Number System
In short: Number types, primes, divisibility, factors, unit digit, remainders — 21 graded models.
Arithmetic — SI › Number System & Simplification · 45 min read · Free to read · English & Telugu
Bilingual study notes for the Andhra Pradesh Police SI & Constable exams (SLPRB AP) by AP Police Hub, a product of Anvil Academy.
Classification, primes, divisibility, factors, unit digits and remainders — the complete SI ladder, fundamental to advanced.
1. Introduction & why it matters
Number System is the foundation of the entire Arithmetic paper. Every topic that follows — HCF-LCM, simplification, powers, averages — rests on a firm grip of how numbers are classified and how they behave under division. In the APSLPRB SI Prelims the section is worth a steady 3–5 marks, and because the questions are short and rule-based, they are among the fastest marks on the paper once the models are drilled.
2. Classification of numbers
Every number sits somewhere in one family tree. The first split is real vs imaginary; real numbers then split into rational and irrational. Knowing exactly which box a number falls in answers a whole class of true/false questions instantly.
The number families| Family | Meaning | Examples |
|---|
| Natural (N) | Counting numbers from 1 | 1, 2, 3, … |
| Whole (W) | Naturals together with 0 | 0, 1, 2, 3, … |
| Integers (Z) | Positive, negative and zero — no fractions | …, −2, −1, 0, 1, 2, … |
| Rational (Q) | Can be written as \frac{p}{q}, q\ne 0; decimal terminates or repeats | \frac{5}{3}, −8, 0.55…, \sqrt{9} |
| Irrational | Not expressible as \frac{p}{q}; decimal non-terminating, non-repeating | \sqrt{2}, \sqrt{5}, \pi |
| Imaginary | Even roots of negatives; i=\sqrt{-1} | \sqrt{-7}, 3i |
| Complex | Real + imaginary, form a+ib | 5+3i |
Decimals decide rational vs irrational
- Terminating — 0.5=\frac{1}{2}, 0.648=\frac{81}{125} — always rational.
- Non-terminating but repeating — 0.333…=\frac{1}{3}, 0.565656…=\frac{56}{99} — still rational.
- Non-terminating and non-repeating — \sqrt{2}=1.414…, \pi=3.1415… — irrational.
Special integer families
| Term | Definition | Note |
|---|
| Even / Odd | Divisible by 2 (2k) / not (2k\pm1) | 0 is even |
| Prime | Exactly two factors: 1 and itself | 2 = smallest & only even prime |
| Composite | More than two factors | 4 = smallest; 9 = smallest odd |
| Co-prime | Two numbers with HCF = 1 | (8, 9), (2, 3) — need not be prime |
| Twin primes | Two primes differing by 2 | (3, 5), (5, 7), (11, 13) |
| Perfect number | Sum of factors (excluding itself) = itself | 6, 28, 496 … |
3. Core identities & formulae
A short list of algebraic identities powers most Number-System shortcuts. Learn each formula with the situation it unlocks — that is what turns a 3-minute sum into a 20-second one.
Identity → where it is used| Identity | Formula | Used for |
|---|
| Difference of squares | a^2-b^2=(a-b)(a+b) | counting pairs, factorising |
| Sum / diff of cubes | a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2) | divisibility of a^n\pm b^n |
| Three-cube identity | if a+b+c=0 then a^3+b^3+c^3=3abc | divisibility puzzles |
| Sum of first n naturals | 1+2+\cdots+n=\dfrac{n(n+1)}{2} | series, unit-digit sums |
| Sum of squares | 1^2+2^2+\cdots+n^2=\dfrac{n(n+1)(2n+1)}{6} | square-series sums |
| Sum of cubes | 1^3+2^3+\cdots+n^3=\left[\dfrac{n(n+1)}{2}\right]^2 | cube-series sums |
| Repunit 10^n-1 | 10^n-1=\underbrace{99\ldots9}_{n}, digit-sum =9n | digit-sum questions |
| Consecutive product | product of any n consecutive integers is divisible by n! | divisibility of expressions |
4. Models — primes, parity & decimals
Model 1 — True/false about number types
Test each option against a definition or a single counter-example. One counter-example kills a universal claim.
