Number system for AFCAT: divisibility, unit digits, remainders and factors

Number system questions in AFCAT are usually quick: a missing digit, a unit digit, a remainder or a count of factors. They reward candidates who know a small set of rules well enough to apply them without writing much. The same rules also help across the section: divisibility tests speed up simplification, and prime factorisation is the heart of HCF and LCM.

This guide covers the rules, explains why each one works (so that you can rebuild it if you forget), and then solves AFCAT-level questions step by step.

Types of numbers

Three facts that setters like: 1 is neither prime nor composite, 2 is the only even prime, and there are 25 primes below 100.

Divisibility rules, and why they work

Why the 3 and 9 rules work. 10 = 9 + 1, so 10, 100, 1,000 and every power of 10 leave remainder 1 when divided by 9 (or 3). So 7,245 = 7 × 1,000 + 2 × 100 + 4 × 10 + 5 leaves the same remainder as 7 + 2 + 4 + 5. A number and its digit sum always leave the same remainder on division by 9.

Why the 4 and 8 rules work. 100 is divisible by 4, and 1,000 is divisible by 8. So everything above the last two digits (for 4) or the last three (for 8) is already divisible, and only the tail matters.

Why the 11 rule works. 10 = 11 − 1, so the powers of 10 leave remainders +1, −1, +1, −1 in turn on division by 11. That is why the digits are added and subtracted alternately.

Combined divisors. To test for 72, test for 8 and 9. To test for 24, test for 3 and 8. This works only when the two parts are co-prime: testing for 4 and 6 does not prove divisibility by 24 (12 passes both but is not divisible by 24).

Unit digits

The unit digit of a power depends only on the unit digit of the base, and it repeats in a short cycle.

Method: divide the power by 4 and use the remainder to pick the position in the cycle. If the remainder is 0, use the fourth (last) digit of the cycle. Since every cycle length divides 4, dividing by 4 works for all bases; for 4 and 9, just write the cycle twice (4, 6, 4, 6).

Remainders

Two rules solve most AFCAT remainder questions.

• Remainders multiply and add. The remainder of a product (or sum) equals the remainder of the product (or sum) of the individual remainders. Work with small remainders instead of big numbers.
• Change of divisor. If a number leaves remainder r on division by d, and m divides d, then the remainder on division by m is the remainder of r on division by m.

Factors and trailing zeros

Number of factors. Write the number as a product of prime powers, add 1 to each power, and multiply. For 72 = 2³ × 3², the count is (3 + 1)(2 + 1) = 12. Why it works: every factor is 2ᵃ × 3ᵇ with a from 0 to 3 (four choices) and b from 0 to 2 (three choices).

Trailing zeros in n! Each zero needs a 10 = 2 × 5, and there are always more 2s than 5s. So count the 5s: n ÷ 5, plus n ÷ 25, plus n ÷ 125, keeping only whole parts.

Useful sums

• Sum of the first n natural numbers = n(n + 1)/2. First 20: 20 × 21 ÷ 2 = 210.
• Sum of the first n odd numbers = n². First 10 odd numbers: 100.
• Sum of the first n even numbers = n(n + 1). First 10 even numbers: 110.

Worked examples

Example 1. What digit should replace x in 7,4x6 to make the number divisible by 9?

• Digit sum = 7 + 4 + x + 6 = 17 + x.
• The next multiple of 9 is 18, so x = 1. (x = 10 is not a digit.) Check: 7,416 = 9 × 824.

Example 2. Is 13,968 divisible by 72?

• Test 8: the last three digits, 968 = 8 × 121. Yes.
• Test 9: 1 + 3 + 9 + 6 + 8 = 27. Yes.
• 8 and 9 are co-prime, so the number is divisible by 72: yes (13,968 = 72 × 194).

Example 3. Find the unit digit of 7⁴⁵ × 3²².

• 45 ÷ 4 leaves 1, so 7⁴⁵ ends in the first digit of the cycle: 7.
• 22 ÷ 4 leaves 2, so 3²² ends in the second digit of its cycle: 9.
• 7 × 9 = 63, so the product ends in 3.

Example 4. Find the remainder when 1,234 × 1,235 × 1,236 is divided by 7.

• 1,232 = 7 × 176, so the remainders are 2, 3 and 4.
• 2 × 3 × 4 = 24, and 24 ÷ 7 leaves 3.

Example 5. A number leaves remainder 17 when divided by 36. What is the remainder when it is divided by 12?

• The number is 36k + 17. Since 12 divides 36, the 36k part leaves 0.
• 17 ÷ 12 leaves 5. (Check with k = 1: 53 = 12 × 4 + 5.)

Example 6. How many factors does 360 have, and how many zeros end 100!?

• 360 = 2³ × 3² × 5, so the count is 4 × 3 × 2 = 24 factors.
• 100! has 100 ÷ 5 = 20 fives, plus 100 ÷ 25 = 4 more: 24 zeros.

Practice set

1. Find the unit digit of 2⁵⁰.
2. What is the smallest number that must be added to 1,000 to make it divisible by 7?
3. How many factors does 36 have?
4. Is 5,832 divisible by 72?
5. What digit should replace x in 3,x52 to make it divisible by 11?
6. Find the remainder when 17 × 23 × 29 is divided by 5.
7. How many zeros are there at the end of 60!?
8. Find the sum of all natural numbers from 21 to 40.

Answers:

1. 4. 50 ÷ 4 leaves 2, and the cycle of 2 is 2, 4, 8, 6.
2. 1. 1,000 ÷ 7 leaves 6 (7 × 142 = 994), so add 1 to reach 1,001 = 7 × 143.
3. 9. 36 = 2² × 3², so 3 × 3 = 9.
4. Yes. 832 = 8 × 104 and 5 + 8 + 3 + 2 = 18, so it is divisible by 8 and 9 (5,832 = 72 × 81).
5. 6. (3 + 5) − (x + 2) = 6 − x must be 0, so x = 6. Check: 3,652 = 11 × 332.
6. 4. Remainders 2, 3 and 4 give 24, which leaves 4 on division by 5.
7. 14. 60 ÷ 5 = 12, and 60 ÷ 25 = 2 (whole part), so 12 + 2 = 14.
8. 610. Sum 1 to 40 = 820, minus sum 1 to 20 = 210.

What to do next

• Write the divisibility table and the unit-digit table from memory, then check them.
• Solve 20 mixed number-system questions in 20 minutes.
• Move on to HCF and LCM, which uses the same prime factorisation.
• Then practise decimals and fractions, the next topic in the numerical ability plan.