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

Digit Problems, Floor Values and Number Construction short notes for CAT: 22 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

The Borrow-Chain Method

When zeros borrow, they become $9$ s

This is the main trick behind the CAT PYQ type.

Template

For small

m

with

d

digits:

1 0^{n} - m = n - d digits 99 \dots 9 (1 0^{d} - m)

Example:

1 0^{5} - 123 = 100000 - 123 = 99877

Digit sum

= 9 + 9 + 8 + 7 + 7 = 40

PYQ 1: Digit Sum of $10^{50}+10^{25}-123$

Direct CAT PYQ Pattern

Find the digit sum of $1 0^{50} + 1 0^{25} - 123$

Group it as:

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

Now,

1 0^{25} - 123 = 21 nines 99 \dots 9 9877

because

9877 = 10000 - 123

Digit sum of

1 0^{25} - 123

21 \times 9 + 9 + 8 + 7 + 7 = 189 + 31 = 220

The extra

1 0^{50}

contributes one digit

1

separately.

Final digit sum = 220 + 1 = 221

Answer:

221

Trap: Digital Root Is Not Digit Sum

Do not confuse these two

Digit Sum

Actual sum of all written digits.

9877 \Rightarrow 9 + 8 + 7 + 7 = 31

Digital Root / mod

9

Only tells the remainder behavior.

9877 \equiv 4 (mod 9)

CAT warning: If options are close, mod

9

may help eliminate choices, but it usually cannot give the exact digit sum alone.

PYQ 2: Digit Sum of $(625)^{65}(128)^{36}$

Direct CAT PYQ Pattern

Find the digit sum of $(625)^{65} \times (128)^{36}$

625 = 5^{4}, 128 = 2^{7}

(625)^{65} (128)^{36} = (5^{4})^{65} (2^{7})^{36}

= 5^{260} 2^{252}

Pair

2^{252}

with

5^{252}

5^{260} 2^{252} = 5^{8} (5^{252} 2^{252})

= 5^{8} \cdot 1 0^{252}

Now,

5^{8} = 390625

Multiplying by

1 0^{252}

only adds zeros.

Digit sum = 3 + 9 + 0 + 6 + 2 + 5 = 25

Answer:

25

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 short notes in this unit

chapter

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

Digit Problems, Floor Values and Number Construction short notes for CAT: 22 study cards covering concepts, formulas, shortcuts and exam traps, plus solved pr

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

The Borrow-Chain Method

When zeros borrow, they become 9s

PYQ 1: Digit Sum of $10^{50}+10^{25}-123$

Find the digit sum of 1050+1025−123

Trap: Digital Root Is Not Digit Sum

Do not confuse these two

PYQ 2: Digit Sum of $(625)^{65}(128)^{36}$

Find the digit sum of (625)65×(128)36

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

Step-by-Step Solution

Step-by-Step Solution

More short notes in this unit

A question from this chapter

Login to view the complete short notes

Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

Revision that hits your weak spots

Questions calibrated to the real exam

Notes written for recall, not for volume

One place for everything

Honest progress

Unlock the whole course

When zeros borrow, they become $9$ s

Find the digit sum of $1 0^{50} + 1 0^{25} - 123$

Find the digit sum of $(625)^{65} \times (128)^{36}$