Model 1
Question: Which one is NOT correct? (a) 1 is neither prime nor composite (b) 0 is neither positive nor negative (c) if p\times q is even, then p and q are both even (d) \sqrt{2} is irrational.
Approach: In a true/false set a single counter-example is enough to kill a universal claim, so instead of proving the safe options we look for the one an example can break. The parity rules make claim (c) the natural target.
- Statements (a), (b) and (d) are standard definitions — 1 is neither prime nor composite, 0 is neither positive nor negative, and \sqrt{2} is irrational — all plainly true.
3\times 4 = 12 \;(\text{even}) — Attack (c) by picking one odd and one even number, since 'the product is even' should not require both to be even:\text{even product} \Rightarrow \text{at least one factor even, not both} — The product 12 is even even though p=3 is odd, so a single even factor is enough — the claim that both must be even is false:
Answer: Option (c) is NOT correct.
Shortcut: For a product to be even you only need ONE even factor — never assume 'both'.
Model 2 — Prime test & the 6k±1 rule
To test if n is prime, trial-divide by primes up to \sqrt{n} only. Also: every prime > 3 is of the form 6k\pm1 — but the converse fails (a 6k\pm1 number need not be prime, e.g. 25=6\times4+1).
Model 2
Question: Is 247 a prime number?
Approach: To test whether a number is prime you only need to trial-divide by primes up to its square root, because any larger factor would already be paired with a smaller one you have checked. For 247 that means primes up to about 15.7.
- The primes at or below 15.7 are 2, 3, 5, 7, 11 and 13, so those are the only divisors worth trying:
- 247 is odd, its digit-sum 13 is not a multiple of 3, and it ends in neither 0 nor 5, so 2, 3 and 5 are ruled out at once:
247 = 13 \times 19 — Testing 13 succeeds — it divides 247 exactly, giving the co-factor 19:- Since a divisor other than 1 and 247 exists, the number is composite, not prime:
Answer: 247 is NOT prime (=13\times19).
Shortcut: Learn the smallest/largest 3-digit primes: 101 and 997.
Model 3 — Prime puzzles using parity
All primes except 2 are odd. So if a sum of primes is even/odd, the presence or absence of the number 2 is the whole trick.
Model 3
Question: The sum of three prime numbers is 100, and one exceeds another by 36. Find the three numbers.
Approach: Every prime except 2 is odd, and three odd numbers always add to an odd total. The target 100 is even, so exactly one of the three primes must be the even prime, 2 — that single parity fact cracks the whole puzzle.
2 + x + (x+36) = 100 — Fix one prime as 2 and let the other two be x and x+36; their total must be 100:2x + 38 = 100 — Collect the terms into a simple linear equation:2x = 62x = 31- So the three primes are 2, 31 and 31+36=67; a quick check confirms all are prime and they add to 100:
Answer: The three primes are 2, 31 and 67.
Shortcut: Whenever an odd count of primes must add to an even total, one of them is forced to be 2 — that observation alone starts most 'sum of primes' puzzles.
Model 4 — Recurring decimal → fraction
For a pure repeating block, put the block over as many 9's as its length. For a mixed decimal (some non-repeating digits first), the denominator is 9's for the repeating part followed by 0's for the non-repeating part.
Model 4
Question: Express 2.8\overline{3} (that is 2.8333\ldots) as a fraction in lowest terms.
Approach: A repeating decimal converts most cleanly if we set the whole-number part aside and handle only the decimal, using the mixed-recurring rule: the numerator is the block minus its non-repeating part, and the denominator carries a 9 for each repeating digit followed by a 0 for each fixed digit.
