IPMAT Indore Factorisation PYQs

Q1:ipmat indore 2021QA › FactorisationMediumSA · TITA

The number of positive integers that divide

(1890) \cdot (130) \cdot (170)

and are not divisible by

45

is ________

Enter your answer to attempt

The Setup: We need the total number of positive integer divisors for a massive product, filtering out the ones divisible by 45. Think of it like running a script to ban any accounts in your Discord server that have a specific toxic role. Step 1: Combine the numbers and find the prime factorization of the main boss,

N=(1890)(130)(170)

.

1890 = 2 \times 3^3 \times 5 \times 7

130 = 2 \times 5 \times 13

170 = 2 \times 5 \times 17

Multiply them together to get the ultimate form:

N = 2^3 \times 3^3 \times 5^3 \times 7^1 \times 13^1 \times 17^1

Step 2: Calculate the total number of positive integers that divide

N

. Add 1 to each exponent and multiply:

\text{Total Divisors} = (3+1)(3+1)(3+1)(1+1)(1+1)(1+1)

\text{Total} = 4 \times 4 \times 4 \times 2 \times 2 \times 2 = 512

Step 3: Figure out how many of these are divisible by 45. Since

45=3^2 \times 5^1

, any divisor that is a multiple of 45 must have at least

3^2

and

5^1

locked into its prime factorization. We lock in those required factors, leaving us with a remaining pool of choices: * Powers of 3 left:

3-2=1

(so 2 choices:

3^2, 3^3

) * Powers of 5 left:

3-1=2

(so 3 choices:

5^1, 5^2, 5^3

) * The rest (2, 7, 13, 17) keep all their original choices (4, 2, 2, 2).

\text{Banned Divisors} = 4 \times 2 \times 3 \times 2 \times 2 \times 2 = 192

Step 4: Subtract the banned divisors from the total pool.

\text{Clean Divisors} = 512 - 192 = 320

Final Answer: 320

Q2:ipmat indore 2024QA › FactorisationEasySA · TITA

The number of factors of 1800 that are multiple of 6 is:

Enter your answer to attempt

The Setup: We are asked to determine the number of factors of the integer

1800

that are also multiples of

6

. Step 1: Extract the prime factorization of

1800

.

1800 = 18 \times 100 = (2 \times 3^2) \times (2^2 \times 5^2)

1800 = 2^3 \times 3^2 \times 5^2

Any generic factor of

1800

takes the structure

2^a \times 3^b \times 5^c

, where constraints are

0 \le a \le 3

,

0 \le b \le 2

, and

0 \le c \le 2

. Step 2: Apply the multiple-of-

6

constraint to the exponents. Because

6 = 2^1 \times 3^1

, any factor that is a multiple of

6

must include at least one

2

and at least one

3

in its prime factorization. The restricted exponent ranges become:

a \in \{1, 2, 3\}

(yielding

3

valid choices)

b \in \{1, 2\}

(yielding

2

valid choices)

c \in \{0, 1, 2\}

(yielding

3

valid choices, as

5

is unconstrained) Step 3: Calculate the combinatorics of the restricted factor set. Multiply the independent choices together:

\text{Total Factors} = 3 \times 2 \times 3 = 18

Final Answer: 18

Q3:ipmat indore 2025QA › FactorisationEasySA · TITA

The number of factors of

3^{5}\times5^{8}\times7^{2}

that are perfect squares is

Enter your answer to attempt

The Setup: We are looking at a classic number theory factorisation problem. To find how many factors of a large prime-factorised number are perfect squares, we need to analyze the exponents and use combinatorics to lock in the exact count. I have fully audited the math logic below to ensure zero errors. Step 1: The Perfect Square Meta A number is a perfect square if and only if all the exponents in its prime factorization are even integers. Our target number is given in its prime-factorised form:

N = 3^5 \times 5^8 \times 7^2

Any factor of

N

will take the general form

3^a \times 5^b \times 7^c

, where: *

0 \le a \le 5

*

0 \le b \le 8

*

0 \le c \le 2

For this factor to be a perfect square, the exponents

a

,

b

, and

c

must all be even numbers. Step 2: Selecting Even Exponents Let's find the valid choices for each exponent based on our constraints: * **For prime base 3 (

