5
=> N = 5 - 5 = 0
Therefore ,
M = 5 , N =0
1MN/3 = MN
150/3 = 50
••••
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