Mathematics - MCQ Practice Questions

Understanding:

We must find the total number of positive divisors of $720$.

Formula:

If $n=p_1^{a_1}\cdot p_2^{a_2}\cdots p_k^{a_k}$, then the number of divisors is:

Step 1: Prime factorisation of $720$

So $720=2^4\cdot 3^2\cdot 5^1$.

Step 2: Apply the divisor formula

Answer:

The number of positive divisors of $720$ is $30$.

Quick Tip:

$720=6!$ — factorising factorials is a common source of divisor-count questions in Indian exams.

Understanding:

We must find $2^{100}(\mathrm{mod}7)$.

Formula:

By Fermat's Little Theorem, for a prime $p$ and $\gcd (a,p)=1$:

Step 1: Identify the period

Since $7$ is prime and $\gcd (2,7)=1$:

Step 2: Reduce the exponent modulo $6$

Step 3: Compute

since $16=2\times 7+2$.

Answer:

The remainder when $2^{100}$ is divided by $7$ is $2$.

Quick Tip:

Always reduce the exponent modulo $(p-1)$ when the divisor is prime — Fermat's Little Theorem is the fastest route.

Understanding:

We must find $\gcd (252,198)$.

Formula:

The Euclidean algorithm repeatedly applies:

Step 1: First application

Step 2: Second application

Step 3: Third application

Step 4: Final step

Verification:

$252=18\times 14$ and $198=18\times 11$, and $\gcd (14,11)=1$ ✓

Answer:

The greatest common divisor of $252$ and $198$ is $18$.

Understanding:

We must count integers in $\{1,2,\dots ,200\}$ not divisible by $3$ or $5$.

Formula:

By the Inclusion–Exclusion Principle:

where $A$ = multiples of $3$, $B$ = multiples of $5$, and $A\cap B$ = multiples of $15$.

Step 1: Count multiples of $3$, $5$, and $15$ up to $200$

Step 2: Apply inclusion–exclusion

Step 3: Count integers divisible by neither

Answer:

The number of integers from $1$ to $200$ divisible by neither $3$ nor $5$ is $107$.

Quick Tip:

A common error is to forget to subtract the double-counted multiples of $\text{lcm}(3,5)=15$.

Understanding:

We must find the second number given $\text{LCM}$, $\text{HCF}$, and one number.

Formula:

For any two positive integers $a$ and $b$:

Step 1: Solve for the second number $b$

Verification:

$\gcd (180,168)$: $180=1\times 168+12$, $168=14\times 12+0$, so $\gcd =12$ ✓

$\text{lcm}(180,168)=\frac{180\times 168}{12}=\frac{30240}{12}=2520$ ✓

Answer:

The other number is $168$.

Understanding:

We must determine for which positive integers $n$ the expression $n^2-n$ is divisible by $6$.

Formula:

Divisibility by $6$ requires divisibility by both $2$ and $3$:

Step 1: Divisibility by $2$

$n(n-1)$ is the product of two consecutive integers. Among any two consecutive integers, one is always even, so:

Step 2: Divisibility by $3$

Among any three consecutive integers, exactly one is divisible by $3$. Since $n-1$, $n$, $n+1$ are three consecutive integers, either $n-1$ or $n$ is divisible by $3$ (or the next one $n+1$ is, but that doesn't affect our pair unless $n\equiv 0$ or $n\equiv 1(\mathrm{mod}3)$).

Check all residues of $n(\mathrm{mod}3)$:

For $n\equiv 2(\mathrm{mod}3)$: $n(n-1)=2\times 1=2$ in residue, so $3\nmid n(n-1)$?

Actual check: $n=2$: $n(n-1)=2$, and $3\nmid 2$. But $6\nmid 2$.

So the claim must be re-examined. $n=2$: $4-2=2$, not divisible by $6$. $n=3$: $9-3=6$, divisible by $6$. $n=4$: $16-4=12$, divisible by $6$. $n=1$: $1-1=0$, divisible by $6$.

