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.

Step-by-step solution

Idea: Step 5 writes the final carry. Without it, a carry made in the leftmost column has nowhere to go, so the answer loses its first digit 1.

1. Step 5 says: if the value of carry is 1 at the end, write 1 to the left of the bottom row. Without it, the final carry is simply thrown away.1 mark
2. Works: 473 + 325. Units 3 + 5 = 8, tens 7 + 2 = 9, hundreds 4 + 3 = 7, carry 0. Result 798 ✓. The last column made no carry, so Step 5 was not needed.1 mark
3. Fails: 473 + 695. Units 3 + 5 = 8, tens 7 + 9 = 16 (write 6, carry 1), hundreds 4 + 6 + 1 = 11 (write 1, carry 1). Without Step 5 the bottom row is 168, but the correct sum is 1168.1 mark
4. So the algorithm without Step 5 works exactly when the leftmost column gives no carry, and fails whenever the leftmost column makes a carry (the answer then has one more digit than the longer number).

Answer to write in the exam

Common mistakes that cost marks

• Giving only a “works” example, or only a “fails” example. The question asks for both.
• Choosing a failing example where the carry happens in a middle column only; Step 5 is needed only for a carry out of the leftmost column.
• Writing the lost digit as 0 instead of 1 (the final carry is always 1, never more).

How this can come in the exam

Try one yourself

Without Step 5, what answer does the algorithm give for 856 + 279? What is the correct answer?

More questions like this

• 1. See if you can complete the argument about grouping by units, tens, hundreds, … to justify why the addition algorithm works.
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• 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?
• Suppose we want to compute gcd(375, 825).

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