IPMAT Indore Divisibility Rules — PYPs

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

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

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:

Final Answer: 96

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: $3^{30}+8$

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: $\frac{1}{2}$

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