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Lots of people submitted solutions to this problem - well done, everyone!
Matthew from West Island School in Hong Kong found the other two sets of numbers adding to $726$:
726 can be made by adding 165, 264 and 363 with their reversals.
Hank and Vacha from Elm Park School in New Zealand gave these answers too, along with a great explanation for how they worked them out:
We knew that the units had to add up to $6$ because we couldn't use decimals. The only ways to make $6$ were $3+3$, $5+1$ and $2+4$ ($2+4$ leads to the sum given in the question).
So the first answer has to be something like this: $3?3 + 3?3 = 726$.
The only ways to make $2$ in the tens column are $1 +1$ and $6+6$. It can't be $1+1$ because then the total will be $626$, so the number in the tens place must be $6$. Then you
repeat this step with $1$ and $5$ in the units column to find $561+165=726$.
The same method also works for finding the three ways to do this for $707$ and $766$. Ashlyn from St. Stephen's School Carramar gave this explanation for $707$:
$3$ ways of forming $707$ by adding a number to its reversal:
Rules:
The $1^{st}$ digit and $3^{rd}$ digit of the number must total $7$ (the $3^{rd}$ digit of
$707$).
$0$ cannot be used as a $1^{st}$ or $3^{rd}$ digit as you need both the number and its reversal to have three digits.
The $2^{nd}$ digit in any three digit number and its reversal must be the same.
We can find which numbers add to give $7$:
$0+7$
$1+6$
$2+5$
$3+4$
We know we can't use $0+7$, so the other three combinations must be the $1^{st}$
and $3^{rd}$ digits.
Now we use a system of trial and error to find the $2^{nd}$ digit. We find that $0$ is the only one that fits.
Therefore the $3$ combinations must be:
$304+403=707$
$502+205=707$
$601+106=707$.
Catherine and Charlotte from Culford School found the three numbers which made $766$ when added to their reversal:
$185+581=766$
$284+482=766$
$383+383=766$
Beth and Emily from Culford school successfully found the $10$ numbers between $700$ and $800$ which can be written as the sum of a number and its reversal, and spotted some patterns:
$706$, $707$, $726$, $727$, $747$, $746$, $767$, $766$, $787$, $786$.
They all have even numbers in the middle. They all end in either $6$ or $7$.
Michael Presland from Sevenoaks School added this explanation for the pattern:
For the numbers ending in $7$ the numbers on the outside (of the numbers
you are adding together) must be $3$ and $4$, $6$ and $1$, or $5$ and $2$.
The number in the middle must be less than $5$ so there is no carry over.
For the numbers ending in $6$, we found that the number in the middle of
the two numbers you are adding together must be $5$ or more, so as to make the digit in the hundreds column be $7$ (by carrying one over) while the digit in the units column is six.
Some people tried using an algebraic method to explain the pattern.
Let's say that our three-digit number is written $ABC$. We can express this algebraically as $100A+10B+C$.
Then the reversal of our number is $CBA$, which we can express as $100C+10B+A$. Adding these together gives $101(A+C) + 20B$.
If $A+C$ were greater than $9$, our total would be bigger than $1000$: as we are looking for numbers between $700$ and $800$, $A+C$ must be a single digit.
Let's think about the sum:
$A$ $B$ $C$
+ $C$ $B$ $A$
The units value in this total is $A+C$, with no carry over.
In the tens column, we calculate $B+B$. If $B$ is less than $5$, the tens value is $2B$ with no carry over. If $B$ is between $5$ and $9$, the tens value is the last digit of $2B$ with a carry over of $1$.
The hundreds value in this total is either $A+C$ (if there is no carry over from the tens column), or $A+C+1$ (if there is a carry over from the tens column).
We want the hundreds value in this total to always be $7$, as we are looking for numbers between $700$ and $800$. So either $A+C=6$ or $A+C=7$. This means the units value in the total is always $6$ or $7$.
Whatever $B$ is, the last digit of $2B$ will always be even. So the second digit in our total will always be $0$, $2$, $4$, $6$, or $8$.
Lots of people correctly spotted that there are $10$ numbers between $300$ and $400$ which can be written as the sum of a three-digit number and its reversal, and $10$ numbers between $800$ and $900$ with this property. Will, Lew, and Nick from All Saints Junior School found the numbers:
Between $300$ and $400$ : $363$,$383$,$323$,$343$,$303$,$322$,$302$,$342$,$362$,$382$
Between $800$ and $900$ : $808$,$828$,$848$,$868$,$888$,$807$,$827$,$847$,$867$,$887$.
Well done to everybody who submitted a solution to this problem!
