Digit Problems, Floor Values and Number Construction Notes for CAT: Concepts, Formulas, Worked Examples & Practice

Digit Problems, Floor Values and Number Construction notes for CAT: 41 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

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 Sum Means the Final Written Digits

Digit sum is not expression-sum

For a number $N$ , the sum of digits means:

N = 50807

, then digit sum

= 5 + 0 + 8 + 0 + 7 = 20

Important: First simplify the number into its actual decimal form. Carries and borrows can completely change the digits.

Example warning:

1000 - 123

is not handled as digits

1, 0, 0, 0, 1, 2, 3

. The actual number is

877

, so the digit sum is

8 + 7 + 7 = 22

Why Powers of 10 Are Digit-Sum Friendly

Powers of 10 create digit positions

A power like $1 0^{n}$ is not scary. It is a place-value marker.

1 0^{3} = 1000

1 0^{6} = 1000000

1 0^{50} = 1 followed by 50 zeros

CAT idea: When zeros are disturbed by subtraction, a long chain of

9

s usually appears.

Digit Problems, Floor Values and Number Construction: Solved Questions with Step-by-Step Explanations (2 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.

More notes in this unit

chapter

Digit Problems, Floor Values and Number Construction Notes for CAT: Concepts, Formulas, Worked Examples & Practice

Digit Problems, Floor Values and Number Construction notes for CAT: 41 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice

A question from this chapter

Question 1

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} ⌋) .

Question 2

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?

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