Digit Problems, Floor Values and Number Construction Practice Questions for CAT: 73+ Solved Questions with Step-by-Step Solutions

Solve 73+ Digit Problems, Floor Values and Number Construction practice questions for CAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Digit Problems, Floor Values and Number Construction

CAT QA · Number System

Digit Problems, Floor Values and Number Construction

A short but clever chapter where place value, digit behavior, and integer structure do most of the work.

🔢

t1 · Selected Now · 2 PYQs · Moderate frequency

Digit Sums and Powers

Master digit sums of huge numbers by using powers of 10, borrowing, and factor conversion.

🔁

t2 · 4 PYQs · Highest within chapter

Digit Constraints and Reversal Problems

Later you will translate digit conditions into equations and handle reversed numbers.

⌊x⌋

t3 · 2 PYQs · Moderate frequency

Floor Functions and Integer Part Sums

Later you will group values where the integer part remains constant.

By the end of this chapter: you will stop expanding large numbers blindly and start seeing their digit structure directly.

Topic Hero: Digit Sums and Powers

Selected Topic · t1

Digit Sums and Powers

The art of finding digit sums without writing the full number.

Direct CAT PYQs

0.37

Base importance hint

Low

Formula load

Core promise: You will learn two CAT weapons:
1. Borrow-chain digit sums like

1 0^{50} + 1 0^{25} - 123

.
2. Power conversion like

(625)^{65} (128)^{36}

into a small prefix times

1 0^{k}

Digit Problems, Floor Values and Number Construction: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Quantitative Ability MCQ

Let $S (M)$ denote the sum of the decimal digits of a positive integer $M$ . Find

$S (k = 1 \sum 12 ⌊ \frac{k \cdot 1 0 ^{24}}{13} ⌋) .$

A.
198
B.
216
C.
225
D.
234

Correct Answer:

Step-by-Step Solution

Key idea: this is a complementary-pair floor sum. The denominator is $13$ and the numerator runs through $k = 1$ to $12$ , so pair $k$ with $13 - k$ . The huge power of $10$ is then handled by a borrow-chain digit sum.

Step 1: Let $N = 1 0^{24}$ . Since $13$ does not divide $N$ , none of $k N$ is divisible by $13$ for $k = 1, 2, \dots, 12$ .

Step 2: For a fixed $k \in {1, 2, \dots, 6}$ , write

$k N = 13 q + r,$

where $1 \leq r \leq 12$ .

Step 3: Then

$(13 - k) N = 13 N - k N = 13 N - (13 q + r) = 13 (N - q) - r .$

Rewrite the last expression as

$13 (N - q - 1) + (13 - r) .$

Since $13 - r$ is between $1$ and $12$ , we get

$⌊ \frac{( 13 - k ) N}{13} ⌋ = N - q - 1.$

Step 4: Therefore each complementary pair gives

$⌊ \frac{k N}{13} ⌋ + ⌊ \frac{( 13 - k ) N}{13} ⌋ = q + (N - q - 1) = N - 1.$

Step 5: There are $6$ pairs: $(1, 12), (2, 11), \dots, (6, 7)$ . Hence the whole sum is

$6 (N - 1) = 6 (1 0^{24} - 1) = 6 \cdot 1 0^{24} - 6.$

Step 6: Find the digit sum of $6 \cdot 1 0^{24} - 6$ . The number $6 \cdot 1 0^{24}$ is a $6$ followed by $24$ zeroes. Subtracting $6$ creates a borrow chain:

$6 \cdot 1 0^{24} - 6 = 5 23 nines 99 \dots 9 4.$

Its digit sum is

$5 + 23 \cdot 9 + 4 = 5 + 207 + 4 = 216.$

Common trap: replacing the sum of floors by the floor of the sum. Floors do not distribute over addition; the fractional parts are exactly what make the pairing work.

Answer: 216

Question 2 · Quantitative Ability MCQ

For a 4-digit number $\overline{ab c d}$ , the sum of the thousands, hundreds and tens digits is $18$ , the sum of the hundreds, tens and units digits is $15$ , and the tens digit is $2$ more than the units digit.

If the number is odd, what is the highest possible value of the number?

