I have found in one of the Russian high-school competitions a relatively easy puzzle & to make it more challenging replaced a certain number
by N.
The modified text runs as follows:
Prove that among N sequential positive integers there always exists a number whose s.o.d. is a multiple of 11.
a. Restore N.
b. Now prove the statement.
(In reply to
Part a. (but no proof) by Larry)
Here's a span of 39:
n sod(n) mod 11
999980 44 0
999981 45 1
999982 46 2
999983 47 3
999984 48 4
999985 49 5
999986 50 6
999987 51 7
999988 52 8
999989 53 9
999990 45 1
999991 46 2
999992 47 3
999993 48 4
999994 49 5
999995 50 6
999996 51 7
999997 52 8
999998 53 9
999999 54 10
1000000 1 1
1000001 2 2
1000002 3 3
1000003 4 4
1000004 5 5
1000005 6 6
1000006 7 7
1000007 8 8
1000008 9 9
1000009 10 10
1000010 2 2
1000011 3 3
1000012 4 4
1000013 5 5
1000014 6 6
1000015 7 7
1000016 8 8
1000017 9 9
1000018 10 10
1000019 11 0
It's unfortunate that a million is within the bounds of this gap.
I can't guarantee there's nothing higher beyond 10 million
prev=0; mx=0
for n=11:9999999
s=sod(n);
if mod(s,11)==0
diff=n-prev;
prev=n;
if diff>mx
disp([n diff])
mx=diff;
end
end
end
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Posted by Charlie
on 2021-05-24 12:11:44 |
re: Part a. (but no proof)
