Digit Problems, Floor Values and Number Construction Previous Year Questions (PYQs) for CAT: 6+ Solved Questions with Step-by-Step Solutions

Solve 6+ Digit Problems, Floor Values and Number Construction previous year 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 NAT

In a 3-digit number N, the digits are non-zero and distinct such that none of the digits is a perfect square, and only one of the digits is a prime number. Then, the number of factors of the minimum possible value of N is

Correct Answer:

Step-by-Step Solution

Pattern: this is a digit-property filtering question combined with minimize-then-compute — recognisable because the digits must satisfy several yes/no filters (non-zero, distinct, not a perfect square, exactly one prime), and then we must build the smallest number and analyse a property of it (its factor count).

Step 1 — List the allowed digit pool.

Digits are from $1$ to $9$ (non-zero). Perfect squares among these are ${1, 4, 9}$ — these are banned since "none of the digits is a perfect square."

Remaining allowed digits: ${2, 3, 5, 6, 7, 8}$ .

Step 2 — Split this pool by primality.

Prime digits in the pool: ${2, 3, 5, 7}$ .

Non-prime digits in the pool: ${6, 8}$ (only two digits!).

Step 3 — Use the "only one prime digit" condition.

$N$ has 3 distinct digits, and exactly one of them is prime — so exactly two digits must be non-prime. But the non-prime pool has only two members: $6$ and $8$ . So both non-prime digits are forced: they must be exactly ${6, 8}$ .

Step 4 — Choose the prime digit to minimize N.

The third digit is any one prime from ${2, 3, 5, 7}$ . To make $N$ as small as possible, pick the smallest available prime digit: $p = 2$ .

So the digit set is ${2, 6, 8}$ .

Step 5 — Arrange the digits to minimize the number.

With all three digits non-zero, the smallest 3-digit number from a given digit set is formed by placing digits in ascending order: hundreds < tens < units.

$N = 268$

Step 6 — Verify all conditions.

Digits $2, 6, 8$ : non-zero ✓, distinct ✓, none is a perfect square (1,4,9 excluded) ✓, exactly one is prime (2 is prime; 6,8 are not) ✓.

Step 7 — Find the number of factors of 268.

$268 = 4 \times 67 = 2^{2} \times 6 7^{1}$ (67 is prime).

Number of factors $= (2 + 1) (1 + 1) = 3 \times 2 = 6$ .

Trap: a student who doesn't notice that the non-prime pool has only two elements might try to freely choose 2 non-prime digits from a larger set, missing that ${6, 8}$ is forced — or might forget that a perfect-square digit (like 1, which looks "small" and tempting for minimizing) is actually banned.

Question 2 · Quantitative Ability NAT

How many three-digit numbers are greater than 100 and increase by 198 when the three digits are arranged in the reverse order?

Correct Answer:

Step-by-Step Solution

Pattern: this is a digit-reversal equation question — recognisable because a number and its digit-reversal are compared through a fixed numeric difference ("increases by 198").

Step 1 — Name the digits.

Let the 3-digit number be $N = \overline{ab c} = 100 a + 10 b + c$ , where $a \in {1, \dots, 9}$ (leading digit, cannot be 0) and $b, c \in {0, \dots, 9}$ .

Step 2 — Write the reversed number.

Reversing the digits gives $M = \overline{c ba} = 100 c + 10 b + a$ .

Step 3 — Translate "increases by 198" into an equation.

The reversed number is 198 more than the original: $M = N + 198$ .

$100 c + 10 b + a = 100 a + 10 b + c + 198$

Step 4 — Simplify.

The $10 b$ terms cancel from both sides:

$100 c + a = 100 a + c + 198$

$99 c - 99 a = 198$

$c - a = 2$

Step 5 — Count the valid digit combinations.

$a$ ranges over natural leading-digit values $1$ to $9$ , but $c = a + 2$ must also stay within $0$ – $9$ , so $a$ can be $1, 2, 3, 4, 5, 6, 7$ (7 values), giving $c = 3, 4, \dots, 9$ .

$b$ has no constraint from the equation, so it can be any of $0$ – $9$ (10 values).

Step 6 — Multiply.

Total numbers $= 7 \times 10 = 70$ .

Every one of these automatically satisfies $N > 100$ , since $a \geq 1$ and $c \geq 3 > 0$ already make $N \geq 103$ .

Common trap: forgetting that $b$ is completely free and only counting the 7 values of $a$ (giving a wrong answer of 7), or making a sign slip and solving $a - c = 2$ instead of $c - a = 2$ , which produces no valid solutions since $a \geq 1, c \geq 0$ would force $a \geq 3$ with c going negative.

