1 Activities Sheet Per Student
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 subtly.
Resources
· Calculators may prove useful.
· 1 Activities sheet per student
1. Mixed-up Multiplication (5 mins)
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:
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=2,730. Expanding this out we get
10×P+8×35=2,730
Isolating P we get
350×P=2,730-8×35=2,450
Hence
P=2,450350=7
So P=7 and Q=5.
2. Austere Addition (10 mins)
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: Note that this solution is quite long, especially compared to solution 2. 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, ie A=1. If B=9, then by looking at the last column, C+D=2 or C+D=12. By trial and error, neither of these work. If B=9, then by looking at the 3rd column (and remembering that the largest carry over we can get from the last column is 2) we can say that C=1. Hence, we can find the solution ABCD=1812. Other values of B don’t work because on inspection, they are too small.
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
Note: Students should be shown both solutions! They are two different techniques (brute force and decimal representation) and one is as important as the other.
3. Playing with digits (5 mins)
a) Find all 3 digit numbers which become 9 times smaller when you erase their middle digit. Hint: Let the 3 digit number be ABC.
Solutions:
1. 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=90×A+9×C
We can simplify this to:
10×A+10×B=10×A+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 153, 252, 351 and 450. Why isn’t 054 an answer?
4. Activities Sheet
· Attempt to answer all the questions on the activities sheet.
Activities Sheet - Superb Sums
1. What is the sum of the digits of 111,111,1112? No calculators allowed!
2. In the following addition, each letter stands for a different digit from 1-9. Find the digit corresponding to each letter.
3. Find the five digit numbers whose digits are reversed on multiplying by 4.
4. Find all five digit numbers whose digits are reversed on multiplying by 9.
5. In the following addition, each letter stands for a different digit from 0-9. Find the digit corresponding to each letter.