IPMAT Indore Divisibility Rules PYQs

Q1:ipmat indore 2019QA › Divisibility RulesHardSA · TITA

You have been asked to select a positive integer N which is less than 1000, such that it is either a multiple of 4, or a multiple of 6, or an odd multiple of 9. The number of such numbers is

Enter your answer to attempt

The Setup: A three-set counting problem that would be messy through full Inclusion-Exclusion. The elegant route is to notice the sets split by parity, which makes the awkward intersections vanish outright. Step 1: Handle the even sets. Let

A

be the multiples of 4 and

B

the multiples of 6, over

1\leq N\leq 999

:

|A|=\left\lfloor\frac{999}{4}\right\rfloor=249, \qquad |B|=\left\lfloor\frac{999}{6}\right\rfloor=166

Their overlap is the multiples of

\text{LCM}(4,6)=12

- the LCM, not the product 24, since 4 and 6 share a factor of 2:

|A\cap B|=\left\lfloor\frac{999}{12}\right\rfloor=83

Step 2: Combine them.

|A\cup B|=249+166-83=332

Step 3: Count the odd set. Let

C

be the odd multiples of 9, i.e.

9k

with

k

odd and

9k\leq 999

, so

k\leq 111

. The admissible

k

are

1,3,5,\ldots,111

:

|C|=\frac{111+1}{2}=56

Step 4: Exploit disjointness. Every multiple of 4 is even and every multiple of 6 is even, so

A\cup B

contains only even numbers. Every element of

C

is

9\times(\text{odd})

, hence odd. A number cannot be both even and odd, so

(A\cup B)\cap C=\varnothing

and the three-set Inclusion-Exclusion collapses to plain addition:

\text{Total}=332+56=388

Step 5: A note on why the 'odd' restriction costs nothing. An *even* multiple of 9 is a multiple of 18, and since

6\mid 18

every such number already sits in

B

. So the answer would be unchanged had the question said *a multiple of 9* instead - the restriction to odd multiples is what makes the sets disjoint and the counting clean, not what makes the total smaller. Final Answer: 388

Q2:ipmat indore 2022QA › Divisibility RulesEasyMCQ · MCQ

If the five-digit number abcde is divisible by 6, then which of the following numbers is not necessarily divisible by 6?

Aedcba
Beee
Cbbadcacede
Dcdbae

Pick an option to attempt

The Setup: We are given a

5

-digit number

abcde

that is fully divisible by

6

. We need to evaluate four altered configurations to see which is *not necessarily* divisible by

6

. Step 1: Establish the divisibility rule for 6. A number is divisible by

6

if and only if it is mathematically divisible by both

2

(its unit digit is even) and

3

(the sum of its digits is a multiple of

3

). For

abcde

: 1.

a+b+c+d+e \equiv 0 \pmod 3

2.

e

is an even digit. Step 2: Test the proposed configurations. * **Option 2 (

eee

):** Sum is

3e

(divisible by

3

). Ends in

e

(even). Always divisible by

6

. * **Option 3 (

bbadcacede

):** Sum is

2a + 2b + 2c + 2d + 2e = 2(\text{sum})

. Since the original sum is divisible by

3

,

2 \times \text{sum}

is also divisible by

3

. Ends in

e

(even). Always divisible by

6

. * **Option 4 (

cdbae

):** Sum is the same as the original (divisible by

3

). Ends in

e

(even). Always divisible by

6

. * **Option 1 (

edcba

):** The sum of the digits remains the same (divisible by

3

). However, the unit digit is now

a

. We only know

e

is even;

a

can be either odd or even. Therefore,

edcba

is not guaranteed to be even, meaning it is not necessarily divisible by

6

. Final Answer: edcba

Q3:ipmat indore 2019QA › Divisibility RulesEasyMCQ · MCQ

Placing which of the following two digits at the right end of

4530

makes the resultant six digit number divisible by

6, 7

and

9

:

A96
B78
C42
D54

Pick an option to attempt

The Setup: Rather than running three separate divisibility side quests, fuse them into one by taking the LCM - a number divisible by 6, 7 and 9 is exactly a number divisible by their least common multiple. Step 1: Find the master divisor. Factorising:

6=2\times 3

,

7=7

,

9=3^2

. The LCM takes the highest power of each prime:

\text{LCM}(6,7,9)=2\times 3^2\times 7=126

Note it is not

6\times 7\times 9=378

- 6 and 9 share a factor of 3, so the product over-counts. Step 2: Build the number. Appending two digits to 4530 gives

453000+x

, where

x

runs from 00 to 99. Step 3: Find the gap. Divide the base by 126:

453000=126\times 3595+30

The base overshoots a multiple of 126 by 30, so to reach the next one we must add:

126-30=96

That gives

453096=126\times 3596

. And since the following candidate would be

96+126=222

- far beyond two digits - 96 is the only possible answer in range, which is worth noting rather than stopping at the first hit. Step 4: Confirm the distractors fail. All three wrong options clear the divisibility by 6, so that test alone separates nothing - 78 and 42 both fall at the 7, and 54 falls at the 9:

Appended	Number	by 6	by 7	by 9
96	453096	yes	yes	yes
78	453078	yes	no	yes
42	453042	yes	no	yes
54	453054	yes	yes	no

Final Answer: 96

Q4:ipmat indore 2025QA › Divisibility RulesMediumMCQ · MCQ

Which of the following numbers is divisible by

3^{10}+2

?

