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3's & 7's (Posted on 2006-04-09) Difficulty: 3 of 5
Find the smallest number comprised of only 3ís and 7ís which fits the following conditions:

1) It has at least one 3;
2) It has at least one 7;
3) It is divisible by 3;
4) It is divisible by 7;
5) The sum of its digits is divisible by 3;
6) The sum of its digits is divisible by 7.

See The Solution Submitted by Jer    
Rating: 3.2500 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: the computer doing the trial and error | Comment 9 of 13 |
(In reply to the computer doing the trial and error by Charlie)

Close but no cigar...

 

<TABLE style="WIDTH: 174pt; BORDER-COLLAPSE: collapse" cellSpacing=0 cellPadding=0 width=232 border=0 x:str> <COLGROUP> <COL style="WIDTH: 174pt; mso-width-source: userset; mso-width-alt: 8484" width=232> <TBODY> <TR style="HEIGHT: 12.75pt" height=17> <TD class=xl22 style="BORDER-RIGHT: #ece9d8; BORDER-TOP: #ece9d8; BORDER-LEFT: #ece9d8; WIDTH: 174pt; BORDER-BOTTOM: #ece9d8; HEIGHT: 12.75pt; BACKGROUND-COLOR: transparent" align=right width=232 height=17 x:num="3333337773">3333337773</TD></TR></TBODY></TABLE>

How to solve

3*7  = 21  but no number of figures of 7's and 3's solve it

so 21*2 = 42, so how many 3's and 7's can make 42 and devive by 7's and 3's  (i.e.  21 + 21)

so 7 3's and 3 7's

 

Make an order and the last digit MUST be 3.

3333337773 added totals 42

/3 = 1111112591

/7 = 476191110

 

Took about 1 hour with pen and paper.

 


  Posted by stewart on 2006-04-10 07:23:03
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