In Step 3, why cannot the value of carry be more than 1?
Step-by-step solution
Idea: Find the largest number a column can ever add up to. Each digit is at most 9 and the carry coming in is at most 1, so the total is at most 19.
- In Step 2 (the rightmost column) we add two digits only. Each digit is at most 9, so the sum is at most 9 + 9 = 18 < 20. So the first carry is 0 or 1.½ mark
- In Step 3 we add two digits and the carry. If the carry coming in is at most 1, the sum is at most 9 + 9 + 1 = 19.½ mark
- A sum of at most 19 is less than 20, so it has at most one ten. The carry passed to the next column is therefore 0 or 1.½ mark
- The same argument then works for the next column, and the next, all the way to the left. So the carry can never be more than 1.½ mark
Answer to write in the exam
Each digit ≤ 9.
Step 2: sum ≤ 9 + 9 = 18 < 20, so carry ≤ 1.
Step 3: sum ≤ 9 + 9 + 1 = 19 < 20, so carry passed on ≤ 1.
Repeating for every column, carry is never more than 1.
∴ The carry is always 0 or 1.
Common mistakes that cost marks
- Writing the largest sum as 9 + 9 = 18 and forgetting the carry in Step 3. The correct largest sum is 19.
- Saying “because we only carry 1” without explaining why the sum can never reach 20.
- Bringing in the three-number case. With three numbers a carry of 2 is possible, but this algorithm adds only two numbers, so the largest column sum is 9 + 9 + 1 = 19.
How this can come in the exam
When two numbers are added column by column, the largest possible sum in a single column (two digits plus the carry) is
- 18
- 19
- 20
- 27
Show answer
(B) 19
9 + 9 + 1 = 19.
Assertion (A): When three numbers are added column by column, the carry can be 2.
Reason (R): Three digits and a carry can add up to more than 19.
- Both A and R are true, and R is the correct explanation of A.
- Both A and R are true, but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
Show answer
(A) Both A and R are true, and R is the correct explanation of A.
With three numbers a column can reach 9 + 9 + 9 + 2 = 29 (for example 9 + 9 + 9 = 27 gives carry 2), so the carry can be 2. R is true and explains A.
Try one yourself
Add 9999 + 9999 by the algorithm. What carry does each column produce, and what is the largest column sum?
Show answer
Units 9 + 9 = 18: write 8, carry 1. Tens, hundreds, thousands: 9 + 9 + 1 = 19: write 9, carry 1 each time. Step 5 writes the last 1. Answer 19998; the largest column sum is 19, so every carry is 1.
More questions like this
- What happens if we do not include the fifth step in the algorithm above? Give examples where the algorithm will work correctly and where it will fail to work.
- 1. See if you can complete the argument about grouping by units, tens, hundreds, … to justify why the addition algorithm works.
2. How would you modify the algorithm to add two decimal fractions? - Algorithm to find the divisors of n: 1. Start with an empty list-of-divisors. 2. For each number j in the sequence 1, 2, 3, …, n – if j divides n, add j to the list-of-divisors. Let us execute this algorithm for a small number, say 18.
- Try to execute the algorithm to compute the divisors of 15, 135, and 775. How does the amount of work increase as the numbers grow?
- If we look through the divisors of 375 and 825 above, we see that the answer is 75. What if the numbers were 54000 and 81000?