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UPSC CSE · CSAT · Numeracy and data interpretation · TOPIC LESSON

Number properties and remainders

Use number structure, factors and cycles to replace large calculations with short, checkable arguments.

UPSC Civil Services45 min study guide8 indexed sections

BY THE END, YOU CAN

  • Classify and order numbers with stated conventions.
  • Use divisibility, prime factors, greatest common divisors and least common multiples.
  • Compute remainders and estimate orders of magnitude safely.

BEFORE YOU START

  • Add, multiply and divide integers.
  • Understand powers as repeated multiplication for positive integer exponents.

Original educational lesson. Work through the examples before revealing the practice answers. This original common-core lesson supports its declared paper scope. Optional subjects, qualifying language papers and the interview are outside this resource.

LEARN THROUGH ORIGINAL PRACTICE

Original UPSC-style worked applications and practice

All teaching examples, cases, prompts and model answers are original. Official exam questions and answer keys are not reproduced. This common-core lesson does not imply complete historical paper coverage or an official score.

01 / UNDERSTAND & APPLY

Domains, signs and magnitude

In this lesson, natural numbers mean 1,2,3,… and whole numbers include 0. Some books use a different convention for natural numbers, so state it when zero matters. Integers include negative whole numbers. Rational numbers can be written as p/q with integers p,q and q nonzero; irrational real numbers cannot. Together rational and irrational numbers form the real numbers.

A finite decimal is rational because it can be written over a power of 10; an eventually repeating decimal is also rational. Not every infinite decimal is irrational. A nonterminating, nonrepeating decimal expansion characterizes an irrational real number in base 10. Examples include √2, whereas √9=3 is rational.

Order and magnitude are different. On the number line,−7 is less than−2, but its absolute value 7 exceeds 2. Absolute value is distance from zero and is never negative. A large negative number in magnitude is not a larger signed value.

ORIGINAL PRACTICE · 1

Order−7,−2,0,3 from least to greatest and state |−7|.

Reveal answer & reasoning

−7<−2<0<3; |−7|=7.

Moving right increases signed value. Absolute value ignores direction and measures the distance to zero.

02 / UNDERSTAND & APPLY

Place value and scale

In base 10, each position represents a power of 10. Thus 345=3×100+4×10+5. Multiplying by 10 shifts place value; it does not merely append a zero to every kind of written number. For decimals,3.45×10=34.5.

Scientific notation writes a positive value as a×10^k with 1≤a<10. Multiplication and division combine coefficients and add or subtract exponents. The scale check often reveals a misplaced thousand or million before exact arithmetic is finished.

“Order of magnitude” is sometimes used informally for scale and sometimes for the nearest power of 10. State which convention you mean. Here an estimate such as “a few thousand” is a scale check, not a claim that every source uses one rounding convention for the phrase.

ORIGINAL WORKED EXAMPLE

Thousands, not millions

Calculate(4.8×10^6)/(1.2×10^3) and first estimate its scale.

  1. The numerator is a few million and the denominator a little over one thousand, so expect a few thousand.
  2. Divide coefficients:4.8/1.2=4.
  3. Subtract exponents:10^6/10^3=10^3.
  4. Combine to get 4×10^3=4000, consistent with the scale estimate.

Answer 4000. Leaving the numerator’s exponent unchanged would confuse the divisor’s thousand-fold scale.

03 / UNDERSTAND & APPLY

Why the familiar tests work

An integer n is divisible by a nonzero integer d when n=dq for an integer q. Zero is divisible by every nonzero integer because 0=d×0. Division by zero is undefined; do not treat it as a special case of a divisibility test.

For divisibility by 2,5 or 10, only the last digit matters because every higher place is a multiple of 10, and 10 is divisible by 2 and 5. Even last digits give divisibility by 2;0 or 5 by 5;0 by 10. For 4, the last two digits suffice because 100 is divisible by 4. For 8, the last three suffice because 1000 is divisible by 8.

For 3 and 9,10 leaves remainder 1, so every power of 10 does too. A number has the same remainder as its digit sum modulo 3 or 9. For 11,10 leaves remainder−1, so alternating place values give an alternating digit sum; divisibility holds when that sum is a multiple of 11, including 0. For example,121 gives 1−2+1=0. These tests establish divisibility of integers, not arbitrary decimals without conversion.

