Apply divisibility tests and reason about factors and multiples.
Free, timed SHSAT practice sets on divisibility, factors & multiples— every question includes a full worked explanation, and there's no account required.
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Use these tests instead of long division:
Counting multiples of \(d\) from \(a\) to \(b\): (multiples up to \(b\)) minus (multiples up to \(a - 1\)). Multiples of 7 from 12 to 78: \(11 - 1 = 10\).
"Divisible by 4 or 6": add the two counts, then subtract the multiples of the LCM (12), because you counted them twice. From 1 to 60: \(15 + 10 - 5 = 20\).
Common traps: missing an endpoint, using \(4 \times 6 = 24\) instead of the LCM 12, and missing the word NOT.
263 has a digit sum of 11, but the rule for 11 uses the alternating sum 2 - 6 + 3 = -1, so 263 \(\div\) 11 leaves a remainder of 10. The other choices are divisible: \(605 = 11 \times 55\), \(286 = 11 \times 26\), \(396 = 11 \times 36\). (Choosing one of them answers which number IS divisible.)
Count multiples of 3 from 1 to 30, then take away the multiples from 1 to 8: \(\lfloor 30 \div 3 \rfloor - \lfloor 8 \div 3 \rfloor = 10 - 2 = 8\). Leaving out the endpoint 30 gives \(7\). Leaving out both endpoints gives \(6\). Counting the multiples from 1 to 30 instead of from 9 gives \(10\).
Divisible by 4: 25. Divisible by 6: 16. Numbers divisible by both are multiples of 12, the LCM: 8. They were counted twice, so subtract them once: \(25 + 16 - 8 = 33\). Adding the two counts without removing the numbers counted twice gives \(41\). Removing the numbers counted twice two times gives \(25\). Removing the multiples of 24 instead of 12 (the LCM) gives \(37\).