Matt Elder
The answer to the substring sum puzzle (Feb. 24) is 2931.
I worked out the algebra on number ABCD to find:
110A + 112B = 77C + 997D
where A,B,C, and D are digits. I listed all the cases and started eliminating possibilities.
D = 1: D<2 because the maximum left-hand value is 1888, and D=0 doesn't leave a four-digit number.
Since the right side must be even, it has five possible values (1074, 1228, 1382, 1536, 1690). For each case, only two values are possible for B (the unit digits of 110A are 0, the unit digits of 112B are the even numbers in two cycles), so you only have to guess and check for ten cases.
I like it. ::)

From: Sudipta Das ( Calcutta , India )
The 4-digit number is 2931.
293+931+29+93+31+2+9+3+1=1392.

Yoah Bar-David

Koshi Arai:
Erich's substrings number is 2931: 293+931+29+93+31+2+9+3+1=1392

WBR, Stas Soumarokov
Unfortunately, there is exist other 3 digit number with same property: 891:
89+91+8+9+1=198.
What about 4 (and more) digit numbers:
2931: 293+931+29+93+31+2+9+3+1=1392
51070: 5107+1070+510+107+070+51+10+07+70+5+1+0+7+0=7015 (=07015
I found only one 8 digit number solution for problem same to E. Friedman 4
digit number puzzle: 75914061. But I solve this, using computer, 'cause it's
more easy (but, of course, I know how to solve this by hand)
Yesterday I post to you solutions for 3? 4 and 5 digit numbers. IMHO there
is no sols for 6 and 7 digit numbers.

Dane Brooke
Ed - I was thinking about the 3 digit case and didn't get 941:
a + b + c + ab + ac + bc is 21a + 12b + 3c = 100c + 10b + a
20a + 2b = 97c
c = 2
10a + b = 97
972 proper substrings are
9 + 7 + 2 + 97 + 92 + 72 = 279
Then I realized that for 941 as exemplared on your page you
apparantly call a substring proper only if the digits are
contiguous, not only in preserved-order. That changes the
consideration of the 4 digit case to
a + b + c + d + ab + bc + cd + abc + bcd or
111a + 122b + 23c + 3d = 1000d + 100c + 10b + a
110a + 112b = 997d + 77c (whence c and d are both even or both odd)
110 (a + b) = 990d + 110c - (33c - 7d + 2b)
a + b = 9d + c - (33c - 7d + 2b))/110
Not too many options:
c d 33c-7d need b forces a
0 0 0 0 0
2 0 66 22 -
4 0 142 39 -
6 0 198 11 -
8 0 264 33 -
1 1 26 42 -
3 1 92 9 2
5 1 158 31 -
7 1 224 53 -
9 1 290 15 -
0 2 -14 7 11
2 2 52 29 -
4 2 118 51 -
6 2 184 18 -
8 2 250 40 -
So its just 2931. Check
2 + 9 + 3 + 1 + 29 + 93 + 31 + 293 + 931 = 1392.
Cheers!
abcd has proper substrings
a, b, c, d, ab, ac, ad, bc, bd, cd, abc, abd, acd and bcd.
Sum these
331a + 142b + 34c + 7d and require this = 1000d + 100c + 10b + a
whence
330a + 132b = 993d + 66c divide by 3
110a + 44b = 331d + 22c
And I need d to be either 0 or 11. Temporarily choose 0 and divide by 11
10a + 4b = 2c. For 2c to exceed 10 we have c = 9, 8, 7, 6 giving a = 1 and b = 2, x, 1, x
So I find 1170 and 1290 have the required property.
Check 1170
1 + 1 + 7 + 11 + 17 + 10 + 17 + 10 + 70 + 117 + 110 + 170 + 170 = 711.
Check 1290
1 + 2 + 9 + 12 + 19 + 10 + 29 + 20 + 90 + 129 + 120 + 190 + 290 = 921.
QED
OTOH if d is 11 we have 10a + 4b - 2c = 331 so d must be at least 22 with
5a - c + 2b = 331 and we don't start finding solutions until a approaches 60, the largest
base I usually am willing to consider. I don't think any base other than 10 was intended

Stephen Kloder
891 and 2931 share this property. If 07 is considered a different
substring as 7 (but with the same value), 51070 and 147970 also have
this property.

Bob Kraus
2931 is the only 4-digit number with the sum of proper substrings = reverse
property.
51070 is the only 5-digit number. (But you have to accept initial zeroes.)

Joseph DeVincentis
The number is 2931. 1392 = 2 + 9 + 3 + 1 + 29 + 93 + 31 + 293 + 931.

I solved the problem by writing out the sum in terms of variables A, B,
C, D
for the digits which led to the diophantine equation 997D + 77C = 110A +
112B.
D is too large to be anything but 0 or 1 or maybe 2 if C is 0. Only D=1
Also note C and D must be both odd or both even since the right side is
even.
Since 997, 110, and 112 are all nearly multiples of 111, C needs to be a
number to make 77C close to a multiple of 111. The only number close
enough
is C=3, which makes the left side of the equation 1228. It is possible
to
make the right side come out to 1228, using A=2 and B=9.
The various other cases for D=0, 1, 2 and C of the same parity do not
give
numbers close enough to multiples of 111 to be formed from 110A + 112B.

Michael Dufour
2931
293 + 931 + 29 + 93 + 31 + 2 + 9 + 3 + 1
1392