0.8\overline{3} = \frac{83 - 8}{90} — Here the 8 is fixed and the 3 repeats, so subtract the non-repeating part from the block and divide by one 9 (for the repeat) and one 0 (for the fixed 8):= \frac{75}{90}= \frac{5}{6}2 + \frac{5}{6} = \frac{17}{6} — Finally add the whole-number 2 that we set aside at the start:
Answer: 2.8\overline{3}=\dfrac{17}{6}.
Shortcut: Verify by dividing back: 17\div6=2.8333\ldots ✓
5. Models — divisibility & digits
Divisibility rules you must know cold| Divisor | Rule |
|---|
| 2, 4, 8, 16 | last 1 / 2 / 3 / 4 digits divisible by it |
| 3, 9 | digit-sum divisible by 3 / 9 |
| 5, 25, 125 | last 1 / 2 / 3 digits divisible by it |
| 11 | (sum of odd-place digits) − (sum of even-place digits) = 0 or a multiple of 11 |
| 7, 11, 13 | group in 3s from the right, alternately add/subtract groups; test the result |
| Composite (6, 72, 88, 90…) | split into co-prime factors and test each (6 = 2×3, 72 = 8×9, 88 = 8×11, 90 = 9×10) |
Model 5 — Composite divisor: split into co-prime factors
Model 5
Question: If the 11-digit number 543247x968y is divisible by 90, find the value of 4x+5y.
Approach: 90 is composite, so rather than one unwieldy rule we split it into the co-prime factors 9\times10 and demand divisibility by each. The 10-rule pins the last digit, and the 9-rule fixes the remaining unknown through the digit-sum.
y = 0 — Divisibility by 10 forces the last digit to be 0, so y=0:5+4+3+2+4+7+x+9+6+8+0 = 48 + x — Divisibility by 9 needs the digit-sum to be a multiple of 9; the known digits already total 48, leaving 48+x:48 + x = 54 \Rightarrow x = 6 — The next multiple of 9 after 48 is 54, which pins the digit x at 6:4x + 5y = 4(6) + 5(0) = 24 — Substitute x=6 and y=0 into the required expression:
Answer: With x=6 and y=0, the value of 4x+5y is 24.
Shortcut: A composite divisor is always broken into its co-prime factors and tested one at a time — never as a single big divisibility rule.
Model 6 — The 7-11-13 grouping test
Because 7\times11\times13=1001, a 6-digit pattern \overline{abcabc}=abc\times1001 is divisible by 7, 11 and 13. For any number, group the digits in threes from the right and take the alternating sum of the groups.
Model 6
Question: Is 6901895 divisible by 7, 11 and 13?
Approach: The key fact is that 7\times11\times13=1001. So rather than dividing by each of the three separately, we split the number into blocks of three digits from the right and take their alternating sum — if that total is 0 or a multiple of 1001, the whole number is divisible by all three at once.
6 \,|\, 901 \,|\, 895 — Starting from the right, mark the digits off three at a time. 6901895 splits into 895, then 901, and a leftover 6 on the left:895 - 901 + 6 = 0 — Now take the alternating sum of those blocks — rightmost minus the next plus the next — exactly the divisibility-by-11 idea, but applied to whole three-digit blocks instead of single digits:- The blocks cancel completely and leave 0. Because 0 is a multiple of 1001, the number is an exact multiple of 1001 = 7\times11\times13, so all three divide it with no remainder:
Answer: Yes — 6901895 is divisible by 7, 11 and 13.
Model 6b — the 11 rule
Question: If 738A6A is divisible by 11, find the digit A.
Approach: The 11-rule compares the digits sitting in odd positions with those in even positions: their difference must be 0 or a multiple of 11. Add each group, set the difference to the simplest allowed value, and solve for the unknown digit A.
