Three-digit 7248

Find all three-digit numbers n with three different non-zero digits divisible by the sum of all three two-digit numbers we get when we delete one digit in the original number.

Correct answer:

n1 =  138
n2 =  294
n3 =  351
n4 =  459

Step-by-step explanation:

n1=138  a1=18 b1=13 c1=38  s1=a1+b1+c1=18+13+38=69  z1=n1/s1=138/69=2
n2=294  a2=24 b2=29 c2=94  s2=a2+b2+c2=24+29+94=147  z2=n2/s2=294/147=2
n3=351  a3=31 b3=35 c3=51  s3=a3+b3+c3=31+35+51=117  z3=n3/s3=351/117=3
n4=459  a4=49 b4=45 c4=59  s4=a4+b4+c4=49+45+59=153  z4=n4/s4=459/153=3

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