11. Shortcuts & Speed Tricks

Digit-sum trick for 3 and 9 remainders: The remainder of any number when divided by 3 or 9 equals the remainder of its digit sum divided by 3 or 9. Saves you from long division on huge numbers. Example: Remainder of 9,87,654 ÷ 9 → digit sum 9+8+7+6+5+4=39 → 39÷9 leaves remainder 3.

Combine rules for composite divisors: For any divisor that factors into co-prime parts (6=2×3, 12=4×3, 15=3×5, 18=2×9, 24=8×3, 40=8×5, 72=8×9), check each co-prime factor separately — much faster than a direct long division. Example: Check 4,320 for divisibility by 40 → last digit 0 & second-last even qualifies for last-3-digits/8 test: last 3 digits =320, 320÷8=40 ✓; last digit 0 → divisible by 5 ✓ → divisible by 40.

1001 = 7×11×13 grouping trick: For a number with 6+ digits, split into groups of 3 from the right and alternately add/subtract. This single number instantly tests divisibility by 7, 11, AND 13 simultaneously.

Last-k-digit trick generalization: For 2^k or 5^k, only the last k digits matter (since 10^k absorbs the rest). Memorize: 2→1 digit, 4→2 digits, 8→3 digits, 16→4 digits; 5→1 digit, 25→2 digits, 125→3 digits.

Osculator memory chart: 7→(−2), 13→(+4), 17→(−5), 19→(+2), 3→(+1), 9→(+1), 11→(−1). Positive osculator means “multiply and add”; negative means “multiply and subtract.”

Never actually multiply out factorials for trailing zeros: Directly apply ⌊ n/5⌋+⌊ n/25⌋+… — this always finishes in 2-3 steps for SSC-level n (≤1000).

For a^n± b^n questions, check parity of n first: Even n → check (a-b) and (a+b) both divide a^n-bn. Odd n → only (a-b) divides a^n-bn; (a+b) divides a^n+bn.

Cyclicity for remainder-of-power questions: Find the remainder cycle of the base modulo the divisor (usually length 1 to 6 for small divisors), then use exponent mod cycle-length to jump directly to the answer instead of computing huge powers.

“Divisible by both A and B” ⇒ divisible by LCM(A,B), NOT A×B, unless A,B co-prime. A frequent trap: 4 and 6 are both factors, but LCM(4,6)=12, not 24 (since they share a factor of 2).

For missing-digit problems with two blanks, solve the “always true regardless of blank” rule first (e.g., last-3-digit tests for 8 that don’t involve the unknown), then use the remaining rule (usually digit-sum) to pin the digit — this often eliminates one variable instantly, as shown in Section 6, Example 3.

Sum/difference divisibility rule for quick elimination: If a is divisible by d and b is divisible by d, then a+b and a-b are also divisible by d. Use this to break large sums into pieces you can each test for divisibility rather than testing the whole sum.

99/999-style grouping for large numbers: Just as 1001 helps with 7, 11, 13, remember 99=9×11 and 999=27×37. Grouping digits in pairs (for 99) or triples (for 999) and summing the groups gives a fast divisibility/remainder check for 99 or 999 without long division — useful when a question directly asks about these divisors or their factors.

When a question gives you a specific factorial and asks for “number of zeros” vs “highest power of 2/3,” don’t confuse the two — zeros always use p=5 regardless of how the question is phrased; highest power of a prime other than 5 requires the general Legendre formula with that prime.

When the “unknown digit” doesn’t uniquely pop out, check whether its coefficient shares a common factor with the modulus. As Section 2.3 Example 5 and Section 2.7 Example 5 show, if the missing digit sits in a place-value position whose coefficient (10, 100, etc.) shares a factor with the divisor (4, 8, etc.), you may legitimately get 2, 3, or 5 valid digits rather than one — don’t force a “unique” answer where the mathematics allows several.

For remainder-of-a-power questions where the base shares a common factor with the modulus (e.g., 6^(75) mod 8), don’t blindly apply the co-prime cyclicity method — compute the first 2–3 powers directly; the remainder frequently collapses to 0 well before the target exponent and stays there.