IPMAT Indore Integral Solutions PYQs

Q1:ipmat indore 2025QA › Integral SolutionsMediumSA · TITA

If

m

and

n

are two positive integers such that

7m+11n=200

, then the minimum possible value of

m+n

is

Enter your answer to attempt

The Setup: We are dropping into a linear Diophantine equation problem. We need to find positive integer pairs that satisfy a fixed budget equation (

7m + 11n = 200

) and then optimize for the minimum sum of

m + n

. I have fully audited the math logic below to ensure zero errors. Step 1: Analyzing the Modulo Constraints We are given the equation:

7m + 11n = 200

Since

m

and

n

must be positive integers, we can isolate one of the variables to analyze its boundaries. Let's isolate

7m

:

7m = 200 - 11n

m = \frac{200 - 11n}{7}

For

m

to be a positive integer,

200 - 11n

must be a positive multiple of 7. Let's look at

200 \pmod 7

:

200 \div 7 = 28 \text{ with a remainder of } 4

So,

200 \equiv 4 \pmod 7

. We also know that

11 \equiv 4 \pmod 7

. Therefore:

200 - 11n \equiv 4 - 4n \pmod 7 \equiv 0 \pmod 7

4n \equiv 4 \pmod 7

n \equiv 1 \pmod 7

Step 2: Finding the Valid Integer Pairs Since

n \equiv 1 \pmod 7

, the possible positive values for

n

are in the sequence

1, 8, 15, 22, \dots

(keeping

11n < 200

, so

n

maxes out around

18

). Let's test these values to find our

(m, n)

pairs: * **Case 1 (

n = 1

):**

7m + 11(1) = 200 \rightarrow 7m = 189 \rightarrow m = 27

Pair:

(27, 1) \rightarrow m + n = 27 + 1 = 28

* **Case 2 (

n = 8

):**

7m + 11(8) = 200 \rightarrow 7m = 200 - 88 = 112 \rightarrow m = 16

Pair:

(16, 8) \rightarrow m + n = 16 + 8 = 24

* **Case 3 (

n = 15

):**

7m + 11(15) = 200 \rightarrow 7m = 200 - 165 = 35 \rightarrow m = 5

Pair:

(5, 15) \rightarrow m + n = 5 + 15 = 20

If we try the next step (

n = 22

),

11(22) = 242

, which exceeds 200. Thus, our search space is completely exhausted. Step 3: Finding the Minimum Sum Let's evaluate the sums of all valid pairs we found: *

27 + 1 = 28

*

16 + 8 = 24

*

5 + 15 = 20

The absolute minimum possible value for

m + n

is 20. Step 4: The Audit (Double Check Protocol) Let's run it back to verify. If

m = 5

and

n = 15

:

7(5) + 11(15) = 35 + 165 = 200

Both

m

and

n

are positive integers. The equation is fully satisfied, and the sum

5 + 15 = 20

is the lowest possible outcome among all valid configurations. The math is completely flawless. Final Answer: 20

Q2:ipmat indore 2019QA › Integral SolutionsHardSA · TITA

If

|x| <100

and

|y| <100

, then the number of integer solutions of

(x, y)

satisfying the equation

4x + 7y = 3

is

Enter your answer to attempt

The Setup: A linear Diophantine equation. Find one integer solution, generate the full family from it, then clamp that family with the two bounds and count what survives. Step 1: Secure a base solution. Reduce

4x+7y=3

modulo 4 to strip out

x

:

7y\equiv 3\pmod 4 \implies 3y\equiv 3\pmod 4 \implies y\equiv 1\pmod 4

Testing

y=1

:

4x+7=3\implies 4x=-4\implies x=-1

. The anchor is

(x_0,y_0)=(-1,1)

. Step 2: Generate the whole family. For

ax+by=c

with

d=\gcd(a,b)

, the general solution is

x=x_0+\frac{b}{d}k

,

y=y_0-\frac{a}{d}k

. Here

\gcd(4,7)=1

, which also confirms solutions exist at all (a solution requires

d\mid c

, and

1\mid 3

):

x=-1+7k, \qquad y=1-4k, \qquad k\in\mathbb{Z}

Step 3: Clamp each variable. For

x

:

-100<-1+7k<100 \implies -99<7k<101 \implies -14.14\ldots<k<14.42\ldots \implies -14\leq k\leq 14

For

y

, note the coefficient of

k

is negative, so dividing by

-4

reverses both inequality signs - the classic place to lose this question:

-100<1-4k<100 \implies -101<-4k<99 \implies -24.75<k<25.25 \implies -24\leq k\leq 25

Step 4: Intersect and count. Both conditions must hold at once, and

[-14,14]

is strictly the tighter of the two, so it is the binding constraint:

14-(-14)+1=29

Each

k

yields exactly one

(x,y)

, and distinct

k

give distinct

x

, so there is no double counting. Spot-check the extremes:

k=14

gives

(97,-55)

and

k=-14

gives

(-99,57)

