Superb Sums
This lesson plan will deal with addition problems of a wide variety, and introduce the
idea of using variables ie 𝑥,𝑦 and such. It also explores decimal representation.
Resources
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The computer mind-reading game is based on an applet which requires Sun's Java VM 2.
Before you start the class, check to see if the applet works. If you can’t make it work,
you can visit Sun's website at http://www.java.com/en/download/index.jsp, download
and install Java VM. A simpler solution is to just download our slides here and play the
one round of the game.
Calculators may prove useful.
1 Activities sheet per student
Superb Sums
1. Predict the Answer:
a)
Rules:
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Circle any number
Then cross off all of the other numbers in
the same column and row as the number
that you circled.
repeat step 1 and 2 until you have 4
numbers
The sum of the 4 circled numbers is …………………
Show your answer sheet to people around you. Have they chosen the same numbers? Have they
gotten the same result? Can you explain what’s going on?
b) If you ask your friends to play the same game with the table below, can you predict what sum
they will get?
2. A Mind-Reader Computer Programme
Play this game a few times:
http://www.cut-the-knot.org/Curriculum/Magic/MindReaderNine.shtml
Can you explain what’s going on?
3. Mixed-up Multiplication
In the product shown to the right, the letters 𝑃 and 𝑄 represent
different digits from 1 to 9. What are 𝑃 and 𝑄?
4. ABCD Addition
In the summation to the right, the letters 𝐴, 𝐵, 𝐶 and 𝐷 are
unknown. If the sum is to work out, what must they be?
5. Different Division
Find all 2 digit numbers 𝐴𝐵 such that
𝑨𝑩 ÷ 𝟗 = 𝑩 remainder 𝑨.
6. Abracaddition
In the following addition, each letter stands for a different digit from 0-9. Find the digit
corresponding to each letter.
7. Dastardly Division
Find all 4 digit numbers 𝐴𝐵𝐶𝐷 such that
𝑨𝑩𝑪𝑫 ÷ 𝑨𝑪𝑪 = 𝟗 remainder 𝑪𝑪𝑪
8. Playing with digits
a) What is the sum of the digits of 111,111,1112 ? No calculators allowed!
b) Find all 3 digit numbers which become 9 times smaller when you erase their middle digit.
c) Find the five digit numbers whose digits are reversed on multiplying by 4.
Superb Sums Solutions
1. Predict the Answer:
Rules:



Circle any number
Then cross off all of the other numbers in
the same column and row as the number
that you circled.
repeat step 1 and 2 until you have 4
numbers
The sum of the 4 circled numbers is …………………
Show your answer sheet to people around you. Have they chosen the same numbers? Have they
gotten the same result? Can you explain what’s going on?
a) Example:
The sum of the 4 numbers here is 4+6+11+13=34. No matter which rows and columns are chosen,
the sum is always the same: 34.
Explanation: The rules make us pick exactly one number from each row and column.
•
The second row is 4+ first row
rd
•
The 3 row is 8+ the first row.
•
The 4 row is 12+ the first row.
th
1
2
3
4
1+4
5
2+4
6
3+4
7
4+4
8
1+8
9
2+8
10
3+8
11
4+8
12
1+12
13
2+12
14
3+12
15
4+12
16
As you are selecting one number from each column and from each row, the sum is:
1+2+3+4 (from the columns) +4+8+12 (from the extras added to the rows) = 34.
Note: In a sum, the order of the numbers doesn’t matter. You can rearrange the numbers in the sum
in any order you like and still you will get the same result. We say that the sum is commutative, from
co (together) and mutatis (move).
b) If you ask your friends to play the same game with the table on the left, can you predict what sum
they will get?
You can think your answer through in the same way as above, but now that we understand what’s
going on, we can also find the answer in a quicker way: Since we know that the answer is the same
no matter how we choose the 5 numbers, as long as we choose one from each row and one from
each column, we can get the answer by summing all the numbers on the diagonal:
1 + 7 + 13 + 19 + 25
= (1 + 19) + (7 + 13) + 25
= 20 + 20 + 25
= 65
2. A Mind-Reader Computer Programme
Play this game a number of times:
http://www.cut-the-knot.org/Curriculum/Magic/MindReaderNine.shtml
Can you explain what’s going on?
In case Java doesn’t work, you can play the game using the slides here.
A Mind-Reader Computer Programme
Think of a 2-digit integer. Subtract from the number the sum of its digits
and find the result in the table below. Note that each cell of the table
contains a number and a geometric shape. Concentrate hard on the
shape that shares a cell with the result of your calculations.
