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Aptitude & Reasoning · A — Quantitative Aptitude

Numbers, Factors & Divisibility

Everything downstream — percentages, ratios, remainders — sits on a confident grip of how numbers are built.

Test weight: HighTime / question: 45–60 secDifficulty: Easy → Medium

Most placement papers open with a number-properties question because it is fast to set and fast to mark. Speed here frees up minutes for the hard problems later, so treat this chapter as a warm-up you must never lose marks on.

Number families

Natural numbers (1, 2, 3…), whole numbers (0 included), integers (… −2, −1, 0, 1…), rationals (any p/q), and primes (exactly two factors). Knowing the family tells you which shortcuts are legal.

Factors, multiples, HCF & LCM

A factor divides a number exactly; a multiple is the number scaled up. HCF (highest common factor) is the biggest number dividing all given numbers. LCM (lowest common multiple) is the smallest number all of them divide into. For any two numbers: HCF × LCM = Product of the two numbers.

Divisibility rules worth memorising
These let you reject wrong options without long division.
DivisorQuick test
2Last digit is even
3Digit sum divisible by 3
4Last two digits divisible by 4
5Ends in 0 or 5
6Passes both the 2-test and the 3-test
8Last three digits divisible by 8
9Digit sum divisible by 9
11Difference of alternate digit sums is 0 or a multiple of 11

How to Approach It

Number questions are won by recognising structure before you calculate. Train yourself to ask 'what kind of number is this, and what property is being tested?' — the answer points you straight to the right tool.

  • Identify what's being asked. Decide whether the question is about divisibility, factors and multiples, HCF/LCM, remainders, or the last digit of a power. Naming the type first stops you from blindly grinding through arithmetic.
  • Reach for the matching tool. For a last digit, use unit-digit cyclicity; for 'largest/smallest number that divides...', think HCF or LCM; for 'how many factors', prime-factorise.
  • Prime-factorise when you're stuck. Breaking a number into its primes exposes its factors, divisibility and HCF/LCM all at once. Whenever the path isn't obvious, factorising almost always reveals it.
  • Bound-check before you commit. An HCF can't exceed the smallest number, an LCM can't be below the largest, and a remainder must be smaller than the divisor. A two-second bound check catches careless errors.

Techniques & Methods

  • Unit-digit cyclicity. Last digits repeat every 4 powers. Take the exponent mod 4 (remainder 0 → use the 4th in the cycle). e.g. 8^25: 25 mod 4 = 1, 8's cycle 8,4,2,6 → last digit 8.
  • HCF × LCM = product. For two numbers, multiply HCF and LCM to recover the product, then divide out the known number. e.g. HCF 6, LCM 36, one number 12 → other = 6×36/12 = 18.
  • Count factors fast. Prime-factorise, add 1 to each power, multiply the results. e.g. 72 = 2^3 × 3^2 → (3+1)(2+1) = 12 factors.
  • Alternating-digit test for 11. Difference of alternate digit sums divisible by 11 → the number is too; it also gives the remainder. e.g. 123456: (6−5+4−3+2−1) = 3 → remainder 3 on division by 11.
The Edge
Powers cycle. The unit digit of any power repeats every 4 steps. To find the last digit of 7^83, divide the exponent by 4: remainder 3, so take the 3rd digit in 7's cycle (7, 9, 3, 1) = 3. A remainder of 0 means use the 4th value in the cycle, not the 1st. This single trick clears almost every "find the unit digit" question in under ten seconds.
Worked example
Find the unit digit of 3 raised to the power 65.
  1. Recognise the type: we only need the last digit of a power, so this is a cyclicity question — no need to compute 3^65 itself.
  2. Watch 3's last digit as the power grows: 3^1 ends in 3, 3^2 in 9, 3^3 in 7, 3^4 in 1, then it repeats. The cycle is 3, 9, 7, 1 — a period of 4.
  3. Divide the exponent by 4 and keep the remainder: 65 ÷ 4 = 16 remainder 1.
  4. A remainder of 1 points to the 1st value in the cycle, which is 3. (Had the remainder been 0, we would take the 4th.)
Worked example
The LCM of two numbers is 2079 and their HCF is 27. If one number is 189, find the other.
  1. Recall the identity: HCF × LCM = product of the two numbers — this finds a missing number without listing factors.
  2. Rearrange: other number = (HCF × LCM) ÷ known = (27 × 2079) ÷ 189.
  3. Compute: 27 × 2079 = 56,133; then 56,133 ÷ 189 = 297.
  4. Sanity-check: 189 = 27 × 7 and 297 = 27 × 11. Since 7 and 11 are coprime, HCF stays 27 and LCM is 27 × 7 × 11 = 2079 — both match.

