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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)

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Solution Puzzle Solution | Comment 12 of 13 |
(In reply to answer by K Sengupta)

Let N be the smallest number which contains precisely p number of 3's and q number of 7's.

Then, s.o.d(N) = 3p+7q. Since (3p+7q) is divisible by 3, it follows that the minimum q must be 3, and alos since (3p+7q) is divisible by 7, it follows that the least p must be 7. Thus N must contain at least three 7's and at least seven 3's. Since the location of 7's does not affect the divisiblity of N by 7, we will determine the remainders of 3*10^q, for q = 0, 1, 2, ..., 9 when divided by 7.

We observe that for q= 1,2,3,4,5,6,7,8,9; the respective residues of 3*(10^n) are 3,2,6,4,5,1,3,2,6,4. Therefore, the remainder when the number 3,333,333,333 is divided by 7 will correspond to 3+2+6+4+5+1+3+2+6+4 = 36 (mod 7) = 1 (mod 7). Thus, 3,333,333,333 leaves a remainder of 1

We will now consider triplets of q such that the sum of the residues is 1 (mod 7) and replacing the 3's in these positions by 7's we will arrive at the desired N.

checking for the possible triplets (0,1,6); (0,2,8); (0,3,5); (1,3,7); (1,4,5);(2,3,4) etc, we observe that for the triplet (2,3,4), the residues are (6,4,5), and since (6+4+5)(mod 7) = 1, we replace the 3's in the third, fourth and the fifth position from the right by 7's in the number 3,333,333,333 to obtain 3,333,377,733.

Consequently, the required smallest number is 3,333,377,733.

Edited on July 8, 2008, 3:56 pm
  Posted by K Sengupta on 2008-07-08 15:55:28

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