Question 3 · Quantitative Ability MCQ

For a 4-digit number (greater than 1000), sum of the digits in the thousands, hundreds, and tens places is 15. Sum of the digits in the hundreds, tens, and units places is 16. Also, the digit in the tens place is 6 more than the digit in the units place. The difference between the largest and smallest possible value of the number is

A.
811
B.
3289
C.
735
D.
4078

Correct Answer:

Step-by-Step Solution

Pattern: this is the "largest minus smallest" variant of digit-construction problems — same equation-writing method as before, but now you must find both extreme values and subtract.

Step 1 — Set up variables.

Let $N = \overline{ab c d}$ , digits $a, b, c, d$ (thousands, hundreds, tens, units), $a \geq 1$ .

Step 2 — Translate conditions.

$a + b + c = 15$ ... (i)
$b + c + d = 16$ ... (ii)
$c = d + 6$ ... (iii)

Step 3 — Link $a$ and $d$ .

(i) $-$ (ii): $a - d = 15 - 16 = - 1 \Rightarrow a = d - 1$ .

Step 4 — Apply digit bounds.

$c = d + 6 \leq 9 \Rightarrow d \leq 3$ . $a = d - 1 \geq 1 \Rightarrow d \geq 2$ . So $d \in {2, 3}$ .

Step 5 — Work out each case fully (every digit is forced once $d$ is chosen).

$d = 2$ : $a = 1$ , $c = 8$ , and from (ii) $b = 16 - c - d = 16 - 8 - 2 = 6$ . Number $= 1682$ .
$d = 3$ : $a = 2$ , $c = 9$ , and $b = 16 - 9 - 3 = 4$ . Number $= 2493$ .

Check (i) for both: $1 + 6 + 8 = 15$ ✓ and $2 + 4 + 9 = 15$ ✓.

Step 6 — Identify largest and smallest.

Since $a$ is the leading digit, $d = 3$ (with $a = 2$ ) gives the larger number, $2493$ ; $d = 2$ (with $a = 1$ ) gives the smaller number, $1682$ . There is only one valid number for each value of $d$ , so these are exactly the largest and smallest possible values.

Step 7 — Subtract.

$2493 - 1682 = 811$ .

Answer: option A, 811.

Question 4 · Quantitative Ability NAT

The sum of digits of the number $(625)^{65} \times (128)^{36}$ is

Correct Answer:

Step-by-Step Solution

Step 1: Convert both bases to prime powers.

$625 = 5^{4} \Rightarrow 62 5^{65} = 5^{260}$

$128 = 2^{7} \Rightarrow 12 8^{36} = 2^{252}$

Step 2: Multiply.

$5^{260} \times 2^{252} = (2^{252} \times 5^{252}) \times 5^{8} = 1 0^{252} \times 5^{8}$

Step 3: Compute the small leftover factor.

$5^{8} = 390625$

Step 4: So the number is $390625$ followed by $252$ zeros — the zeros contribute nothing to the digit sum.

Step 5: Digit sum

$3 + 9 + 0 + 6 + 2 + 5 = 25$

Question 5 · Quantitative Ability MCQ

The sum of all the digits of the number $1 0^{50} + 1 0^{25} - 123$ , is

A.
212
B.
221
C.
324
D.
255

Correct Answer:

Step-by-Step Solution

Step 1: Write $1 0^{50} + 1 0^{25} - 123 = 1 0^{50} + (1 0^{25} - 123)$ , since $1 0^{25} - 123$ is positive and much smaller than $1 0^{50}$ , addition doesn't cause interaction between the two parts.

Step 2: Compute $1 0^{25} - 123$ . In general, $1 0^{n} - ab c$ (a 3-digit subtractor) equals $(n - 3)$ nines followed by $(1000 - ab c)$ :

$1 0^{25} - 123 = 22 nines 9 \dots 9 877$

(Check with small case: $1 0^{4} - 123 = 9877$ — one nine then 877; here $n = 25$ gives $25 - 3 = 22$ nines.)

Step 3: Add $1 0^{50}$ . Since $1 0^{25} - 123$ is a 25-digit number, and $1 0^{50}$ is $1$ followed by $50$ zeros, adding them just places the 25-digit number in the last 25 digit positions:

$1 0^{50} + (1 0^{25} - 123) = 1 25 zeros 0 \dots 0 22 nines 9 \dots 9 877$

Step 4: Digit sum

$= 1 + 0 + (22 \times 9) + (8 + 7 + 7) = 1 + 198 + 22 = 221$

Digit Problems, Floor Values and Number Construction Previous Year Questions (PYQs) for CAT: 6+ 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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