Enigma 1535 — Back to front
by Bob Walker
There are two six digit numbers for which reversal of the digits delivers a multiple of the original number.
Find the samller of the two — or even both of them!
Don’t miss the footnote.
a b c d e f
x
−−−−−−−−
f e d c b a
If x is even then a is even
ax ≤ 9
fx mod 10 = a
If x ≥ 5 then
a = 1
x ≤ f ≤ 9
fx mod 10 = 1
x = 9 and f = 9 is the only candidate
If x = 4 then
a = 2
ax = 8 ≤ f ≤ 9
fx mod 10 = 2
f = 8 is the only candidate
If x = 3 then
a = 1, 2 or 3
If a = 1
ax = 3 ≤ f ≤ 5
fx mod 10 = 1
there are no candidates for f
If a = 2
ax = 6 ≤ f ≤ 8
fx mod 10 = 2
there are no candidates for f
If a = 3
ax = 9 ≤ f ≤ 9
fx mod 10 = 3
there are no candidates for f
If x = 2 then
a = 2 or 4
If a = 2
ax = 4 ≤ f ≤ 5
fx mod 10 = 2
there are no candidates for f
If a = 4
ax = 8 ≤ f ≤ 9
fx mod 10 = 4
there are no candidates for f
Switch attention to 2nd and 5th digits
call the carry digit from fx z.
call the carry digit from eb y.
fx = 10z + a
y = f − ax
e = bx − 10y + q where 0 ≤ q < x
(ex + z) mod 10 = b
If x = 9 and f = 9 and a = 1
z = 8
y = 0
e = 9b + q
b = 0 or 1
If b = 0
e = q = 0,1 ... 8
e = 8 gives (ex + z) mod 10 = 0 = b *****
If b = 1
e = 9 gives (ex + z) mod 10 = 9 != 1
If x = 4 and f = 8 and a = 2
z = 3
y = 0
e = 4b + q where 0 ≤ q < 4
b = (ex + z) mod 10 must be odd
b = 1
e = 4, 5, 6 or 7, but only 7 satisfies
(ex + z) mod 10 = b
Our smallest candidate is:
a = 1 b = 0 e = 8 f = 9 x = 9
1 0 c d 8 9
9
−−−−−−−−
9 8 d c 0 1
9d + 8 = c + 10z
9c + z = d + 10y
y = 8
9c + z = d + 80
9d + 8 = c + 10z
9c = 80 + d − z ≥ 72
c = 8 or 9
If c = 8
d = 0 and z = 8
If c = 9
d − z = 1
9d + 8 = 9 + 10d − 10
d = 9
1 0 9 9 8 9
9
−−−−−−−−
9 8 9 9 0 1
Our other candidate is:
a = 2 b = 1 e = 7 f = 8 x = 4
2 1 c d 7 8
4
−−−−−−−−
8 7 d c 1 2
4d + 3 = c + 10z
4c + z = d + 10y
4 + y = 7
4c + z = d + 30
4c = 30 + d − z ≥ 27
c = 7, 8 or 9
If c = 7
z = d + 2
4d + 3 = c + 10z = 7 + 10d + 20
0 = 6d + 24 i.e. c != 7
If c = 8
z = d − 2
4d + 3 = c + 10z = 8 + 10d − 20
0 = 6d − 15 i.e. c != 8
If c = 9
z = d − 6
4d + 3 = c + 10z = 9 + 10d − 60
0 = 6d − 54
c = 9 and d = 9
2 1 9 9 7 8
4
−−−−−−−−
8 7 9 9 1 2
// Confirmed by
#include
#include
#include
main(int argc, char **argv)
{ int n, prod, x;
char nchars[10];
char pchars[10];
n = 1000000;
while ( --n >= 100000 )
{ x = 1;
sprintf(nchars, "%d", n);
while ( (prod = (++x)*n) < 1000000 )
{ sprintf(pchars, "%d", prod);
if ( nchars[0] == pchars[5]
&& nchars[1] == pchars[4]
&& nchars[2] == pchars[3]
&& nchars[3] == pchars[2]
&& nchars[4] == pchars[1]
&& nchars[5] == pchars[0]
)
printf("%d times %s = %s\n", x, nchars, pchars);
}
}
}
In fact, you can have a many nines in the middle as you like,
and in both solutions the sums of corresponding digits in product and original number
are the same.