Hint: Since we can’t seriously believe that the computer is a mind reader, it must be that the
programme assigns the same shape to all the possible answers to the problem. Play again and check
the numbers having the same shape as your solution. Notice any special property? Now try to prove
it!
Solution: All possible answers to the problem are multiples of 9. So the answer above is a pentagon.
Let’s suppose that the number is written 𝐴𝐵, with 𝐴 the tens digits and 𝐵 the unit digit.
Then the number is equal to 10𝐴 + 𝐵. (Here 10𝐴 means 10 × 𝐴, and can be read as 10𝐴-s, like in
10 Apples). For example, 59 = 10 × 5 + 9 is made of 10 fives and a nine.
Similarly 𝐵𝐴, the number read backwards, is 10𝐵 + 𝐴, that is 10𝐵-s and an 𝐴.
The difference between the number and the number read backwards is
𝐴𝐵
− 𝐵𝐴
Which in decimal notation is
or, after swapping terms in second row:
Or, in one line:
10 𝐴 + 𝐵
− 10 𝐵 − 𝐴
10 𝐴 + 𝐵
− 𝐴 − 10𝐵
9𝐴 – 9𝐵
10𝐴 + 𝐵 − (10𝐵 + 𝐴) = 10𝐴 + 𝐵 − 10𝐵 − 𝐴 = 9𝐴 − 9𝐵 = 9(𝐴 − 𝐵),
Which is always a multiple of 9.
Note: As it is expected that at this stage, the students may not be used with abstract algebraic
computations, we can explain the above subtraction as
(10 Apples + 1 Blueberry) - (1 Apple + 10 Blueberry) =
= (10 Apples - 1 Apple) + (1 Blueberry - 10 Blueberries) =
= 9 Apples – 9 Blueberries.
3. Mixed-up Multiplication
a) In the product shown to the right, the letters 𝑃 and 𝑄 represent
different digits from 1 to 9. What are 𝑃 and 𝑄?
Solution I:
a) If we expand out the multiplication (exactly as if we were doing long multiplication), we
see that 8 × 𝑄 must end in a zero. As 𝑄 cannot be a zero, this
means that 𝑄 must equal 5. So now our multiplication is
𝑃8
× 35
2730
We get the same result
if we swap the two numbers
and do long multiplication:
35
× 𝑃8
280
+? ? ?
2730
Here by the rules of long multiplication, the mystery number ? ? ? stands for 𝑃 × 35.
From the second sum we can deduce
? ? ? = 2,730 − 280 = 245. Hence 𝑃 × 35 = 245 so 𝑃 = 245 ÷ 35 = 7.
Solution II: This solution only works for people who have studied distributivity. If this lesson
comes before Distributivity, this solution can be omitted.
Expanding this out we get
(10 × 𝑃 + 8) × 35 = 2,730
Isolating 𝑃 we get
350 × 𝑃 = 2,730 − 8 × 35 = 2,450
Hence
2,450
𝑃=
=7
350
So 𝑃 = 7 and 𝑄 = 5.
4. ABCD Addition
a) In the summation to the right, the letters 𝐴, 𝐵, 𝐶 and 𝐷
are unknown. If the sum is to work out, what must they
be?
Solutions:
a) Solution 1: Trial and Error: By looking at the first column, we see that 𝐴 must be 1 or 2.
However we know that a 1 must “carry over” from the second column, otherwise both 𝐴
and 𝐵 would have to be zero. So 𝐴 can’t be 2, i.e. 𝐴 = 1.
We now have:
1BCD
1BC
1B
+
1
2 0 1 2
Taking away 1000 from the first row, 100 from the second row, 10 from the third row
and 1 from the fourth row amounts to taking away 1111 from the answer:
2012 − 1111 = 901.
So we have
BCD
BC
+ B
9 0 1
By looking at the middle column, we see that either B = C = 0 or B+C carries 1 over in the
first column. As B, C can’t be 0, this means that B=8 and he have
8CD
8C
+ 8
9 0 1
Taking away 800 from the first row, 80 from the second row and 8 from the third row
amounts to taking away 888 from the total: 901 − 888 = 13.
Hence, we can find
CD
+C
13
Which clearly stands for 12+1=13. Hence the solution is 𝐴𝐵𝐶𝐷 = 1812.
