1. Divisibility Rules
A divisibility rule lets you answer "is this divisible by N?" by inspecting the digits, not by dividing. Nine of these are worth having completely automatic:
By 2: last digit is even (0,2,4,6,8)
By 3: digit sum is divisible by 3
By 4: last TWO digits form a number divisible by 4
By 5: last digit is 0 or 5
By 6: divisible by BOTH 2 and 3
By 7: double the last digit, subtract from the rest,
repeat until small enough to check directly
e.g. 371: 37 - (2*1) = 35 -> divisible by 7
By 8: last THREE digits form a number divisible by 8
By 9: digit sum is divisible by 9
By 10: last digit is 0
By 11: (sum of digits in odd positions) - (sum of digits in
even positions) is divisible by 11 (including 0)
e.g. 9,163: (9+6) - (1+3) = 15-4 = 11 -> divisible
Rule 7 is the one students skip and later regret — it's the only one that isn't a
single glance, but it's still dramatically faster than long division once practiced.
Rules 4 and 8 generalize the same idea: for divisibility by 2^k, only
the last k digits matter, because everything before them is already a
multiple of a high enough power of 2.
Divisibility by 12 = divisible by both 3 and 4. By 15 = divisible by both 3 and 5. By 18 = divisible by both 2 and 9. Whenever a divisor factors into two coprime pieces (no shared factors), test each piece separately instead of memorizing a new rule — this covers almost every composite divisor a test will ask about.
2. Remainder Theorems
Remainder questions on large numbers (especially large powers) look intimidating but almost always reduce to one repeated trick: reduce the base modulo the divisor first, then find a repeating pattern in the powers.
Find the remainder when 7^45 is divided by 5.
Step 1: reduce the base mod 5 first: 7 mod 5 = 2
so this is the same question as: remainder of 2^45 / 5
Step 2: find the pattern of 2^n mod 5:
2^1 = 2 mod 5 = 2
2^2 = 4 mod 5 = 4
2^3 = 8 mod 5 = 3
2^4 = 16 mod 5 = 1 <-- pattern repeats every 4 powers
2^5 = 32 mod 5 = 2 (same as 2^1 -- confirmed)
Step 3: 45 mod 4 = 1 (45 = 4*11 + 1)
so 2^45 mod 5 behaves like 2^1 mod 5 = 2
Answer: remainder is 2This three-step shape — reduce the base, find the cycle length, use the exponent mod the cycle length — solves the overwhelming majority of "remainder of a huge power" questions. It's also exactly the mechanism behind Section 3's last-digit shortcut, which is this same idea specialized to divisor 10.
If reducing gives you a negative-looking result (e.g. "the remainder is -1 mod 7"), that's the same as remainder 6 — a negative remainder just means you're one step past zero going the other way around the cycle. Converting to the positive equivalent (divisor + negative remainder) avoids picking a negative-number answer that isn't actually one of the options.
3. Cyclicity & Last Digits
"What's the last digit of 7^45?" is Section 2's remainder trick applied specifically to divisor 10 — and every digit's last-digit pattern has a cycle length of either 1, 2, or 4.
0,1,5,6: last digit never changes (cycle length 1)
4,9: cycle length 2 (4,6,4,6... / 9,1,9,1...)
2,3,7,8: cycle length 4 (2,4,8,6,2... / 3,9,7,1,3...
/ 7,9,3,1,7... / 8,4,2,6,8...)7 has cycle length 4: 7^1=7, 7^2=9, 7^3=3, 7^4=1, then repeats
45 mod 4 = 1 -> behaves like 7^1
Last digit of 7^45 is 7One trap worth naming: when the exponent mod cycle-length comes out to 0 (not 1), that corresponds to the last position in the cycle, not a "zeroth" one — e.g. for a cycle length of 4, an exponent that's an exact multiple of 4 uses the 4th value in the cycle, not a nonexistent 0th value.