Thus $n=2$ fails. The correct characterisation: $6\mid n(n-1)$ if and only if $n\not\equiv 2(\mathrm{mod}3)$, i.e. $n\equiv 0$ or $1(\mathrm{mod}3)$.

Step 3: Conclude

Since $n=2$ gives $n(n-1)=2$ which is not divisible by $6$, the condition does NOT hold for all positive integers. The smallest $n$ for which $n(n-1)$ is divisible by $6$ (and $n\ge 1$) is $n=1$ (giving $0$, divisible by everything), and the next is $n=3$ (giving $6$).

Among the options, the correct statement is: divisibility holds for $n=1$ (trivially, as $1\times 0=0$) and the options ask us to identify which is smallest. $n=1$ gives $0$, which is divisible by $6$.

Answer:

The smallest positive integer $n$ such that $n^2-n$ is divisible by $6$ is $n=1$, since $1^2-1=0$ and $6\mid 0$.

Quick Tip:

$0$ is divisible by every positive integer by definition ($0=6\times 0$), so $n=1$ is always the trivial solution to such problems.

Understanding:

We are given $\gcd (a,b)=1$ (i.e., $a$ and $b$ are coprime) and must identify which statement must always be true.

Formula:

A standard result in number theory states:

Step 1: Verify option A

If a prime $p$ divided both $a^2$ and $b$, then since $p$ is prime and $p\mid a^2$, we get $p\mid a$. Combined with $p\mid b$, this means $p\mid \gcd (a,b)=1$, a contradiction. Therefore:

Step 2: Check the other options with counterexamples

Step 3: Conclude

Only $\gcd (a^2,b)=1$ is guaranteed whenever $\gcd (a,b)=1$.

Answer:

If $\gcd (a,b)=1$, then $\gcd (a^2,b)=1$ must be true.

Quick Tip:

More generally, if $\gcd (a,b)=1$ then $\gcd (a^m,b^n)=1$ for all positive integers $m,n$.

Understanding:

Trailing zeros in $n!$ come from factors of $10=2\times 5$. Since factors of $2$ are more abundant than factors of $5$, we count the number of times $5$ divides $50!$.

Formula:

The exponent of a prime $p$ in $n!$ is given by Legendre's formula:

Step 1: Apply the formula for $p=5$, $n=50$

Step 2: Confirm factors of $2$ exceed $12$

So the limiting factor is indeed $5$, and the number of trailing zeros equals $12$.

Answer:

The number of trailing zeros in $50!$ is $12$.

Quick Tip:

Multiples of $25$ contribute two factors of $5$ each — forgetting this is the most common error in trailing-zero questions.

Understanding:

We are given that the sum of all positive divisors of $496$ is $992$, and we must classify $496$ accordingly.

Formula:

A positive integer $n$ is classified as:

Step 1: Compare $\sigma (496)$ with $2\times 496$

Step 2: Classify

Since $\sigma (496)=2n$, the number $496$ is a perfect number.

Step 3: Distinguish from the other options

Answer:

$496$ is a perfect number.

Quick Tip:

The first four perfect numbers are $6,28,496,8128$ — all of the form $2^{p-1}(2^p-1)$ where $2^p-1$ is a Mersenne prime.

Understanding:

We need to count divisors of $720$ that are divisible by $4$ but NOT by $8$.

Formula:

First find the prime factorisation of $720$, then count divisors with exactly the right power of $2$.

Step 1: Prime factorise $720$

Step 2: Identify the condition

A divisor $d$ of $720$ has the form $d=2^a\cdot 3^b\cdot 5^c$ where $0\le a\le 4$, $0\le b\le 2$, $0\le c\le 1$.

Divisible by $4=2^2$ but NOT by $8=2^3$ means exactly $a=2$.

Step 3: Count valid divisors

With $a=2$ fixed, count choices for $b$ and $c$:

Total:

Verification:

The $6$ divisors are: $4\cdot 1=4,4\cdot 3=12,4\cdot 9=36,4\cdot 5=20,4\cdot 15=60,4\cdot 45=180$. Each is divisible by $4$ but not $8$. ✓

Answer:

The number of required divisors is $6$.