Solution 2: Decimal Representation: By using decimal representation, we see that
(1000 × 𝐴 + 100 × 𝐵 + 10 × 𝐶 + 𝐷) + (100 × 𝐴 + 10 × 𝐵 + 𝐶) + (10 × 𝐴 + 𝐵) + (𝐴)
= 2012
In other words,
1,111 × 𝐴 + 111 × 𝐵 + 11 × 𝐶 + 𝐷 = 2012
Hence, 𝐴 = 1, otherwise the left hand side is too large. Subtracting:
111 × 𝐵 + 11 × 𝐶 + 𝐷 = 901
So 𝐵 is at most 8. Bringing B over we see that
11 × 𝐶 + 𝐷 = 901 − 111 × 𝐵
If 𝐵 < 8, this is the same as saying 𝐵 ≤ 7. Then:
11 × 𝐶 + 𝐷 ≥ 124
which is impossible as the largest the left hand side can be is 108. So 𝐵 = 8. Hence:
11 × 𝐶 + 𝐷 = 13
So 𝐶 = 1 and 𝐷 = 2
5. Different Division
Find all 2 digit numbers 𝐴𝐵 such that
𝑨𝑩 ÷ 𝟗 = 𝑩 remainder 𝑨.
Solution: First we need to understand what we mean by division with remainder: how can
we write the division above as an equation where we don’t have to use words like
remainder?
Example: 17 ÷ 5 = 3 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 2 means 17
-s can be organized as
17 = 5 × 3 + 2
So the equation 𝐴𝐵 ÷ 9 = 𝐵 remainder 𝐴 can be written as 𝐴𝐵 = 9 × 𝐵 + 𝐴.
But 𝐴𝐵 = 10 × 𝐴 + 𝐵 so the equation above can be rewritten
10 × 𝐴 + 𝐵 = 9 × 𝐵 + 𝐴.
Taking away an 𝐴 and a 𝐵 from both sides we get:
10 × 𝐴 − 𝐴 = 9 × 𝐵 − 𝐵 .
9 × 𝐴 = 8 × 𝐵.
Hence 𝐴 = 8 and 𝐵 = 9. Indeed, 89 ÷ 9 = 9 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 8.
6. Abracaddition
In the following addition, each letter stands for a different digit from 0-9. Find the digit
corresponding to each letter.
Solution: As only two numbers are added and 9+9+1=19<20, at most 1 unit can be carried
over from one column to the next. So 𝑃 = 1. Then 𝐻 + 1 ≥ 9 so 𝐻 = 8 or 9.
If 𝐻 = 8 then 𝑅 = 0 and 1 is carried over from 𝑂 + 𝑂, hence 𝑂 > 4. But 𝑂 is an even
number since it’s the last digit of 𝑆 + 𝑆 so it can only be 𝑂 = 6 since 8 is already taken.
Then 𝑆 + 𝑆 = 6 or 16. However 𝐻 = 8 so 𝑆 = 3 in this case. Then 𝐶 + 𝐶 = 12 or 2.
Unfortunately, both 6 and 1 are taken so this case doesn’t work.
If 𝐻 = 9 then 𝑅 = 0 (since already 𝑃 = 1) and no digit is carried over from 𝑂 + 𝑂 hence
𝑂 ≤ 4. But 𝑂 is an even number since it’s the last digit of 𝑆 + 𝑆 so it can only be 𝑂 = 2 or 4
since 0 is already taken.
If 𝑂 = 2 then 𝑆 + 𝑆 = 12 or 2. However, 𝑃 = 1 so it must be 𝑆 = 6. Then 𝐶 + 𝐶 = 6 or 16,
hence 𝐶 = 3 or 8 while 𝐸 = 4 or 5, respectively. Also 𝑈 + 𝑈 < 10 since it does not
contribute to the next column, thus 𝑈 < 5. The only available options are 𝑈 = 3 or 𝑈 = 4,
which are incompatible with the case 𝐶 = 3, 𝐸 = 4. Hence it must be that 𝐶 = 8, 𝐸 = 5.
We also have 𝑈 + 𝑈 + 1 = 𝑇. If 𝑈 = 4, then 𝑇 = 9 which conflicts with 𝐻 = 9. Hence the
only option is 𝑈 = 3, T=7. This works indeed.
The case 𝑂 = 4 on the other hand doesn’t work. In this case 𝑆 = 7 or 2, while 𝐸 = 8 so 𝐶 <
5, while 𝐶 + 𝐶 = 𝑆 or 𝑆 − 1. Hence 𝐶 = 3 and 𝑆 = 7. Then 𝑇 must be odd and the only
available digit is 𝑇 = 5, hence 𝑈 = 2. But then 𝐶 + 𝐶 = 6 rather than 7. Contradiction.