4. Worked Examples
Is 47,592 divisible by 11?
Odd positions (from right): 2, 5, 9 -> sum = 16
Even positions (from right): 9, 7, 4 -> sum = 20
16 - 20 = -4 -> NOT divisible by 11 (-4 isn't 0 or a
multiple of 11)Find the remainder when 23 x 41 is divided by 7.
Don't multiply first -- reduce each factor mod 7 FIRST:
23 mod 7 = 2
41 mod 7 = 6
Remainder of the product = remainder of (2 x 6) mod 7
= 12 mod 7
= 5
(Reducing before multiplying keeps the numbers small --
this generalizes to any number of factors.)Find the last two digits of 3^102.
Cycle of 3^n mod 100 has length 20 (a longer cycle than the
last-DIGIT cycle of length 4, since we're now tracking two
digits' worth of information, mod 100 instead of mod 10):
3^1=03, 3^2=09, 3^3=27, 3^4=81, ... 3^20 mod 100 = 01,
then it repeats.
102 mod 20 = 2 -> behaves like 3^2 mod 100 = 09
Last two digits: 095. Practice Problems
30 problems, no calculator — Easy, Medium & Tough
Work these on paper, under time pressure. Easy problems should take under 30 seconds each once you've reviewed the sections above; Medium under 60 seconds; Tough under 90 seconds.
Easy (1–10)
- Is 4,536 divisible by 4?
- Is 7,209 divisible by 9?
- Is 3,410 divisible by 5?
- Find the digit sum of 58,392 and state whether it's divisible by 3.
- Is 8,712 divisible by 8?
- What is the last digit of 3^7?
- Find the remainder when 29 is divided by 6.
- Is 96,349 divisible by 11?
- What is the last digit of 9^15?
- Find the remainder when 100 is divided by 7.
1) Last two digits 36, 36÷4=9 → yes. 2) Digit sum 7+2+0+9=18 → yes. 3) Last digit 0 → yes. 4) 5+8+3+9+2=27, divisible by 3 → yes. 5) Last three digits 712, 712÷8=89 → yes. 6) 3's cycle 3,9,7,1 (len 4); 7 mod 4=3 → 3rd value → 7. 7) 29 mod 6 = 5. 8) Odd positions (9,3,6)=18, even (4,9)=13, 18-13=5 → not divisible. 9) 9's cycle 9,1 (len 2); 15 mod 2=1 → 1st value → 9. 10) 100 mod 7 = 2.
Medium (11–20)
- Is 47,736 divisible by 12?
- Find the remainder when 2^20 is divided by 5.
- What is the last digit of 6^234?
- Is 6,930 divisible by 18?
- Find the remainder when 5^41 is divided by 6.
- What is the last digit of 4^57?
- Find the remainder when 17 x 23 is divided by 5.
- Is 123,456 divisible by 8?
- Find the remainder when 3^100 is divided by 4.
- What is the last digit of 7^100?
11) Digit sum 27 → div 3; last two digits 36÷4=9 → div 4; both → yes, divisible by 12. 12) 2's cycle mod 5: 2,4,3,1 (len 4); 20 mod 4=0 → last value → 1. 13) 6's last digit is always 6 (cycle length 1) → 6. 14) Last digit 0 → div 2; digit sum 18 → div 9; both → yes. 15) 5's cycle mod 6: 5,1 (len 2); 41 mod 2=1 → 1st value → 5. 16) 4's cycle 4,6 (len 2); 57 mod 2=1 → 1st value → 4. 17) 17 mod 5=2, 23 mod 5=3; 2×3=6 mod 5=1. 18) Last three digits 456÷8=57 → yes. 19) 3's cycle mod 4: 3,1 (len 2); 100 mod 2=0 → last value → 1. 20) 7's cycle 7,9,3,1 (len 4); 100 mod 4=0 → last value → 1.
Tough (21–30)
- Find the remainder when 2^100 is divided by 7.
- Find the last two digits of 7^45.
