How can I check my answers quickly?
Getting the right answer fast is only half the skill. The other half is catching the wrong one before you commit to it. A few cheap checks can flag most slips in under a second, which matters when you are answering under SIXTY's time pressure.
None of these checks prove an answer is correct. They prove it is plausible. Use them as quick filters: estimation first, then a digit-level check if you want more confidence. With practice they run in the background while you solve.
What is the fastest first check I can do?
Estimate before you trust the exact number. Round each value to something easy, do the rough sum or product, and confirm your answer lands in the right order of magnitude.
For 47 × 6, round to 50 × 6 = 300, so expect something near 300. If your working gave 1692 or 28, you know instantly it is wrong. Estimation will not catch small errors, but it kills the large ones, which are the costly ones.
How does the last-digit check work?
The units digit of a sum or product depends only on the units digits of the inputs. Compute just that one digit and compare it to your answer's last digit. If they differ, the answer is wrong.
Example, addition: 238 + 145. The units are 8 + 5 = 13, ending in 3, so the total must end in 3. The real total is 383, which checks out. Example, multiplication: 47 × 6. Units are 7 × 6 = 42, ending in 2, so the product must end in 2 — and 282 does.
When is the even/odd (parity) check useful?
Parity is a one-bit version of the last-digit check, and it is even faster. A product is odd only when every factor is odd; if any factor is even, the product is even. A sum is odd only when an odd count of its terms are odd.
So 47 × 6 must be even because 6 is even — an odd answer like 281 is impossible. And 13 + 8 + 4 has one odd term, so it must be odd; the answer 25 is odd, which passes.
How do I cast out nines to check a result?
Casting out nines reduces each number to a single digit by repeatedly summing its digits, then checks that the operation holds on those reduced digits. It catches most random errors because the digit sum of a number equals the number's remainder when divided by 9.
It will not catch a swapped pair of digits (a transposition), since that does not change the digit sum. Treat a pass as good evidence, not proof.
- Take 47 × 6 = 282 and reduce each part.
- 47: 4 + 7 = 11, then 1 + 1 = 2.
- 6: stays 6.
- Multiply the reduced digits: 2 × 6 = 12, then 1 + 2 = 3.
- Reduce the result 282: 2 + 8 + 2 = 12, then 1 + 2 = 3.
- Both sides reduce to 3, so the answer passes the check.
Which quick divisibility checks should I keep handy?
Divisibility rules let you sanity-check answers that should land on a round factor. A number is divisible by 2 if it ends in an even digit, by 5 if it ends in 0 or 5, and by 10 if it ends in 0.
For 3, sum the digits and see if that is divisible by 3 — 282 gives 2 + 8 + 2 = 12, which is divisible by 3, so 282 is too. For 9, the same digit sum must be divisible by 9. For 4, check the last two digits as a pair: 282 ends in 82, which is not divisible by 4, so 282 is not.
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