Number system for RRB NTPC: rules, shortcuts and practice

Number system questions test whether you understand how numbers behave. They rarely need long calculation. A candidate who knows the rules can answer most of them in 30 to 40 seconds, which makes this one of the best-value topics in the Maths section.

It is also the base for other topics. LCM and HCF, simplification and remainders in series questions all depend on it. Study it first, and study it properly.

Types of numbers

Divisibility rules, and why they work

For a composite divisor, split it into co-prime factors. For 12, test 3 and 4, not 2 and 6, because 2 and 6 share a factor.

Unit digits of powers

The unit digit of a product depends only on the unit digits of the numbers multiplied. So powers of any number ending in 3 behave exactly like powers of 3, and they repeat in a cycle.

Method for 2, 3, 7 and 8: divide the power by 4 and use the remainder to pick the term. A remainder of 0 means the last term of the cycle. For 4 and 9: an odd power gives the first digit (4 or 9), an even power the second (6 or 1).

Factors

Write the number as a product of prime powers, for example 72 = 2³ × 3². Then:

• Number of factors: add 1 to each power and multiply. For 72, (3 + 1)(2 + 1) = 12. Why: every factor picks a power of 2 from 0 to 3 (four choices) and a power of 3 from 0 to 2 (three choices).
• Sum of factors: for each prime, add 1 and every power of it up to the highest, then multiply the results. For 72, (1 + 2 + 4 + 8)(1 + 3 + 9).

Remainders

• The remainder of a sum or product equals the remainder of the sum or product of the separate remainders. For example, 17 × 23 divided by 5: remainders 2 and 3, and 2 × 3 = 6, which leaves 1. Check: 391 = 5 × 78 + 1.
• If a number leaves remainder r when divided by d, and k is a factor of d, then the remainder on division by k is the remainder of r divided by k.

Two more question types worth knowing

• Trailing zeros of n! (n factorial): count the 5s. For 50!: 50 ÷ 5 = 10, and 50 ÷ 25 = 2, so 12 zeros. Why: each zero needs a 2 × 5 pair, and there are always more 2s than 5s.
• Recurring decimals to fractions: 0.3636… = 36/99 = 4/11. For a decimal like 0.2333…, subtract the non-recurring part: (23 − 2)/90 = 21/90 = 7/30.

Useful sums

• First n natural numbers: n(n + 1)/2. So 1 + 2 + … + 50 = 50 × 51 ÷ 2 = 1,275.
• First n odd numbers: n². First n even numbers: n(n + 1).
• Squares of the first n natural numbers: n(n + 1)(2n + 1)/6. For n = 10, that is 10 × 11 × 21 ÷ 6 = 385.

Worked examples

Example 1. Find the unit digit of 3²³.

The cycle of 3 is 3, 9, 7, 1. 23 ÷ 4 leaves remainder 3. The third term is 7.

Example 2. Find the unit digit of 2⁴¹ × 7³⁰.

41 ÷ 4 leaves 1, so 2⁴¹ ends in 2. 30 ÷ 4 leaves 2, so 7³⁰ ends in 9 (second term of 7, 9, 3, 1).
Unit digit of 2 × 9 = 18 is 8.

Example 3. How many factors does 72 have, and what is their sum?

72 = 2³ × 3². Number of factors: (3 + 1)(2 + 1) = 12.
Sum: (1 + 2 + 4 + 8)(1 + 3 + 9) = 15 × 13 = 195.
Check by listing: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72. Twelve factors, and they add to 195.

Example 4. If the four-digit number 73x4 is divisible by 9, find the digit x.

Digit sum: 7 + 3 + x + 4 = 14 + x. The next multiple of 9 after 14 is 18, so x = 4. (The one after that, 27, would need x = 13, which is not a digit.)
Check: 7,344 ÷ 9 = 816.

Example 5. Is 4,57,391 divisible by 11?

Digits from the right: 1, 9, 3, 7, 5, 4. Sum of 1st, 3rd and 5th: 1 + 3 + 5 = 9. Sum of 2nd, 4th and 6th: 9 + 7 + 4 = 20.
Difference: 20 − 9 = 11, divisible by 11. So yes. Check: 11 × 41,581 = 4,57,391.

Example 6. Find the remainder when 2¹⁰ is divided by 7.

2³ = 8 leaves remainder 1 when divided by 7. 2¹⁰ = (2³)³ × 2, so the remainder is 1 × 1 × 1 × 2 = 2.
Check: 2¹⁰ = 1,024 = 7 × 146 + 2.

Common mistakes

• Using the cycle of the whole number instead of its unit digit. For 23⁷, use the cycle of 3.
• Forgetting that a remainder of 0 means the last term of the cycle.
• Treating 1 as prime.
• Testing divisibility by 12 with 2 and 6. Use 3 and 4, which share no factor.
• Counting trailing zeros with only n ÷ 5, forgetting the extra 5s in 25, 50, 75 and 100.

Practice set

1. Find the unit digit of 7¹⁴.
2. How many factors does 36 have?
3. Is 5,39,152 divisible by 8?
4. What is the smallest number that must be added to 1,000 to make it divisible by 9?
5. Find the sum of the first 15 odd numbers.
6. How many zeros are there at the end of 50!?
7. Write 0.3636… as a fraction in lowest terms.
8. A number leaves remainder 29 when divided by 36. What is the remainder when it is divided by 12?

Answers:

1. 9. 14 ÷ 4 leaves 2; the second term of 7, 9, 3, 1.
2. 9. 36 = 2² × 3², so (2 + 1)(2 + 1) = 9.
3. Yes. The last three digits, 152, give 152 ÷ 8 = 19.
4. 8. The digit sum of 1,000 is 1; adding 8 gives 1,008, whose digit sum is 9.
5. 225. 15² = 225.
6. 12. 50 ÷ 5 = 10 and 50 ÷ 25 = 2; 10 + 2 = 12.
7. 4/11. 36/99, dividing both by 9.
8. 5. 12 is a factor of 36, so divide the remainder: 29 = 12 × 2 + 5.

What to do next

• Write the unit-digit cycles and divisibility rules on one card and revise it daily for a week.
• Solve 30 past NTPC number system questions with a timer, aiming for 40 seconds each.
• Move on to LCM and HCF, which builds directly on prime factors.
• Keep the topic in your mixed practice, as set out in the maths plan.