If 3n zeros are placed between the digits 3 and 7, then the number formed is divisible by 37. In addition, if 3n+1 zeros are placed between the digits 7 and 3, the number formed is also divisible by 37.
It means 3000...7 is divisible by 37. Here,the zeroes are in 3n form.
It means 70000...3 is also divisible by 73. Here zeroes are in 3n+1 form.
Prove it.
Quick proof by induction.
A=300...07 with 3n zeroes. The next term is 1000(A-7)+7 = 1000A-6993 = 1000A-37*89. Since 37 is the first term, the result follows.
Similarly, if B=700...003 with (3n+1) zeroes, the next term is 1000(B-3)+3 = 1000B-2997 = 1000B-37*81.
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Posted by xdog
on 2019-06-19 09:08:34 |
solution