A $3^{20}+4$
B $3^{30}+2$
C $3^{20}+8$
D $3^{30}+8$

Pick an option to attempt

The Setup: We can use algebraic identities and the Factor Theorem to solve this quickly without dealing with massive exponents. By substituting a variable for the base term, we can test the options as simple polynomials. Step 1: Define the variable. Let

x = 3^{10}

. The divisor given in the question is

x + 2

. We are looking for an option that is completely divisible by

(x + 2)

. According to the Factor Theorem, if a polynomial

P(x)

is divisible by

(x + 2)

, then

P(-2)

must equal

0

. Step 2: Translate the options into polynomials. * (A)

3^{20} + 4 = (3^{10})^2 + 4 = x^2 + 4

* (B)

3^{30} + 2 = (3^{10})^3 + 2 = x^3 + 2

* (C)

3^{20} + 8 = (3^{10})^2 + 8 = x^2 + 8

* (D)

3^{30} + 8 = (3^{10})^3 + 8 = x^3 + 8

Step 3: Apply the Factor Theorem (Test

x = -2

). Let's plug

x = -2

into our options to see which one zeros out: * (A)

(-2)^2 + 4 = 4 + 4 = 8 \neq 0

* (B)

(-2)^3 + 2 = -8 + 2 = -6 \neq 0

* (C)

(-2)^2 + 8 = 4 + 8 = 12 \neq 0

* (D)

(-2)^3 + 8 = -8 + 8 = 0

Since option (D) evaluates exactly to

0

, it is perfectly divisible by

(x + 2)

. Step 4: The Algebraic Proof (Double Check Protocol). We can confirm this using the sum of cubes identity:

a^3 + b^3 = (a + b)(a^2 - ab + b^2)

. Let

a = 3^{10}

and

b = 2

.

a^3 + b^3 = (3^{10})^3 + 2^3 = 3^{30} + 8

3^{30} + 8 = (3^{10} + 2)((3^{10})^2 - 2(3^{10}) + 2^2)

Since

(3^{10} + 2)

is explicitly pulled out as a factor,

3^{30} + 8

is strictly divisible by it. The logic is fully locked in. Final Answer:

3^{30}+8

Q5:ipmat indore 2023QA › Divisibility RulesEasyMCQ · MCQ

If a three-digit number is chosen at random, what is the probability that it is divisible neither by 3 nor by 4?

A $\frac{1}{4}$
B $\frac{1}{2}$
C $\frac{2}{3}$
D $\frac{1}{3}$

Pick an option to attempt

The Setup: We are asked for the probability that a randomly chosen

3

-digit number is divisible by *neither*

3

nor

4

. Step 1: Calculate the size of the sample space. The 3-digit numbers run inclusively from

100

to

999

.

\text{Total 3-digit numbers} = 999 - 100 + 1 = 900

Step 2: Use the Inclusion-Exclusion principle to find the complement. We first find the number of integers divisible by

3

OR

4

:

|A \cup B| = |A| + |B| - |A \cap B|

. * Divisible by 3: The sequence is

102, 105 \dots 999

.

\text{Count} = \frac{999 - 102}{3} + 1 = 297/3 + 1 = 300

* Divisible by 4: The sequence is

100, 104 \dots 996

.

\text{Count} = \frac{996 - 100}{4} + 1 = 896/4 + 1 = 225

* Divisible by 12 (Intersection): The sequence is

108, 120 \dots 996

.

\text{Count} = \frac{996 - 108}{12} + 1 = 888/12 + 1 = 75

\text{Divisible by 3 or 4} = 300 + 225 - 75 = 450

Step 3: Calculate the target probability. The target numbers are those *not* divisible by

3

or

4

.

\text{Target count} = \text{Total} - (\text{Divisible by 3 or 4}) = 900 - 450 = 450

\text{Probability} = \frac{450}{900} = \frac{1}{2}

Final Answer:

\frac{1}{2}

Q6:ipmat indore 2022QA › Divisibility RulesMediumMCQ · MCQ

The number of four-digit integers which are greater than 1000 and divisible by both 2 and 3, but not by 5, is

A1333
B1666
C1200
D1500

Pick an option to attempt

The Setup: We are asked to count the quantity of four-digit integers logically bounded by specific divisibility rules (divisible by

2

and

3

, but strictly not by

5

). Step 1: Establish the numerical boundaries. The domain represents four-digit integers greater than

1000

. The inclusive range is mathematically defined as

1001 \le N \le 9999

. Step 2: Apply the primary inclusion divisibility rule. A number mathematically divisible by both

2

and

3

is inherently divisible by their LCM, which is

6

. We calculate the count of numbers divisible by

6

within the domain. * Smallest multiple of

6

:

1002

(

6 \times 167

). * Largest multiple of

6

:

9996

(

6 \times 1666

). Count

= \frac{\text{Largest} - \text{Smallest}}{6} + 1 = \frac{9996 - 1002}{6} + 1 = 1499 + 1 = 1500

. Step 3: Apply the exclusion divisibility rule. We must exclude numbers within this pool that are also divisible by

5

. A number divisible by both

6

and

5

is divisible by their LCM, which is

30

. We calculate the count of numbers divisible by

30

within the domain to subtract them from our pool. * Smallest multiple of

30

:

1020

(

30 \times 34

). * Largest multiple of

30

:

9990

(

30 \times 333

). Count

= \frac{9990 - 1020}{30} + 1 = \frac{8970}{30} + 1 = 299 + 1 = 300

. Step 4: Calculate the final target count. Subtract the excluded intersecting pool from the primary included pool:

\text{Final Count} = 1500 - 300 = 1200

Final Answer: 1200

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