A number N consisting of a string of threes i.e. N=33333...3333, is divisible by 499.
Find the last 5 digits of the quotient. No programming, please.
499*7 = 3493
to drop the tens place by 6, add 499*6*10
499*67 = 33433
to drop the hundreds place by 1, add 499*1*100
499*167 = 83333 (fourth 3 is spurious)
to drop the thousands place by 10, add 499*10*1000
499*10167 = 5073333
etc
499*50167 = 25033333
etc
499*750167 = 373333333
eventually the product will be all 3's.
The last five digits of the multiplier will be 50167.
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Posted by Jer
on 2025-02-11 11:46:00 |
Solution
