Consider all integer pairs (
a,
b) such that (b-a) = sod(a) + sod(b) (with a < b)
Now consider Set A, the set of all possible values for "a", the smaller member of each pair.
Part 1:
Prove that every "a" is a multiple of 9, or find a counterexample.
Part 2: Set M contains every element of Set A divided by 9.
But not every natural number appears in M.
Can you find a pattern to predict which positive integers do not appear in M?
Inspired by a problem on the YouTube channel Dr Peyam Show which asked "Are you smarter than a first grader?" which was itself inspired by Nob Yoshigahara's masterpiece
*** "sod" denotes "sum of digits"
I believe that the following program finds all successful a and lists all unsuccessful ones divided by 9 (up to a limit of 501 listings).
clc
list=double.empty;
for a= 0:9:99999999
hit=false;
soda=sod(a);
for diff=soda+1:9999999
b=a+diff;
if sod(b)==diff-soda
m=a/9;
% if ~ismember(m,list)
hit=true;
break
% disp([a b soda sod(b) diff])
% disp(list);
% end
end
ds=char(string(b));
if 9*length(ds)<diff-soda
break
end
end
if hit==false
list=[list a/9];
if length(list) > 500
for i=1:length(list)
disp(list(i))
end
break
end
end
end
9
20
41
52
63
74
85
96
107
108
120
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152
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185
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207
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285
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1007
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1094
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Edited on January 29, 2022, 7:23 am
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Posted by Charlie
on 2022-01-28 22:24:45 |
FWIQ