Quick Tip:

When counting divisors with an exact power of a prime, fix that prime's exponent and freely choose exponents for all remaining primes.

Understanding:

We are given:

Formula:

For any two positive integers $a$ and $b$:

Step 1: Apply the formula

Step 2: Verify

Answer:

The other number is $60$.

Understanding:

Count integers from $1$ to $300$ divisible by neither $3$ nor $5$ using the Inclusion–Exclusion Principle.

Formula:

Step 1: Count multiples of $3$

Step 2: Count multiples of $5$

Step 3: Count multiples of $15$ (both $3$ and $5$)

Step 4: Apply Inclusion–Exclusion

Answer:

The number of integers divisible by neither $3$ nor $5$ is $160$.

Quick Tip:

Alternatively, use Euler's product: $300\times \left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)=300\times \frac{2}{3}\times \frac{4}{5}=160$.

Understanding:

Find prime numbers $p$ and $q$ such that $p^2-2q^2=1$.

Formula:

Rearrange and test small primes:

Step 1: Test small prime values of $q$

Step 2: Check parity for $q>2$ (odd primes)

If $q$ is odd, $q^2$ is odd, so $2q^2$ is even, $2q^2+1$ is odd. $p^2$ is odd, so $p$ is odd — consistent. But checking shows no further prime solutions for small $q$.

For large $q$, $p=\sqrt{2q^2+1}$ is between $q\sqrt{2}$ and $q\sqrt{2}+1$, making integer solutions rare (Pell equation). The only small prime solution is $p=3,q=2$.

Step 3: Compute the sum

Answer:

The value of $p+q$ is $5$.

Quick Tip:

Whenever a Diophantine equation involves primes, always test $2$ first since it is the only even prime — it often yields the only solution.

Understanding:

The number of trailing zeros in $100!$ equals the highest power of $10$ dividing $100!$, which equals the highest power of $5$ (since there are always more factors of $2$ than $5$).

Formula:

Legendre's formula for the largest power of a prime $p$ dividing $n!$:

Step 1: Apply with $p=5$, $n=100$

Step 2: Sum the terms

Step 3: Confirm factor of $2$ is not limiting

So the number of trailing zeros $=24$.

Answer:

The number of trailing zeros in $100!$ is $24$.

Quick Tip:

Multiples of $25$ contribute two factors of $5$ each, which is why the second term $\lfloor 100/25\rfloor =4$ adds $4$ extra zeros beyond the naive count of $20$.

Understanding:

Find the sum of all two-digit integers $n$ satisfying $n\equiv 3(\mathrm{mod}7)$.

Formula:

Sum of an arithmetic progression with first term $a$, last term $l$, and $k$ terms:

Step 1: Identify the first term

The sequence $3,10,17,24,\dots$ consists of numbers $\equiv 3(\mathrm{mod}7)$.

Verify: $10=7\times 1+3$, so the first two-digit term is $a=10$.

Step 2: Identify the last term

Check downward from $99$:

So the last term is $l=94$.

Step 3: Count the number of terms

Step 4: Compute the sum

Answer:

The sum of all two-digit positive integers leaving remainder $3$ on division by $7$ is $676$.

Quick Tip:

The common difference in such a sequence is always equal to the divisor ($7$ here). Identifying first and last terms in range is the only careful step needed.

Understanding:

Find $(13^{99}+7^{99})(\mathrm{mod}20)$.

Formula:

Use the algebraic identity: for odd exponent $n$,

and compute directly mod $20$.

Step 1: Apply the sum-of-powers divisibility rule

Since $13+7=20$ and the exponent $99$ is odd:

This follows from the factorisation $a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+\cdots +b^{n-1})$ when $n$ is odd.

Step 2: Verify by computing mod $20$

So both $13$ and $7$ have order $4$ mod $20$.

Answer:

The remainder when $13^{99}+7^{99}$ is divided by $20$ is $0$.

Quick Tip:

Whenever you see $a^n+b^n$ with $a+b$ being the divisor and $n$ odd, the remainder is immediately $0$. This is one of the most useful number theory shortcuts in competitive exams.