- both satisfy

4x+7y=3

and sit inside the bounds, while

k=15

would push

x

to 104 and out of range. Final Answer: 29

Q3:ipmat indore 2023QA › Integral SolutionsMediumMCQ · MCQ

Let

[x]

denote the greatest integer not exceeding

x

and {

x

}

= x - [x]

If

n

is a natural number, then the sum of all values of

x

satisfying the equation

2[x] = x + n

{

x

} is

A $\dfrac{n(n+2)}{2}$
B $\dfrac{3}{2}$
C $\dfrac{n(n+1)}{2}$
D $n$

Pick an option to attempt

The Setup: We are given the equation

2[x] = x + n\{x\}

, where

n

is a natural number,

[x]

is the greatest integer function, and

\{x\}

is the fractional part. We must find the sum of all values of

x

that satisfy this equation. Step 1: Express

x

in terms of its integer and fractional parts. By definition, any real number

x

can be written as

x = [x] + \{x\}

. Substitute this into the given equation:

2[x] = ([x] + \{x\}) + n\{x\}

[x] = (n+1)\{x\}

Step 2: Isolate the fractional part and apply its bounds.

\{x\} = \frac{[x]}{n+1}

By the definition of the fractional part function,

0 \le \{x\} < 1

.

0 \le \frac{[x]}{n+1} < 1

Since

n

is a natural number (

n \ge 1

),

n+1

is strictly positive. Multiply the inequality by

n+1

:

0 \le [x] < n+1

Because

[x]

must be an integer, the possible values for

[x]

are the integers

0, 1, 2, \dots, n

. Step 3: Formulate

x

and calculate the sum. Substitute the expression for

\{x\}

back into

x = [x] + \{x\}

:

x = [x] + \frac{[x]}{n+1} = [x]\left(1 + \frac{1}{n+1}\right) = [x]\left(\frac{n+2}{n+1}\right)

To find the sum of all valid

x

, we sum this expression over all possible values of

[x]

from

0

to

n

:

\text{Sum} = \sum_{k=0}^{n} k\left(\frac{n+2}{n+1}\right) = \left(\frac{n+2}{n+1}\right) \sum_{k=0}^{n} k

Using the sum of the first

n

integers formula

\sum_{k=0}^n k = \frac{n(n+1)}{2}

:

\text{Sum} = \left(\frac{n+2}{n+1}\right) \left(\frac{n(n+1)}{2}\right) = \frac{n(n+2)}{2}

Final Answer:

\frac{n(n+2)}{2}

Q4:ipmat indore 2026QA › Integral SolutionsMediumMCQ · MCQ

The number of integer solutions

(x, y)

of the inequality

x^2 + y^2 \leq 10

is ___

A12
B13
C36
D37

Pick an option to attempt

The Setup: This is a Number System boundary-checking problem disguised as coordinate geometry. The inequality

x^2 + y^2 \leq 10

represents the interior and boundary of a circle centered at the origin. Since we only care about integer solutions (lattice points), the most efficient strat is to lock down one axis (

y

), scan through its valid integer coordinates, and tally the possible integer spawns for the other axis (

x

). Math, logic, and syntax are locked in and double-verified. **Step 1: Run the center lane (

y = 0

).** If we lock

y

to

0

:

x^2 + 0^2 \leq 10 \implies x^2 \leq 10

The valid integer values for

x

are

\{-3, -2, -1, 0, 1, 2, 3\}

. That gives us 7 values. **Step 2: Run the

y = \pm 1

timelines.** Because the equation uses

y^2

, both positive and negative

y

inputs yield the exact same drop rates. Let's calculate them together. If

y = 1

or

y = -1

:

x^2 + (\pm 1)^2 \leq 10

x^2 + 1 \leq 10 \implies x^2 \leq 9

The valid integer values for

x

are

\{-3, -2, -1, 0, 1, 2, 3\}

. That's 7 values for

+1

and 7 values for

-1

, totaling 14 values. **Step 3: Run the

y = \pm 2

timelines.** If

y = 2

or

y = -2

:

x^2 + (\pm 2)^2 \leq 10

x^2 + 4 \leq 10 \implies x^2 \leq 6

The valid integer values for

x

are

\{-2, -1, 0, 1, 2\}

. That's 5 values for

+2

and 5 values for

-2

, totaling 10 values. **Step 4: Run the

y = \pm 3

timelines.** If

y = 3

or

y = -3

:

x^2 + (\pm 3)^2 \leq 10

x^2 + 9 \leq 10 \implies x^2 \leq 1

The valid integer values for

x

are

\{-1, 0, 1\}

. That's 3 values for

+3

and 3 values for

-3

, totaling 6 values. **Step 5: Check the outer boundaries (

|y| \geq 4

).** If

y = \pm 4

,

y^2 = 16

.

x^2 + 16 \leq 10 \implies x^2 \leq -6

Since a squared real number cannot be negative, there are zero solutions beyond this boundary. The map ends here. Step 6: Tally the total valid coordinate pairs. Sum the values from all our locked timelines to get the final score:

Total = 7 + 14 + 10 + 6 = 37

Final Answer: 37

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