7 + 8 + 6 = 21 — Add the digits in the odd positions (counting from the left): 7, 8 and 6:3 + A + A = 3 + 2A — Add the even-position digits, two of which are the unknown A:21 - (3 + 2A) = 0 \text{ or } 11 — Set the difference of the two group-sums to a valid value — 0 or a multiple of 11:18 - 2A = 0 \Rightarrow A = 9 — Taking the difference as 0 leaves a single-digit answer for A:
Answer: The digit is A = 9.
Model 7 — Digital sum to pick the right option
The digital sum (repeatedly add the digits until one digit) is preserved through +, − and ×. Compute the digital sum of the given expression and match it against the digital sums of the four options — a fast way to identify a big product or power.
Model 7
Question: 1024^2 = ? (a) 1034316 (b) 1052316 (c) 1153724 (d) 1048576
Approach: The digital sum is preserved through multiplication, so the square of a number has the digital sum of its base squared. Work that out once, then reject any option whose own digital sum disagrees — usually a single option survives.
\text{DS}(1024^2) = 7^2 = 49 \to 4+9 = 13 \to 4 — The digital sum of 1024 is 7, so the square must reduce to 7^2 collapsed to one digit, which is 4:(a)\,0,\;(b)\,0,\;(c)\,5,\;(d)\,1+0+4+8+5+7+6 = 31 \to 4 — Now find the digital sum of each of the four options:- Only option (d) carries the required digital sum of 4, so it must be the correct value:
Answer: 1048576 (option d).
Shortcut: Digital sums 3, 6, 9 can collide — if two options share the DS, fall back to unit digit (4^2=16\Rightarrow ends in 6).
Model 8 — Divisibility of a long number-string
Model 8
Question: The number 123456789101112\ldots is written until it has 79 digits. What is the remainder when it is divided by 8?
Approach: A number's remainder on division by 8 depends only on its last three digits, so the vast 79-digit string is irrelevant apart from its tail. The whole task reduces to finding which number the string is writing when it reaches its 79th digit.
79 - 9 = 70 \text{ digits left} — Writing 1 through 9 uses 9 single digits, leaving 70 of the 79 to come from two-digit numbers:70 \div 2 = 35 \text{ numbers} \Rightarrow \text{up to } 44 — Each two-digit number contributes 2 digits, so 70 digits are exactly 35 numbers, carrying the string up to 44:344 \div 8 = 43 \text{ remainder } 0 — So the string ends …42 43 44, whose last three digits 344 divide by 8 exactly:
Answer: The last three digits 344 are a multiple of 8, so the remainder is 0.
Shortcut: For 4, 8 or 16 never divide the whole giant number — read only the last two, three or four digits respectively.
6. Models — squares, identities & series
Model 9 — Difference of squares → counting pairs
For a^2-b^2=k write it as (a-b)(a+b)=k. A pair of natural numbers exists for each factorisation k=d_1\times d_2 with d_1<d_2 and d_1,d_2 of the same parity (both odd or both even), so that a=\frac{d_1+d_2}{2} is a whole number.
Model 9
Question: How many pairs of natural numbers have the difference of their squares equal to 45?
Approach: Write the condition as (a-b)(a+b)=45. Each way of writing 45 as d_1\times d_2 gives a natural-number pair, provided the two factors share the same parity so that a=\tfrac{d_1+d_2}{2} is whole. Since 45 is odd, every factor pair is odd×odd, so all of them qualify.
(a-b)(a+b) = 45 — Rewrite the difference of squares as the product of the difference and the sum:1\times45,\quad 3\times15,\quad 5\times9 — List the factor pairs of 45 — each is a candidate for (a-b,\ a+b):a = \tfrac{1+45}{2} = 23,\quad b = \tfrac{45-1}{2} = 22 — Each pair, being odd×odd, gives whole a and b; for example 1\times45 gives:- The three factor pairs give three valid natural-number pairs: (23, 22), (9, 6) and (7, 2):
Answer: 3 pairs.