A.
9186
B.
8375
C.
6753
D.
5173

Correct Answer:

Step-by-Step Solution

Key idea: This is a maximum digit-construction question with overlapping digit sums. The trigger is "highest possible" plus sums of place-value groups.

Step 1: Let the number be $\overline{ab c d}$ , where $a$ is thousands, $b$ hundreds, $c$ tens, and $d$ units.

Step 2: Translate the conditions:

$a + b + c = 18$

$b + c + d = 15$

$c = d + 2$

Step 3: Subtract the second equation from the first:

$(a + b + c) - (b + c + d) = 18 - 15$

$a - d = 3$

$a = d + 3.$

Step 4: Use $c = d + 2$ in $b + c + d = 15$ :

$b + (d + 2) + d = 15$

$b + 2 d + 2 = 15$

$b = 13 - 2 d .$

Step 5: Apply digit bounds. Since $b \geq 0$ ,

$13 - 2 d \geq 0 \Rightarrow d \leq 6.$

Since $a = d + 3 \leq 9$ ,

$d \leq 6.$

Also $b \leq 9$ gives

$13 - 2 d \leq 9 \Rightarrow d \geq 2.$

So $d \in {2, 3, 4, 5, 6}$ .

Step 6: Use the added condition: the number is odd, so $d$ must be odd. Hence

$d \in {3, 5} .$

Step 7: Check both cases.

If $d = 5$ , then $a = 8$ , $c = 7$ , and $b = 13 - 10 = 3$ , giving $8375$ .

If $d = 3$ , then $a = 6$ , $c = 5$ , and $b = 13 - 6 = 7$ , giving $6753$ .

The higher number is $8375$ .

Answer: $8375$ .

Trap: Ignoring the odd-number condition gives $d = 6$ and $9186$ , but that number is even and therefore invalid.

Question 3 · Quantitative Ability MCQ

The sum of the digits of the number

$(125)^{10} \times (16)^{10}$

is:

A.
7
B.
9
C.
10
D.
16

Correct Answer:

Step-by-Step Solution

Key idea: This is a power-product digit-sum question. The trigger is that the bases $125$ and $16$ are powers of $5$ and $2$ , so they can be paired into powers of $10$ .

Step 1: Rewrite the bases:

$125 = 5^{3}, 16 = 2^{4} .$

Step 2: Apply the exponents:

$(125)^{10} = (5^{3})^{10} = 5^{30},$

$(16)^{10} = (2^{4})^{10} = 2^{40} .$

Step 3: Multiply:

$5^{30} \times 2^{40} .$

Pair $30$ fives with $30$ twos:

$2^{30} \times 5^{30} = 1 0^{30} .$

This leaves

$2^{40 - 30} = 2^{10} .$

Step 4: Compute the leftover factor:

$2^{10} = 1024.$

So the full number is $1024$ followed by $30$ zeros.

Step 5: Zeros do not affect the digit sum. Thus the digit sum is

$1 + 0 + 2 + 4 = 7.$

Answer: $7$ .

Trap: The leftover factor is a power of $2$ , not a power of $5$ , because there are more twos than fives.

Question 4 · Quantitative Ability MSQ

For a four-digit positive integer $N = \overline{ab c d}$ , where digits may repeat, define

$F (N) = ⌊ \frac{N}{10} ⌋ + ⌊ \frac{N}{100} ⌋ + ⌊ \frac{N}{1000} ⌋$

and let $S (N)$ be the sum of its digits. Consider all four-digit numbers $N$ satisfying

$F (N) = 13 S (N) .$

Which of the following statements is/are true?

A.
The thousands digit of every such number is either 1 or 2.
B.
The largest such number is 2794.
C.
The units digit of every such number is even.
D.
There are exactly six such numbers.

Correct Answer:

["A","C"]

Step-by-Step Solution

Key idea: floors by powers of $10$ truncate digits. The trigger is $F (N)$ built from $⌊ N /10 ⌋$ , $⌊ N /100 ⌋$ and $⌊ N /1000 ⌋$ . Convert those floors into place-value expressions, then compare with the digit sum.

Step 1: Let $N = 1000 a + 100 b + 10 c + d$ , where $a \in {1, \dots, 9}$ and $b, c, d \in {0, \dots, 9}$ .

