Find the smallest number N with the property that its first
3 multiples (i.e. N,2N,3N) contain the digit
7.
Analytical solution preferred.
Bonus: Same for the first 4 multiples.
If N has 7 in units, 2N has to have the 7 in tens. For N=37 (2N=74). But it failed for 3N.
If N has a 7 in tens, 2N is always even so the 7 is in tens or hundreds.
In tens: a7b*2=c7d implies that 2b=d (<20), as 2*7=14 we should need 30<d<40. So, no way.
In hundreds: 2N=7ef implies N has 7 in units or tens
In units: 2N=7e4 so N=3g7 and 3N=[9+(1,2)][3g+2][1] and then 3g+2=7 which is impossible.
In tens: 2N=7ef so N=37h so e=[4,5]. But then 3N=[11][1+(0,1,2)][3h]. So the only possible 7 is with 3h if h=9.
Then 3N=1137, 2N=758, N=379.
Edited on April 1, 2016, 9:36 am
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Posted by armando
on 2016-04-01 09:34:52 |
the rough way