Shortcut: If k is odd → count all factor pairs. If k\equiv0\ (\text{mod }4) → count only even-even pairs. If k\equiv2\ (\text{mod }4) → NO pair exists.
Model 10 — Consecutive integers & the three-cube identity
Model 10a
Question: The product of three consecutive positive integers is 15600. Find the sum of their squares.
Approach: Three consecutive integers multiply to roughly the cube of their middle one, so estimating \sqrt[3]{15600}\approx25 points straight at the central term. Confirm the run around 25, then square each and add.
24 \times 25 \times 26 = 15600 — Testing 25 as the centre, the run 24\times25\times26 hits 15600 exactly:24^2 + 25^2 + 26^2 = 576 + 625 + 676 — Now square each of the three and add them:= 1877
Answer: The three integers are 24, 25, 26 and their squares add to 1877.
Model 10b
Question: 73^3-40^3-33^3 is divisible by which number: 146, 196 or 126?
Approach: The identity a^3+b^3+c^3=3abc holds precisely when a+b+c=0. Read the expression as a sum of three cubes with bases 73, -40 and -33; if those bases add to zero the whole thing collapses into a single product whose factors we can inspect.
73 + (-40) + (-33) = 0 — Read the terms as cubes of 73, -40 and -33, and check their sum:73^3+(-40)^3+(-33)^3 = 3\times73\times(-40)\times(-33) — Because the bases sum to 0, the identity turns the expression into 3abc:= 3 \times 73 \times 40 \times 33- The product carries 2\times73=146 among its factors, so 146 divides it exactly:
Answer: The expression equals 3\times73\times40\times33, which is divisible by 146.
Model 11 — Telescoping (special) series
When each term is \dfrac{1}{n(n+1)}, split it as \dfrac{1}{n}-\dfrac{1}{n+1}. Almost every middle term cancels, leaving just the first and last.
Model 11
Question: Evaluate \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\cdots+\dfrac{1}{99\cdot100}.
Approach: Each term \frac{1}{n(n+1)} splits into \frac1n-\frac1{n+1}. Writing the whole sum in this split form makes consecutive pieces cancel in pairs, so only the very first and very last fragments are left standing.
\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} — Split the general term into a difference of two simple fractions:= \left(1-\tfrac{1}{2}\right)+\left(\tfrac{1}{2}-\tfrac{1}{3}\right)+\cdots+\left(\tfrac{1}{99}-\tfrac{1}{100}\right) — Write out the whole chain in this split form:= 1 - \frac{1}{100} — Every interior fraction is added once and subtracted once, so the chain collapses to the first term minus the last:= \frac{99}{100}
Answer: \dfrac{99}{100}.
Shortcut: General telescoping: sum =\dfrac{1}{\text{common diff}}\left(\dfrac{1}{\text{first denom}}-\dfrac{1}{\text{last denom}}\right).
7. Models — unit digit & cyclicity
The unit digit of a power repeats in a cycle. Digits 0, 1, 5, 6 keep the same unit digit for every power. Digits 4 and 9 have cycle length 2. Digits 2, 3, 7, 8 have cycle length 4 — divide the exponent by 4 and use the remainder.
Cyclicity table| Unit digit of base | Cycle | Cycle length |
|---|
| 0, 1, 5, 6 | same digit always | 1 |
| 4 | 4, 6, 4, 6… | 2 (odd power→4, even→6) |
| 9 | 9, 1, 9, 1… | 2 (odd power→9, even→1) |
| 2 | 2, 4, 8, 6 | 4 |
| 3 | 3, 9, 7, 1 | 4 |
| 7 | 7, 9, 3, 1 | 4 |
| 8 | 8, 4, 2, 6 | 4 |
Model 12 — Unit digit of a single power
Model 12
Question: What is the unit digit of 2^{51}?
Approach: The unit digit of a power repeats in a fixed cycle. A base ending in 2 runs through 2, 4, 8, 6 with period 4, so the remainder of the exponent on division by 4 tells us which of the four to read off.