Step 2: Compute each floor:

$⌊ \frac{N}{10} ⌋ = 100 a + 10 b + c,$

$⌊ \frac{N}{100} ⌋ = 10 a + b,$

$⌊ \frac{N}{1000} ⌋ = a .$

Therefore

$F (N) = 111 a + 11 b + c .$

Step 3: The digit sum is

$S (N) = a + b + c + d .$

The condition $F (N) = 13 S (N)$ becomes

$111 a + 11 b + c = 13 (a + b + c + d) .$

Simplify:

$98 a = 2 b + 12 c + 13 d .$

Step 4: Test statement A. The right-hand side is at most

$2 (9) + 12 (9) + 13 (9) = 18 + 108 + 117 = 243.$

Hence $98 a \leq 243$ , so $a \leq 2$ . Since $N$ is four-digit, $a \geq 1$ . Thus $a$ is $1$ or $2$ . Statement A is true.

Step 5: Test statement C. The left-hand side $98 a$ is even. The terms $2 b$ and $12 c$ are even. Therefore $13 d$ must be even, so $d$ is even. Statement C is true.

Step 6: Test statements B and D by solving the digit equation. Since $d$ is even, check $d = 0, 2, 4, 6, 8$ .

For $a = 1$ :

$2 b + 12 c + 13 d = 98.$

This gives the solutions

$1180, 1770, 1062, 1562, 1534, 1416.$

For $a = 2$ :

$2 b + 12 c + 13 d = 196.$

This gives the solutions

$2596, 2478.$

So there are $8$ such numbers, not $6$ . Statement D is false.

The largest is $2596$ , not $2794$ . Also, directly checking $2794$ :

$F (2794) = 279 + 27 + 2 = 308,$

while

$13 S (2794) = 13 (2 + 7 + 9 + 4) = 13 (22) = 286.$

So $2794$ is not a solution. Statement B is false.

Common trap: because $d$ does not appear in $F (N)$ , students may think $d$ is irrelevant. It is still present in $S (N)$ , so it strongly affects the equation.

Answer: A and C

Question 5 · Quantitative Ability NAT

How many three-digit numbers increase by $297$ when their digits are reversed, and also have digit sum $12$ ?

Correct Answer:

Step-by-Step Solution

Key idea: This is a reversal-equation question with an added digit-sum constraint. The trigger is "increase by ... when the digits are reversed".

Step 1: Let the number be $\overline{ab c} = 100 a + 10 b + c$ , where $a \neq = 0$ .

Step 2: The reversed number is $\overline{c ba} = 100 c + 10 b + a$ .

Step 3: "Increases by $297$ " means

$100 c + 10 b + a = 100 a + 10 b + c + 297.$

Step 4: Simplify. The $10 b$ terms cancel:

$100 c + a = 100 a + c + 297$

$99 c - 99 a = 297$

$99 (c - a) = 297$

$c - a = 3.$

Step 5: Use the digit-sum condition:

$a + b + c = 12.$

Since $c = a + 3$ ,

$a + b + (a + 3) = 12$

$2 a + b + 3 = 12$

$b = 9 - 2 a .$

Step 6: Apply digit bounds.

Since $a$ is the hundreds digit, $a \geq 1$ .

Since $c = a + 3 \leq 9$ , $a \leq 6$ .

Since $b = 9 - 2 a \geq 0$ , $a \leq 4$ .

Therefore

$a \in {1, 2, 3, 4} .$

Step 7: Count the valid values of $a$ . There are $4$ values, and each gives exactly one $b$ and $c$ :

$a = 1, 2, 3, 4.$

Answer: $4$ .

Trap: Without the digit-sum condition, $b$ would be free and the count would be much larger. Here the digit sum fixes $b$ .

Digit Problems, Floor Values and Number Construction Practice Questions for CAT: 73+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Digit Problems, Floor Values and Number Construction

Digit Problems, Floor Values and Number Construction

Digit Sums and Powers

Digit Constraints and Reversal Problems

Floor Functions and Integer Part Sums

Topic Hero: Digit Sums and Powers

Digit Sums and Powers

Digit Problems, Floor Values and Number Construction: Solved Questions with Step-by-Step Explanations (5 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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