Worked Drills

Worked example
Find the unit digit of 17 raised to the power 200.
  1. 17's unit digit follows 7's cycle: 7, 9, 3, 1 (period 4).
  2. 200 ÷ 4 leaves remainder 0 → take the 4th value in the cycle = 1.
Worked example
The largest number that divides 70 and 125 leaving remainders 5 and 8 is:
  1. The number must exactly divide 70 − 5 = 65 and 125 − 8 = 117.
  2. Take HCF(65, 117) = 13.
Worked example
The greatest 4-digit number exactly divisible by 88 is:
  1. 9999 ÷ 88 = 113.6, so use the whole part 113.
  2. 113 × 88 = 9944.
Worked example
The product of two numbers is 2028 and their HCF is 13. The number of such pairs is:
  1. Write the numbers as 13a and 13b with a, b coprime. Then 169ab = 2028 → ab = 12.
  2. Coprime pairs giving 12: (1,12) and (3,4) → 2 pairs.
Worked example
How many trailing zeros are there in 25! ?
  1. Count factors of 5: floor(25/5) = 5.
  2. Add higher powers: floor(25/25) = 1. Total = 5 + 1 = 6.
Worked example
The remainder when 2 raised to the power 31 is divided by 7 is:
  1. 2^3 = 8 leaves remainder 1 mod 7, so powers cycle with period 3.
  2. 31 mod 3 = 1 → 2^1 = 2.
Worked example
The greatest number that divides 43, 91 and 183 leaving the same remainder is:
  1. It must divide the differences: 91 − 43 = 48 and 183 − 91 = 92 (and 183 − 43 = 140).
  2. HCF(48, 92, 140) = 4.
Worked example
The number of factors of 360 is:
  1. 360 = 2^3 × 3^2 × 5.
  2. Add 1 to each power and multiply: (3+1)(2+1)(1+1) = 24.
Worked example
The unit digit of 13^15 + 17^13 is:
  1. 13 → 3's cycle (3,9,7,1); 15 mod 4 = 3 → 7.
  2. 17 → 7's cycle (7,9,3,1); 13 mod 4 = 1 → 7. Sum of units 7 + 7 = 14 → unit digit 4.
Worked example
The greatest number that divides 1023 and 750 leaving remainders 3 and 2 respectively is:
  1. Subtract the remainders: it divides 1023 − 3 = 1020 and 750 − 2 = 748.
  2. HCF(1020, 748) = 68.
⚠ Watch out
  • Remainder 0 in cyclicity means the last position in the cycle, never the first.
  • 1 is not a prime number. 2 is the only even prime.
  • HCF can never be larger than the smallest number; LCM can never be smaller than the largest.
Takeaways
  • Name the type first. Divisibility, HCF/LCM, factor-count, or last-digit each has its own fast tool.
  • Cyclicity rules units. Exponent mod 4 picks the digit; remainder 0 means the 4th value.
  • HCF × LCM = product recovers any missing number for a pair instantly.
  • Prime-factorise to count factors — add 1 to each power and multiply.
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