51 \div 4 = 12 \text{ remainder } 3 — Divide the exponent by the cycle length 4 and keep only the remainder:2^3 = 8 — A remainder of 3 points to the third entry of the cycle 2, 4, 8, 6, which is 8:
Answer: Unit digit is 8.
Shortcut: If the remainder is 0, use the LAST digit of the cycle (here 6), not the first.
Model 13 — Unit digit of a product / sum
Model 13
Question: What is the unit digit of (2153)^{167}\times(8267)^{153}?
Approach: Only the last digit of each base affects the unit digit of the answer, so reduce each base to its unit digit, apply the 4-cycle to each exponent, and multiply the two resulting digits at the end.
167 \div 4 = 41 \text{ r } 3 \Rightarrow 3^3 = 27 \to 7 — The first base ends in 3, and 167 leaves remainder 3 mod 4, so its unit digit is that of 3^3=27, namely 7:153 \div 4 = 38 \text{ r } 1 \Rightarrow 7^1 = 7 — The second base ends in 7, and 153 leaves remainder 1 mod 4, so its unit digit is that of 7^1, namely 7:7 \times 7 = 49 \to 9 — Multiply the two unit digits and keep the last digit of the product:
Answer: Unit digit is 9.
Shortcut: Any product containing a factor ending in 5 and an even factor ends in 0 (5×2 pairs make trailing zeros).
8. Models — factors of a number
Everything about factors starts from the prime factorisation N=a^{x}\times b^{y}\times c^{z} (with a,b,c prime). From it, all the factor questions become one-line formulas.
Factor formulae (for N=a^{x}b^{y}c^{z})| Quantity | Formula |
|---|
| Number of factors (NOF) | (x+1)(y+1)(z+1) |
| Sum of factors (SOF) | \dfrac{a^{x+1}-1}{a-1}\cdot\dfrac{b^{y+1}-1}{b-1}\cdot\dfrac{c^{z+1}-1}{c-1} |
| Product of factors | N^{\,\text{NOF}/2} |
| Odd factors | drop the power of 2, then NOF of the rest |
| Perfect-square factors | count even exponents 0,2,4,\ldots for each prime, multiply |
Model 14 — Number & sum of factors
Model 14
Question: Find the number of factors and the sum of all factors of 2160.
Approach: Every fact about factors flows from the prime factorisation. Once 2160 is written as a product of prime powers, the count of factors is one plus each exponent multiplied together, and the sum of factors is a product of one short geometric series per prime.
2160 = 2^4 \times 3^3 \times 5^1 — First break 2160 into its prime powers:\text{NOF} = (4+1)(3+1)(1+1) = 5\times4\times2 = 40 — The number of factors adds one to each exponent and multiplies:\text{SOF} = \frac{2^5-1}{2-1}\cdot\frac{3^4-1}{3-1}\cdot\frac{5^2-1}{5-1} — The sum of factors multiplies one geometric-series term for each prime:= 31 \times 40 \times 6= 7440
Answer: NOF = 40, SOF = 7440.
Model 14b — product of factors
Question: What is the product of all the factors of 72?
Approach: The factors of a number come in pairs whose product is the number itself, so multiplying all of them raises N to half the total factor count. All we need first is how many factors there are.
72 = 2^3 \times 3^2 \Rightarrow \text{NOF} = (3+1)(2+1) = 12 — Prime-factorise 72 and count its factors first:\text{Product} = 72^{\,12/2} = 72^{6} — The 12 factors pair into 6 pairs, each multiplying to 72, so the product is 72 raised to 6:
Answer: 72^{6}.
Model 15 — Perfect-square factors
Model 15
Question: How many factors of 10800 are perfect squares?
Approach: A factor is a perfect square exactly when every prime in it appears to an even power. So for each prime we count how many even exponents are available and multiply those counts together.