- Find the remainder when 31^31 is divided by 5.
- A number leaves remainder 4 when divided by 7. What remainder does the SQUARE of that number leave when divided by 7?
- Find the remainder when 19 x 23 x 29 is divided by 7.
- Find the last two digits of 3^101.
- Find the remainder when 6^83 is divided by 8.
- Is 15! (15 factorial) divisible by 11? Explain without computing 15! directly.
- Find the remainder when 2^50 is divided by 9.
- Find the last digit of 13^13 × 17^17.
21) 2's cycle mod 7: 2,4,1 (len 3); 100 mod 3=1 → 1st value → 2. 22) 7's cycle mod 100: 07,49,43,01 (len 4); 45 mod 4=1 → 1st value → 07. 23) 31 mod 5=1; 1 raised to anything is 1 → remainder 1. 24) 4²=16, 16 mod 7=2. 25) 19 mod 7=5, 23 mod 7=2, 29 mod 7=1; 5×2×1=10 mod 7=3. 26) 3's cycle mod 100 has length 20 (3^20 mod 100=01); 101 mod 20=1 → 1st value → 03. 27) 6^1=6, 6^2=36 mod 8=4, 6^3=216 mod 8=0, and every power from n=3 onward stays 0 mod 8 → remainder 0. 28) Yes — 11 is prime and appears directly as one of the factors (1×2×...×11×...×15), so 15! is divisible by 11 without needing the full product. 29) 2's cycle mod 9: 2,4,8,7,5,1 (len 6); 50 mod 6=2 → 2nd value → 4. 30) 13's last digit 3, cycle 3,9,7,1, 13 mod 4=1 → 3; 17's last digit 7, cycle 7,9,3,1, 17 mod 4=1 → 7; last digit of 3×7=21 → 1.
6. Knowledge Check
Four quick questions. Expand each to check your answer.
Q1
Why does testing divisibility by 12 reduce to testing divisibility by both 3 and 4 separately, but testing divisibility by 6 by testing "divisible by 2 and 3" (not "divisible by 1 and 6")?
Why does testing divisibility by 12 reduce to testing divisibility by both 3 and 4 separately, but testing divisibility by 6 by testing "divisible by 2 and 3" (not "divisible by 1 and 6")?
The trick only works when you split the divisor into two factors that share no common factors (coprime) — 3 and 4 share none, and neither do 2 and 3. Testing "divisible by 1 and 6" is meaningless since every number is divisible by 1; the split has to actually use both prime-power components of the original number.
Q2
When finding the remainder of a large power divided by a small number, why do you reduce the base first instead of computing the full power?
When finding the remainder of a large power divided by a small number, why do you reduce the base first instead of computing the full power?
The remainder of a number depends only on its value modulo the divisor, not on its exact magnitude — so a huge base can be replaced with its much smaller remainder from the start, and the rest of the calculation stays small and manageable instead of requiring the actual (enormous) power to ever be computed.
Q3
Why does every digit's last-digit cycle have a length of exactly 1, 2, or 4 — never, say, 3?
Why does every digit's last-digit cycle have a length of exactly 1, 2, or 4 — never, say, 3?
The last digit is governed by arithmetic modulo 10, and the possible cycle lengths are constrained by the structure of that modulus (related to Euler's totient of 10, which is 4) — the observed cycle lengths of 1, 2 and 4 are exactly the divisors of that number, which is a consequence of number theory rather than a coincidence worth re-deriving under exam time pressure, just worth trusting.
Q4
If a number's exponent mod its cycle length comes out to exactly 0, which value in the cycle does that correspond to?
If a number's exponent mod its cycle length comes out to exactly 0, which value in the cycle does that correspond to?
It corresponds to the LAST value in the cycle, not a nonexistent "0th" one — an exponent that's an exact multiple of the cycle length has completed whole cycles and lands exactly where the cycle finishes, which is the same position as the cycle's final entry.