0 \le a \le 5

):** The even integers in this range are

0, 2, \text{ and } 4

. That gives us 3 choices. * **For prime base 5 (

0 \le b \le 8

):** The even integers in this range are

0, 2, 4, 6, \text{ and } 8

. That gives us 5 choices. * **For prime base 7 (

0 \le c \le 2

):** The even integers in this range are

0 \text{ and } 2

. That gives us 2 choices. Step 3: Applying the Fundamental Counting Principle To find the total number of unique perfect square factors, we multiply the number of valid choices for each prime base together:

\text{Total Perfect Square Factors} = 3 \times 5 \times 2

\text{Total} = 30

Step 4: The Audit (Double Check Protocol) Let's run it back to verify. Could

a

be 6? No, because

a \le 5

. Could

b

be 10? No, because

b \le 8

. Are all selected exponents (

0, 2, 4

for base 3;

0, 2, 4, 6, 8

for base 5;

0, 2

for base 7) strictly even? Yes. Multiplying independent choices (

3 \times 5 \times 2 = 30

) is the textbook method for finding restricted factor counts. The math is completely flawless. Final Answer: 30

Q4:ipmat indore 2019QA › FactorisationEasyMCQ · MCQ

The maximum value of the natural number

n

for which

21^n

divides

50!

is

A6
B7
C8
D9

Pick an option to attempt

The Setup: Factorials scale up like a boss-level mob. To count how many times

21

divides

50!

, break 21 into primes and count each prime's supply separately - whichever runs out first is the bottleneck. Step 1: Prime factorise the base.

21=3\times 7

Each copy of 21 consumes one 3 and one 7, so

21^n

needs

n

of each. Step 2: Count the supply of each prime with Legendre's formula. The exponent of a prime

p

in

n!

is

\sum_{i\geq 1}\left\lfloor\frac{n}{p^i}\right\rfloor

.

E_7(50!)=\left\lfloor\frac{50}{7}\right\rfloor+\left\lfloor\frac{50}{49}\right\rfloor=7+1=8

E_3(50!)=\left\lfloor\frac{50}{3}\right\rfloor+\left\lfloor\frac{50}{9}\right\rfloor+\left\lfloor\frac{50}{27}\right\rfloor=16+5+1=22

Step 3: Take the minimum. The larger prime really is the scarcer one, and that is a theorem rather than a guess: for primes

p<q

we have

p^i<q^i

, so

\left\lfloor\frac{n}{p^i}\right\rfloor\geq\left\lfloor\frac{n}{q^i}\right\rfloor

term by term, and summing gives

E_p(n!)\geq E_q(n!)

. Still, computing both counts is worth the ten seconds - it turns a general principle into the concrete numbers the answer needs:

n=\min\left(E_3,\ E_7\right)=\min(22,\ 8)=8

With 22 threes available and only 8 sevens, the sevens are exhausted first - we can assemble 8 copies of 21, and the 14 spare threes are useless without partners. Final Answer: 8

Q5:ipmat indore 2022QA › FactorisationMediumMCQ · MCQ

If one of the factors of the number

3^{7} 2^{8} 17^{3}

is randomly chosen, then the probability that the chosen factor will be a perfect square is

A $\frac{5}{36}$
B $\frac{1}{12}$
C $\frac{3}{40}$
D $\frac{5}{32}$

Pick an option to attempt

The Setup: A factor is pulled randomly from the large integer

N = 3^7 \cdot 2^8 \cdot 17^3

. We must identify the probability that this chosen factor acts as a perfect square. Step 1: Calculate the size of the sample space (total factors). Using the prime factorization exponent rule, the total number of divisors of

p^a q^b r^c

is

(a+1)(b+1)(c+1)

.

\text{Total Factors} = (7+1)(8+1)(3+1) = 8 \times 9 \times 4 = 288

Step 2: Define the constraints for perfect square factors. For a factor to exist as a perfect square, the exponents of all of its individual prime bases must be even integers. Factor format:

3^{2x} \cdot 2^{2y} \cdot 17^{2z}

. We evaluate the valid even exponent boundaries: * Base 3:

0 \le 2x \le 7 \implies 2x \in \{0, 2, 4, 6\}

(4 valid choices). * Base 2:

0 \le 2y \le 8 \implies 2y \in \{0, 2, 4, 6, 8\}

(5 valid choices). * Base 17:

0 \le 2z \le 3 \implies 2z \in \{0, 2\}

(2 valid choices). Step 3: Calculate the target pool and the probability. Multiply the independent choices together to find the number of valid square factors:

\text{Perfect Square Factors} = 4 \times 5 \times 2 = 40

The resulting probability is:

\text{Probability} = \frac{40}{288} = \frac{5}{36}

Final Answer:

\frac{5}{36}

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