10800 = 2^4 \times 3^3 \times 5^2 — Start from the prime factorisation of 10800:3 \times 2 \times 2 = 12 — Count the even exponents available for each prime — 2^4 offers {0,2,4}=3, 3^3 offers {0,2}=2, 5^2 offers {0,2}=2 — then multiply:
Answer: 12 perfect-square factors.
Model 16 — Trailing zeros / highest power of a prime in n!
The number of trailing zeros in n! equals the highest power of 5 it contains (5's are scarcer than 2's). Use Legendre's rule: keep dividing n by the prime and add the quotients (ignoring remainders).
Model 16
Question: How many trailing zeros does 100! have?
Approach: A trailing zero comes from a factor of 10=2\times5, and in a factorial the 5's are always rarer than the 2's, so the number of zeros is just the highest power of 5. Legendre's rule counts that power by dividing repeatedly by 5 and adding the quotients.
\left\lfloor\tfrac{100}{5}\right\rfloor = 20 — Count the multiples of 5 up to 100 — each supplies one factor of 5:\left\lfloor\tfrac{100}{25}\right\rfloor = 4 — Add the multiples of 25, since each of those carries a second factor of 5:20 + 4 = 24 — The next term \lfloor100/125\rfloor=0, so stop and add the counts:
Answer: 24 trailing zeros.
Shortcut: Same rule gives the highest power of any prime, e.g. power of 3 in 50! is 16+5+1=22.
9. Models — remainder theorems
Model 17 — Dividend = DQ + R puzzles
Model 17
Question: A number x is divided by a divisor. The divisor is 4 times the quotient and 2 times the remainder. If the remainder is 80, find x.
Approach: The two relations pin down the divisor and quotient from the known remainder, and then the division law \text{Dividend}=\text{Divisor}\times\text{Quotient}+\text{Remainder} assembles the number itself.
\text{Divisor} = 2 \times 80 = 160 — The divisor is twice the remainder, and the remainder is 80:\text{Quotient} = 160 \div 4 = 40 — The divisor is also 4 times the quotient, so divide back to get the quotient:x = 160 \times 40 + 80 — Feed divisor, quotient and remainder into the division law:x = 6400 + 80 = 6480
Answer: x = 6480.
Model 18 — Remainder of a product / sum
Replace each factor by its remainder, and — the big time-saver — use negative remainders when a factor is just below the divisor (e.g. 89\equiv-1\pmod{90}).
Model 18a
Question: Find the remainder when 1998\times1999\times2000 is divided by 7.
Approach: The remainder of a product equals the product of the remainders of its factors, so we reduce each of the three numbers modulo 7 first and multiply those small remainders instead of the huge product.
1998 \to 3,\quad 1999 \to 4,\quad 2000 \to 5 — Since 1995=7\times285 is a multiple of 7, the three numbers leave remainders 3, 4 and 5:3 \times 4 \times 5 = 60 — Multiply the three small remainders:60 \div 7 = 8 \text{ remainder } 4 — Reduce that product modulo 7 for the final remainder:
Answer: The product leaves remainder 4 on division by 7.
Model 18b — negative remainder
Question: Find the remainder when 88\times89 is divided by 90.
Approach: When a factor sits just below the divisor, it is cleaner to treat it as a small negative remainder. Here 88 and 89 are only 2 and 1 short of 90, so replacing them by -2 and -1 turns a scary product into a trivial one.
88 \equiv -2,\quad 89 \equiv -1 \pmod{90} — Read each factor as its small negative remainder below 90:(-2)\times(-1) = 2 — Multiply the two negative remainders — the two minus signs cancel:- The answer 2 is already positive and less than 90, so it is the remainder as it stands:
Answer: The remainder is 2.
Model 19 — Remainder of a^{n} (cyclicity, binomial, Fermat)
Three tools: (i) reduce the base to (\text{divisor}\pm1) so (ax+1)^n\to1 and (ax-1)^n\to(-1)^n; (ii) Fermat: if p is prime and a is co-prime to p, then a^{p-1}\equiv1\pmod p; (iii) plain cyclicity of the remainder.
