IPMAT Indore 2025 (MCQ) - A natural number n lies between 100 and 400, and the sum of its digits is 10. The probability that n is divisible by 4, is | PYQs + Solutions

IPMAT Indore 2025

Modern Math

Probability

Medium

A natural number $n$ lies between $100$ and $400$ , and the sum of its digits is $10$ . The probability that $n$ is divisible by $4$ , is

\frac{1}{4}

\frac{7}{27}

\frac{1}{3}

\frac{2}{9}

✅ Correct Option: 2

Let's denote a three-digit number as

abc

where:

\newline

a

is the hundreds digit (1, 2, or 3)

\newline

b

is the tens digit (0-9)

\newline

c

is the units digit (0-9)

We need

a + b + c = 10

When

a = 1

, we need

b + c = 9

\newline

Possible numbers: 109, 118, 127, 136, 145, 154, 163, 172, 181, 190

When

a = 2

, we need

b + c = 8

\newline

Possible numbers: 208, 217, 226, 235, 244, 253, 262, 271, 280

When

a = 3

, we need

b + c = 7

\newline

Possible numbers: 307, 316, 325, 334, 343, 352, 361, 370

In total, there are 10 + 9 + 8 = 27 numbers between 100 and 400 with digit sum 10.

A number is divisible by 4 if its last two digits form a number divisible by 4.

Checking each number:

Divisible by 4: 136, 172, 208, 244, 280, 316, 352

That gives us 7 numbers divisible by 4 out of 27 total numbers.

Therefore, the probability that

n

is divisible by 4 is:

\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{7}{27}