7. Different Division
Find all 4 digit numbers 𝐴𝐵𝐶𝐷 such that
𝑨𝑩𝑪𝑫 ÷ 𝑨𝑪𝑪 = 𝟗 remainder 𝑪𝑪𝑪
Solution: The equation can be rewritten as
𝐴𝐵𝐶𝐷 = 9 × 𝐴𝐶𝐶 + 𝐶𝐶𝐶
or, in decimal representation:
1000𝐴 + 100𝐵 + 10𝐶 + 𝐷 = 9(100𝐴 + 10𝐶 + 𝐶) + 100𝐶 + 10𝐶 + 𝐶.
After simplifying: 1000𝐴 + 100𝐵 + 10𝐶 + 𝐷 = 900𝐴 + 99𝐶 + 111𝐶,
100𝐴 + 100𝐵 + 𝐷 = 200𝐶.
Hence 𝐷 = 0 and 𝐴 + 𝐵 = 2𝐶.
Solutions are: 1320, 1530, 1740, 1950, 2010, 2220, 2430, 2640, 2850, 3120, 3330, 3540,
3750, 4020, 4230, 4440, 4650, 5130, 5340, 5550, 6030, 6240, 6450, 7140, 7350, 8040, 8250,
9150.
8. Playing with digits
a) What is the sum of the digits of 111,111,1112 ? No calculators allowed!
Solution: 111,111,1112 =12,345,678,987,654,321 so the sum of its digits is
9 + 2(1 + 2 + ⋯ + 8) = 9 + 8 × 9 = 81.
b) Find all 3 digit numbers which become 9 times smaller when you erase their middle digit.
Hint: Let the 3 digit number be 𝐴𝐵𝐶.
Solution:
The trick is to rephrase our question into an equation, like we did with the last question.
Think about it carefully and you should get
100 × 𝐴 + 10 × 𝐵 + 𝐶 = 9 × (10 × 𝐴 + 𝐶)
100 × 𝐴 + 10 × 𝐵 + 𝐶 = 90 × 𝐴 + 9 × 𝐶
We can simplify this to:
10 × 𝐴 + 10 × 𝐵 = 8 × 𝐶
Dividing across by 2:
5 × (𝐴 + 𝐵) = 4 × 𝐶
As 𝐶 must be a 1 digit number and 𝐶 must be a multiple of 5 (by observing the right
hand side of the above equation), then C must equal 5. Hence
𝐴+𝐵 =4
So to get our values of 𝐴 and 𝐵, we simply run through the different possible
expressions. Our 4 answers are 135, 225, 315 and 405. Why isn’t 045 an answer?
c) Find the five digit numbers whose digits are reversed on multiplying by 4.
Solution:
When we multiply the number by 4 we get a 5 digit number, so all in all our number must be
less than 1,000,000 ÷ 4 = 250,000.
The digits satisfy 𝐴 = 1 or 2 and
4 × (100,000𝐴 + 10,000𝐵 + 1,000𝐶 + 100𝐷 + 10𝐸 + 𝐹) =
= 100,000𝐹 + 10,000𝐸 + 1,000𝐷 + 100𝐶 + 10𝐵 + 𝐴
Simplify and group digits:
399,999𝐴 + 39,990𝐵 + 3,900𝐶 = 600𝐷 + 9,960𝐸 + 99,996𝐹.
𝐴 must be even as all other numbers are even. So 𝐴 = 2.
799,998 + 39,990𝐵 + 3,900𝐶 = 600𝐷 + 9,960𝐸 + 99,996𝐹.
So 𝐹 ≥ 8 to make up for the large number on the left hand side. Also, the last digit on the
left-hand-side must be 8 so 𝐹 = 8.
30 + 39,990𝐵 + 3,900𝐶 = 600𝐷 + 9,960𝐸.
It turns out that we can simplify by 30:
1 + 1333𝐵 + 130𝐶 = 20𝐷 + 332𝐸.
B must be odd and not too large, so let’s try
B=1. Then
1334 = 20𝐷 − 130𝐶 + 332𝐸,
1334 − 332𝐸 = 20𝐷 − 130𝐶.
Looking at the last digit, E=2 or 7. If E=2 then 20𝐷 − 130𝐶 = 1334 − 664 = 670, which is
impossible since 20𝐷 ≤ 180. If E=7 then 20𝐷 − 130𝐶 = 1334 − 2324 = −990,
Or 13𝐶 − 2𝐷 = 99, which works for C=9 and D=9.
We got ABCDEF=219978.
On the other hand, it’s easy to see that B=3 is already too large, so that’s the only solution.