Model 19a — binomial reduction
Question: Find the remainder when 25^{68} is divided by 24.
Approach: Write the base as one more than the divisor. Then in the binomial expansion of (24+1)^{68} every term except the last carries a factor of 24 and vanishes on division, leaving only 1^{68} behind.
25^{68} = (24+1)^{68} — Rewrite the base 25 as 24+1:(24+1)^{68} \equiv 1^{68} = 1 \pmod{24} — Every term of the expansion except 1^{68} contains a factor of 24, so only 1 survives modulo 24:
Answer: Remainder = 1.
Model 19b — Fermat's little theorem
Question: Find the remainder when 5^{16} is divided by 17.
Approach: Fermat's little theorem says that for a prime p and a base co-prime to it, a^{p-1}\equiv1\pmod p. Here p=17 is prime and 5 shares no factor with 17, and the exponent is exactly p-1=16, so the theorem gives the remainder in one line.
5^{17-1} = 5^{16} \equiv 1 \pmod{17} — Apply Fermat with p=17 and a=5, and the exponent is p-1=16:
Answer: By Fermat, the remainder is 1.
Model 20 — Successive division
In successive division the quotient of one step becomes the dividend of the next. To find the smallest number from given successive remainders, work backwards from the last quotient (start with 0).
Model 20
Question: A number, when successively divided by 6, 5 and 7, leaves remainders 3, 1 and 4. Find the smallest such number.
Approach: In successive division the quotient of one step is the dividend of the next, so the smallest number is built by working backwards: start from the final quotient of 0 and undo each division in turn using \text{dividend}=\text{divisor}\times\text{quotient}+\text{remainder}.
7 \times 0 + 4 = 4 — Begin at the last division (÷7, remainder 4) with quotient 0:5 \times 4 + 1 = 21 — Carry that 4 up as the quotient of the middle division (÷5, remainder 1):6 \times 21 + 3 = 129 — Carry 21 up as the quotient of the first division (÷6, remainder 3) to get the number:
Answer: The smallest number is 129.
Shortcut: The general answer is (6\times5\times7)k + 129 = 210k + 129.
Model 21 — Divisibility of a^{n}\pm b^{n}
Model 21
Question: 17^{21}+29^{21} is exactly divisible by which number?
Approach: For an odd exponent n, the sum a^n+b^n always has (a+b) as a factor. The exponent here is 21, which is odd, so we simply add the two bases to name a guaranteed divisor.
17 + 29 = 46 — Since n=21 is odd, (a+b) divides the sum, and a+b here is:
Answer: So 17^{21}+29^{21} is exactly divisible by 46.
Shortcut: Also 64^{n}-16^{n}=(2^6)^n-(4^2)^n is divisible by 64-16=48 for every n.
10. Shortcuts & exam tips
11. At a glance — formula sheet
Quick-revision formula sheet| Topic | Key result |
|---|
| Terminating decimal | denominator has only 2's and 5's |
| Prime form | every prime > 3 is 6k\pm1 (converse false) |
| Divisibility 7-11-13 | group in 3s from right; alternating sum |
| Cyclicity | 0/1/5/6 → 1; 4/9 → 2; 2/3/7/8 → 4 |
| Number of factors | (x+1)(y+1)(z+1) |
| Product of factors | N^{\text{NOF}/2} |
| Trailing zeros of n! | \lfloor n/5\rfloor+\lfloor n/25\rfloor+\cdots |
| Division law | \text{Dividend}=Dq+R, R<D |
| Binomial remainder | (ax+1)^n\to1, (ax-1)^n\to(-1)^n |
| Fermat | a^{p-1}\equiv1\pmod p (p prime, \gcd(a,p)=1) |
| a^n\pm b^n | odd n: a^n+b^n\div(a+b); a^n-b^n\div(a-b) always |