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Look for 3 sevens (Posted on 2016-04-01) Difficulty: 2 of 5
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.

No Solution Yet Submitted by Ady TZIDON    
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the rough way | Comment 1 of 2
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
  Posted by armando on 2016-04-01 09:34:52

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