In the number 6,54,321, what is the place value of the digit 5?

50,000. The digit 5 sits in the ten-thousands place, so its place value is 5 10,000 = 50,000. (Its face value is just 5.)

Which of the following numbers is divisible by 11?

83,721. For divisibility by 11, take the alternating sum of the digits from the right. For 83,721 this is 1 - 2 + 7 - 3 + 8 = 11, a multiple of 11. The other numbers give alternating sums of 6, 3 and 13, none divisible by 11.

What is the unit digit of 7^95?

3. Unit digits of powers of 7 repeat in the cycle 7, 9, 3, 1 (period 4). Since 95 = 4 23 + 3, the unit digit is the third term of the cycle, which is 3.

When a certain number is divided by 5 it leaves a remainder of 3. What is the remainder when twice that number is divided by 5?

1. Write the number as 5k + 3. Twice the number is 10k + 6 = 5(2k + 1) + 1, so dividing by 5 leaves a remainder of 1.

How many whole numbers between 100 and 400 are divisible by both 6 and 8?

12. A number divisible by both 6 and 8 is divisible by their LCM, 24. The multiples of 24 in this range run from 24 5 = 120 to 24 16 = 384, giving 16 - 5 + 1 = 12 numbers.

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Quantitative Aptitude · Numbers & Simplification

Number System — Questions & Answers

The number system is the foundation of quantitative aptitude: place value, types of numbers, divisibility rules, factors and multiples, unit digits and remainders. These solved questions show the shortcuts examiners expect you to use.

5 solved questions
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Key formulas included

Key formulas

Sum of first $n$ natural numbers: $\dfrac{n(n+1)}{2}$
Sum of first $n$ odd numbers: $n^2$ ; sum of first $n$ even numbers: $n(n+1)$
Divisibility by 3 or 9: digit sum divisible by 3 or 9; by 11: alternating digit sum divisible by 11
$(a+b)^2 = a^2 + 2ab + b^2$ , $\;(a-b)^2 = a^2 - 2ab + b^2$ , $\;a^2 - b^2 = (a+b)(a-b)$

5 solved Number System questions

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01
In the number 6,54,321, what is the place value of the digit 5?
4 steps · reveal one at a time
1. Step 1: Understand Place Value
  Place value tells us the value of a digit based on its position in a number. Each position represents a power of $10$ . For example, the digit in the ones place is multiplied by $1$ , the tens place by $10$ , and so on.
2. Step 2: Locate the Digit
  Let's look at the number $6,54,321$ . We need to find the digit $5$ . Counting from the right, the positions are: $1$ (ones), $2$ (tens), $3$ (hundreds), $4$ (thousands), and $5$ (ten-thousands).
3. Step 3: Calculate Place Value
  Since the digit $5$ is in the ten-thousands place, its value is found by multiplying the digit by the value of that place. So, the place value of $5$ is $5 \times 10,000 = 50,000$ .
4. Step 4: The answer
  So the answer is (A) 50,000. This is because the digit 5 occupies the ten-thousands place in the number $6,54,321$ , making its value $50,000$ .
02
Which of the following numbers is divisible by 11?
4 steps · reveal one at a time
1. Step 1: Divisibility Rule
  To check if a number is divisible by 11, we use a special rule. Find the alternating sum of its digits, starting from the rightmost digit. You alternate between adding and subtracting digits.
2. Step 2: Apply to 83,721
  Let's apply this rule to the number 83,721. Starting from the right, we calculate the alternating sum: $1 - 2 + 7 - 3 + 8$ . This sum equals $11$ .
3. Step 3: Check the Result
  Since the alternating sum, $11$ , is a multiple of 11, the number 83,721 is divisible by 11. If the sum were 0 or any other multiple of 11, it would also be divisible.
4. Step 4: The answer
  So the answer is (A) 83,721 because its alternating sum of digits, $1 - 2 + 7 - 3 + 8 = 11$ , is a multiple of 11.
03
What is the unit digit of $7^{95}$ ?
5 steps · reveal one at a time
1. Step 1: Find the Pattern
  Let's look at the unit digits of the first few powers of 7. We have: $7^1 = 7$ , $7^2 = 49 \rightarrow 9$ , $7^3 = 343 \rightarrow 3$ , $7^4 = 2401 \rightarrow 1$ , and $7^5 = 16807 \rightarrow 7$ .
2. Step 2: Identify the Cycle
  The unit digits repeat in a cycle of 7, 9, 3, 1. This cycle has a length of 4. This means every 4th power will have a unit digit of 1, and the pattern restarts.
3. Step 3: Use the Exponent
  To find the unit digit of $7^{95}$ , we need to see where 95 falls in this cycle. We divide the exponent, 95, by the cycle length, 4. This tells us how many full cycles occur and what the remainder is.
4. Step 4: Calculate the Remainder
  Dividing 95 by 4 gives $95 \div 4 = 23$ with a remainder of 3. This means there are 23 full cycles, and then we look at the 3rd digit in the cycle.
5. Step 5: The answer
  So the answer is (D) 3. The third digit in the cycle (7, 9, 3, 1) is 3, which is the unit digit of $7^{95}$ .
04
When a certain number is divided by 5 it leaves a remainder of 3. What is the remainder when twice that number is divided by 5?
4 steps · reveal one at a time
1. Step 1: Represent the number
  We are told that when a certain number is divided by 5, it leaves a remainder of 3. We can write this number in the form $5k + 3$ , where $k$ is an integer.
2. Step 2: Double the number
  The question asks about twice that number. Let's multiply our expression for the number by 2: $2 \times (5k + 3)$ . This simplifies to $10k + 6$ .
3. Step 3: Find the remainder
  We need to find the remainder when $10k + 6$ is divided by 5. We can rewrite $10k + 6$ as $10k + 5 + 1$ . This can be factored as $5(2k + 1) + 1$ .
4. Step 4: The answer
  So the answer is (A) 1. Since $5(2k + 1)$ is a multiple of 5, when $5(2k + 1) + 1$ is divided by 5, the remainder is 1.
05
How many whole numbers between 100 and 400 are divisible by both 6 and 8?
5 steps · reveal one at a time
1. Step 1: Find Common Divisor
  If a number is divisible by both 6 and 8, it must also be divisible by their Least Common Multiple (LCM). Finding the LCM simplifies the problem.
2. Step 2: Calculate the LCM
  The LCM of 6 and 8 is 24. This means we are looking for multiples of 24 that fall within the specified range. Any number divisible by 24 will also be divisible by both 6 and 8.
3. Step 3: Identify the Range
  We need to find whole numbers between 100 and 400. This means we are looking for multiples of 24 that are greater than 100 and less than 400.
4. Step 4: Count the Multiples
  The first multiple of 24 greater than 100 is $24 \times 5 = 120$ . The last multiple of 24 less than 400 is $24 \times 16 = 384$ . To count them, we calculate $16 - 5 + 1 = 12$ .
5. Step 5: The answer
  So the answer is (B) 12. There are 12 whole numbers between 100 and 400 that are divisible by both 6 and 8.

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