ORIGINAL PRACTICE · 1

Which of 3,4,5,8 and 9 divide 3456?

Reveal answer & reasoning

3,4,8 and 9 divide it;5 does not.

Digit sum 3+4+5+6=18 is divisible by 3 and 9. Last two digits 56 are divisible by 4. Last three 456=8×57. The last digit 6 is neither 0 nor 5.

04 / UNDERSTAND & APPLY

Prime factors and divisor choices

A prime integer exceeds 1 and has exactly two positive divisors:1 and itself. One is not prime, and 0 is not prime. Two is the only even prime, because every even integer greater than 2 has 2 as a proper divisor. Composite positive integers greater than 1 have additional positive divisors.

Every integer greater than 1 has a unique prime factorization apart from factor order. A positive divisor chooses an exponent from 0 up to the available exponent of each prime. If n=p^a q^b for distinct primes, its divisor count is(a+1)(b+1). The multiplication counts independent exponent choices; it does not add the primes themselves.

ORIGINAL WORKED EXAMPLE

Counting the divisors of 360

How many positive divisors does 360 have?

  1. Factor 360=36×10=2³×3²×5¹.
  2. A divisor chooses the power of 2 from 0,1,2,3: four choices.
  3. It chooses the power of 3 from 0,1,2: three choices, and the power of 5 from 0,1: two choices.
  4. Each exponent triple gives one distinct positive divisor, so multiply 4×3×2=24.

Answer 360 has 24 positive divisors, including 1 and 360.

ORIGINAL PRACTICE · 1

Why is 1 not prime, while 2 is prime and every larger even positive integer is not?

Reveal answer & reasoning

1 has only one positive divisor;2 has exactly two; larger even integers have a proper divisor 2.

The definition requires exactly two positive divisors and a value greater than 1. It is not enough to be divisible by 1 and oneself, because every positive integer has those divisors, with the two coinciding for 1.

05 / UNDERSTAND & APPLY

Common factors, common multiples and wording

The greatest common divisor takes the minimum exponent of each shared prime factor; the least common multiple takes the maximum exponent needed to include every factor of both numbers. For two positive integers a,b, gcd(a,b)×lcm(a,b)=ab because each prime’s minimum and maximum exponents sum to its two original exponents. Do not apply that two-number identity unchanged to three numbers.

Interpret the object being maximized. The greatest number of identical groups using every red and blue bead must divide both colour totals; the GCD gives the number of groups. In a different problem asking the largest equal-sized single-colour packets, the same arithmetic can refer to packet size. Name the unit of the GCD result.

An LCM finds the first positive alignment when cycles share the same starting event. If cycles have different starting offsets, solve the remainder conditions instead of simply taking the LCM from zero.

ORIGINAL WORKED EXAMPLE

Identical bead groups

Use all 84 red beads and 126 blue beads to form the greatest possible number of identical groups. Every group must contain the same red count and the same blue count. How many groups and how many beads of each colour per group?

  1. The group count must divide both 84 and 126.
  2. 84=2²×3×7 and 126=2×3²×7, so their GCD is 2×3×7=42.
  3. Divide each total by 42: red 84/42=2 and blue 126/42=3.
  4. A count greater than 42 cannot divide both totals, so no larger identical-group count is possible.

Answer 42 groups, each containing 2 red and 3 blue beads.

ORIGINAL WORKED EXAMPLE

Aligned and offset signals

Signal A repeats every 18 minutes and signal B every 24 minutes. First, both ring at 09:00: when do they next ring together? Separately, take time 0 as A’s ring and suppose B rings at 6 minutes and every 24 minutes after that. Find the first nonnegative common ring time in this offset case.

  1. Aligned case:18=2×3² and 24=2³×3, so LCM=2³×3²=72 minutes.09:00+72 minutes=10:12.
  2. Offset case: A times are 0,18,36,54,…; B times are 6,30,54,… .
  3. The first common time is 54 minutes. Check 54 is divisible by 18 and 54−6=48 is divisible by 24.
  4. The common cycle repeats every 72 minutes, but its offset is 54, not 0.

Answer Aligned signals next coincide at 10:12. In the offset model their first common ring is 54 minutes after time 0.

