Ned asked each of his 4 children to think of a 4-digit number.
"Now transfer the last digit to the front and add the new number to the old one.For example, 1234+4123= 5357. Now tell me the results.
The results were told by each of the four children as follows:
Barry: 2348
Maury: 7847
Jaypee: 11847
Derrick: 9846
"Everyone except one is wrong" Ned told the gathering.
Who was it and how did Ned know?
clc,clearvars
for n=1000:9999
ns=char(string(n));
ns2=[ns(4) ns(1:3)];
n2=str2double(ns2);
if ismember(n2+n,[2348 7847 11847 9846])
disp([n,n2+n])
end
end
finds that, of the given numbers, only 11847 can be the result of the given process:
original result
2589 11847
3498 11847
4407 11847
5316 11847
6225 11847
7134 11847
8043 11847
8952 11847
9861 11847
I thought maybe it had to do with the parity of the sum of its digits, but that's not the case, even considering 4-digit results and 5-digit results separately.
What does work is that the given sum must be a multiple of 11.
If a is the number represented by the firtst 3 digits of the original number and b is the last digit of th original, the total that's asked for is
10a+b + 1000b+a
which simplifies to
11a + 1001b
and this is divisible by 11 as 1001 is a multiple of 11.
The only multiples of 11 that are not possible are
1001
1012
1023
1034
1045
1056
1067
1078
1089
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Posted by Charlie
on 2023-09-02 09:21:30 |
solution