Forwards Add Backwards
Age 11 to 14
Challenge Level 
Lots of people submitted solutions to this problem - well done, everyone!
Matthew from West Island School in Hong Kong found the other two sets of numbers adding to $726$:
726 can be made by adding 165, 264 and 363 with their reversals.
Hank and Vacha from Elm Park School in New Zealand gave these answers too, along with a great explanation for how they worked them out:
We knew that the units had to add up to $6$ because we couldn't use decimals. The only ways to make $6$ were $3+3$, $5+1$ and $2+4$ ($2+4$ leads to the sum given in the question).
So the first answer has to be something like this: $3?3 + 3?3 = 726$.
The only ways to make $2$ in the tens column are $1 +1$ and $6+6$. It can't be $1+1$ because then the total will be $626$, so the number in the tens place must be $6$. Then you
repeat this step with $1$ and $5$ in the units column to find $561+165=726$.
The same method also works for finding the three ways to do this for $707$ and $766$. Ashlyn from St. Stephen's School Carramar gave this explanation for $707$:
$3$ ways of forming $707$ by adding a number to its reversal:
Rules:
The $1^{st}$ digit and $3^{rd}$ digit of the number must total $7$ (the $3^{rd}$ digit of
$707$).
$0$ cannot be used as a $1^{st}$ or $3^{rd}$ digit as you need both the number and its reversal to have three digits.
The $2^{nd}$ digit in any three digit number and its reversal must be the same.
We can find which numbers add to give $7$:
$0+7$
$1+6$
$2+5$
$3+4$
We know we can't use $0+7$, so the other three combinations must be the $1^{st}$
and $3^{rd}$ digits.
Now we use a system of trial and error to find the $2^{nd}$ digit. We find that $0$ is the only one that fits.
Therefore the $3$ combinations must be:
$304+403=707$
$502+205=707$
$601+106=707$.
Catherine and Charlotte from Culford School found the three numbers which made $766$ when added to their reversal:
$185+581=766$
$284+482=766$
$383+383=766$
Beth and Emily from Culford school successfully found the $10$ numbers between $700$ and $800$ which can be written as the sum of a number and its reversal, and spotted some patterns:
$706$, $707$, $726$, $727$, $747$, $746$, $767$, $766$, $787$, $786$.
They all have even numbers in the middle. They all end in either $6$ or $7$.
Michael Presland from Sevenoaks School added this explanation for the pattern:
For the numbers ending in $7$ the numbers on the outside (of the numbers
you are adding together) must be $3$ and $4$, $6$ and $1$, or $5$ and $2$.
The number in the middle must be less than $5$ so there is no carry over.
For the numbers ending in $6$, we found that the number in the middle of
the two numbers you are adding together must be $5$ or more, so as to make the digit in the hundreds column be $7$ (by carrying one over) while the digit in the units column is six.
Some people tried using an algebraic method to explain the pattern.
Let's say that our three-digit number is written $ABC$. We can express this algebraically as $100A+10B+C$.
Then the reversal of our number is $CBA$, which we can express as $100C+10B+A$. Adding these together gives $101(A+C) + 20B$.
If $A+C$ were greater than $9$, our total would be bigger than $1000$: as we are looking for numbers between $700$ and $800$, $A+C$ must be a single digit.
Let's think about the sum:
$A$ $B$ $C$
+ $C$ $B$ $A$
The units value in this total is $A+C$, with no carry over.
In the tens column, we calculate $B+B$. If $B$ is less than $5$, the tens value is $2B$ with no carry over. If $B$ is between $5$ and $9$, the tens value is the last digit of $2B$ with a carry over of $1$.
The hundreds value in this total is either $A+C$ (if there is no carry over from the tens column), or $A+C+1$ (if there is a carry over from the tens column).
We want the hundreds value in this total to always be $7$, as we are looking for numbers between $700$ and $800$. So either $A+C=6$ or $A+C=7$. This means the units value in the total is always $6$ or $7$.
Whatever $B$ is, the last digit of $2B$ will always be even. So the second digit in our total will always be $0$, $2$, $4$, $6$, or $8$.
Lots of people correctly spotted that there are $10$ numbers between $300$ and $400$ which can be written as the sum of a three-digit number and its reversal, and $10$ numbers between $800$ and $900$ with this property. Will, Lew, and Nick from All Saints Junior School found the numbers:
Between $300$ and $400$ : $363$,$383$,$323$,$343$,$303$,$322$,$302$,$342$,$362$,$382$
Between $800$ and $900$ : $808$,$828$,$848$,$868$,$888$,$807$,$827$,$847$,$867$,$887$.
Well done to everybody who submitted a solution to this problem!
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