ORIGINAL PRACTICE · 1

Find gcd(48,180) and lcm(48,180).

Reveal answer & reasoning

GCD 12 and LCM 720.

48=2⁴×3;180=2²×3²×5. Minimum shared exponents give 2²×3=12. Maximum exponents give 2⁴×3²×5=720. Check 12×720=48×180=8640.

06 / UNDERSTAND & APPLY

Use the allowed remainder range

For integer n and positive divisor d, Euclidean division writes n=dq+r with integer q and 0≤r<d. This convention also applies when n is negative. For−17 divided by 5, choosing q=−4 gives r=3; a negative leftover−2 is not an allowed Euclidean remainder.

If a and b have remainders r and s modulo d, their sum and product have the same remainders as r+s and rs after reducing again. Terms containing a factor d contribute no remainder. This lets you work with small numbers rather than huge products.

For example, numbers leaving remainders 4 and 5 modulo 7 have sum remainder 2 because 4+5=9≡2, and product remainder 6 because 20≡6. Division needs more care: one cannot generally divide remainders by an arbitrary number without checking invertibility, so this lesson does not use that shortcut.

ORIGINAL PRACTICE · 1

Find the Euclidean remainder when−17 is divided by 5.

Reveal answer & reasoning

3.

−17=5×(−4)+3 and 0≤3<5. Writing−17=5×(−3)−2 is an identity, but−2 does not meet the specified remainder range.

07 / UNDERSTAND & APPLY

Parity and power cycles

Even integers have form 2 k; odd integers have form 2 k+1. Adding two odds gives an even number, multiplying two odds gives an odd, and adding an even to an odd stays odd. These restrictions can eliminate impossible answers without determining the exact integer.

Last digits are remainders modulo 10. Powers often cycle because each step multiplies the current remainder by the same base. List a complete repeating cycle and map the exponent carefully. If the exponent remainder is 0, choose the final position of the cycle, not a nonexistent zeroth entry.

ORIGINAL WORKED EXAMPLE

The last digit of a large power

Find the last digit of 7^103.

  1. Compute successive last digits:7¹→7,7²→9,7³→3,7⁴→1. Multiplying once more returns to 7.
  2. The cycle length is 4. Divide 103 by 4:103=4×25+3.
  3. Use the third cycle entry,3.
  4. For comparison, exponent 104 would use the fourth entry 1 because its remainder is 0.

Answer The last digit is 3.

ORIGINAL PRACTICE · 1

What is the parity of odd×odd+even?

Reveal answer & reasoning

Odd.

Write the odd factors as 2 a+1 and 2 b+1. Their product is 2(2 ab+a+b)+1, which is odd. Adding 2 c preserves that form.

08 / UNDERSTAND & APPLY

Check bounds and the requested domain

A numerical answer should satisfy the original domain and all constraints. Consecutive squares can locate a square root without decimal approximation:8²=64<80<81=9², so√80 lies strictly between 8 and 9. The smallest integer greater than it is 9.

For divisibility in an interval, find the relevant common multiple and inspect only its multiples within the endpoints. Inclusive and exclusive endpoints can change the answer. Also distinguish a greatest integer below a value from a least integer above it, especially when the value itself is an integer.

Finish by checking signs, units and whether zero is permitted. A technically correct factor calculation can answer the wrong question if it labels a group count as a group size or ignores an excluded boundary.

ORIGINAL PRACTICE · 1

What is the smallest integer strictly greater than√80?

Reveal answer & reasoning

9.

64<80<81 implies 8<√80<9. Thus 9 is the first integer above it; no decimal expansion is needed.

ORIGINAL PRACTICE · 2

Find the integers from 100 to 120 inclusive that are divisible by both 6 and 9.

Reveal answer & reasoning

108 only.

A number divisible by both is a multiple of lcm(6,9)=18. Nearby multiples are 90,108,126, and only 108 lies within the stated interval.

KEEP THE BIG IDEAS

Quick revision

  • State number domains and zero conventions when they matter.
  • Use place value to justify divisibility and scale checks.
  • GCD and LCM answer different common-factor and common-cycle questions.
  • Euclidean remainders satisfy 0≤r<d even for negative dividends.
  • Check power-cycle indexing and the units of the requested result.
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