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The following is taken from 'Amusements in Mathematics' by Henry Ernest >Dudeney.
Sometimes a very simple question in elementary arithmetic will cause a
good deal of perplexity. For example, if I want to divide the four
numbers, 701, 1,059, 1,417 and 2,312, by the largest number possible
that will leave the same remainder in every case. How am I to set to
work? Of course, by a laborious system of trial one can in time discover
the answer, but there is quite a simple method of doing it if you can
only find it.
On Sat, 6 Sep 2025 09:06:17 +0100, David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:
The following is taken from 'Amusements in Mathematics' by Henry Ernest
Dudeney.
Sometimes a very simple question in elementary arithmetic will cause a
good deal of perplexity. For example, if I want to divide the four
numbers, 701, 1,059, 1,417 and 2,312, by the largest number possible
that will leave the same remainder in every case. How am I to set to
work? Of course, by a laborious system of trial one can in time discover
the answer, but there is quite a simple method of doing it if you can
only find it.
.
.
.
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I think that it is sufficient to consider just the two smallest
numbers as we are looking for the larget divisor that leaves
the same remained.
What we have is a (mod n) = b (mod n) = ........
where a, b, .... are known and we need to find the largest
n that will satisfy the above equation.
Focusing on the first pair we have (note b > a)
(b - a) (mod n) = 0
With b = 1059 and a = 701 we get
358 (mod n) = 0
The largest n we can choose in 358 to satisfy
this equation is 358.
.
.
.
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I think that it is sufficient to consider just the two smallest
numbers as we are looking for the larget divisor that leaves
the same remained.
What we have is a (mod n) = b (mod n) = ........
where a, b, .... are known and we need to find the largest
n that will satisfy the above equation.
Focusing on the first pair we have (note b > a)
(b - a) (mod n) = 0
With b = 1059 and a = 701 we get
358 (mod n) = 0
The largest n we can choose in 358 to satisfy
this equation is 358.
That looks much nicer than the route I took. My route may not be rigorous.
If we let g be the greatest common divisor of all of the numbers, then >......
...... So g is
the greatest common divider of the differences of the numbers...
On Mon, 8 Sep 2025 12:23:21 +0100, David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:
That looks much nicer than the route I took. My route may not be rigorous. >>
If we let g be the greatest common divisor of all of the numbers, then
......
...... So g is
the greatest common divider of the differences of the numbers...
I think we both did the same thing. Your method is certainly
"rigorous". The only difference is the notation that we used.
A nice problem. Had to think about it for a minute, bit it was
writing down the problem as I saw it that immediately led to
the solution. If I had tried to do it in my head, it would have
taken longer. Funny how seeing something in writing sometimes
triggers thought processes .....
Thanks for coming up with these posers.
So, for completeness, what's the solution?
Mike.
So, for completeness, what's the solution?
On 08/09/2025 18:13, Mike Terry wrote:
So, for completeness, what's the solution?
Good question. Although I've made some comments, it may not have been an >explicit solution. Here is what is included in the solutions section of
the book. I'd claim, without evidence, that I had arrived at that solution.
The answer provided by Dudeney is given below.
Solution provided by Dudeney.
olution provided by Dudeney.
lution provided by Dudeney.
ution provided by Dudeney.
tion provided by Dudeney.
ion provided by Dudeney.
on provided by Dudeney.
n provided by Dudeney.
provided by Dudeney.
provided by Dudeney.
rovided by Dudeney.
ovided by Dudeney.
vided by Dudeney.
ided by Dudeney.
ded by Dudeney.
ed by Dudeney.
ed by Dudeney.
d by Dudeney.
by Dudeney.
by Dudeney.
y Dudeney.
Dudeney.
Dudeney.
udeney.
deney.
eney.
ney.
ey.
y.
.
Subtract every number in turn from every other number, and we get 358 >(twice), 716, 1,611, 1,253, and 895. Now we see at a glance that, as 358 >equals 2 x179, the only number that can divide in every case without a >remainder will be 179. On trial we find that this is such a divisor. >Therefore, 179 is the divisor we want, which always leaves a remainder
164 in the case of the original numbers given.
On Mon, 8 Sep 2025 19:44:29 +0100, David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:
On 08/09/2025 18:13, Mike Terry wrote:
So, for completeness, what's the solution?
Good question. Although I've made some comments, it may not have been an
explicit solution. Here is what is included in the solutions section of
the book. I'd claim, without evidence, that I had arrived at that solution. >>
The answer provided by Dudeney is given below.
Solution provided by Dudeney.
olution provided by Dudeney.
lution provided by Dudeney.
ution provided by Dudeney.
tion provided by Dudeney.
ion provided by Dudeney.
on provided by Dudeney.
n provided by Dudeney.
provided by Dudeney.
provided by Dudeney.
rovided by Dudeney.
ovided by Dudeney.
vided by Dudeney.
ided by Dudeney.
ded by Dudeney.
ed by Dudeney.
ed by Dudeney.
d by Dudeney.
by Dudeney.
by Dudeney.
y Dudeney.
Dudeney.
Dudeney.
udeney.
deney.
eney.
ney.
ey.
y.
.
Subtract every number in turn from every other number, and we get 358
(twice), 716, 1,611, 1,253, and 895. Now we see at a glance that, as 358
equals 2 x179, the only number that can divide in every case without a
remainder will be 179. On trial we find that this is such a divisor.
Therefore, 179 is the divisor we want, which always leaves a remainder
164 in the case of the original numbers given.
Shows that one should check out all the cases! My error was assuming
that if it worked for the first pair, it should work for all the
numbers .... which erroneously forgot that the "n" in my (mod n)
can be factored. Even though I realised that 358 is 2 x 179, I
failed to test the fourth number.
Mea culpa.