Number System — BSSC Quantitative Aptitude

**Step-by-step solution:**

• Let $x = \frac{p}{q}$ where $p$ and $q$ are non-zero integers (since $x$ is a non-zero rational number).

• Suppose the product $xy = r$ is a rational number.

• Since $x$ is non-zero, we can divide both sides by $x$ to isolate $y$: $y = \frac{r}{x}$.

• Since both $r$ and $x$ are rational numbers, their quotient $\frac{r}{x}$ must also be a rational number.

• This implies that $y$ is rational, which directly contradicts the given condition that $y$ is irrational.

• Therefore, our assumption that $xy$ is rational must be false, meaning $xy$ must always be irrational. **Exam Trick:**

• To solve this quickly, pick simple values: Let rational $x = 2$ and irrational $y = \sqrt{3}$. Their product is $2\sqrt{3}$, which is clearly irrational. This immediately eliminates all other options.

**Step-by-step solution:**

• Identify the position of the digit $7$ in the number $937642$. It is at the thousands place, so its place value is $7 \times 1000 = 7000$.

• Identify the face value of $4$ in $937642$. The face value of any digit is the digit itself, which is $4$.

• Calculate the difference: $7000 - 4 = 6996$. **Exam Trick:**

• Units digit check: When you subtract $4$ from a number ending in $0$ (like $7000$), the unit digit of the result must be $6$ ($10 - 4 = 6$). Only option $6996$ ends with $6$, allowing you to solve it in under 3 seconds.

**Step-by-step solution:**

• Since the number is divisible by $72$, it must be divisible by both its co-prime factors $8$ and $9$.

• For divisibility by $8$, the last three digits $4x2$ must be divisible by $8$. Testing natural numbers for $x$: - If $x=1$, $412$ is not divisible by $8$. - If $x=2$, $422$ is not divisible by $8$. - If $x=3$, $432$ is divisible by $8$ ($432 = 8 \times 54$). Hence, the minimum natural number value of $x$ is $3$.

• For divisibility by $9$, the sum of all digits must be divisible by $9$. - Sum of digits $= 5 + 4 + 3 + 2 + y + 1 + 7 + 4 + x + 2 = 28 + x + y$. - Substitute $x = 3$: Sum $= 28 + 3 + y = 31 + y$. - For $31 + y$ to be divisible by $9$, the single digit $y$ must be $5$ (since $31 + 5 = 36$, which is divisible by $9$).

• Calculate the value of $(4x - y)$: $4(3) - 5 = 12 - 5 = 7$. **Exam Trick:**

• Divisibility by $8$ shortcut: Since $400$ is a multiple of $8$, the 3-digit number $4x2$ is divisible by $8$ if and only if the 2-digit number $x2$ is divisible by $8$. The multiples of $8$ ending in $2$ are $32$ and $72$, so $x$ must be $3$ or $7$. For the minimum value, $x = 3$ is chosen instantly. Sum of digits digit-sum rule makes finding $y = 5$ effortless.

**Step-by-step solution:**

• Find the prime factorization of $360$: $360 = 2^3 \times 3^2 \times 5^1$

• To find the number of even factors, we can use the formula where we keep the exponent of $2$ as it is and add $1$ to the exponents of other prime factors, then multiply them: Number of even factors $= 3 \times (2 + 1) \times (1 + 1)$ $= 3 \times 3 \times 2 = 18$

• Alternatively, calculate the total factors and subtract odd factors: - Total factors $= (3 + 1)(2 + 1)(1 + 1) = 4 \times 3 \times 2 = 24$ - Odd factors (ignoring power of $2$) $= (2 + 1)(1 + 1) = 3 \times 2 = 6$ - Even factors $= 24 - 6 = 18$ **Exam Trick:**

• Instead of writing out the entire factor tree or subtracting, directly apply the shortcut: even factors are those containing at least one factor of $2$. Therefore, the choice of the power of $2$ can be $2^1$, $2^2$, or $2^3$ ($3$ options), while $3$ has $(2+1)=3$ options and $5$ has $(1+1)=2$ options. Multiply directly: $3 \times 3 \times 2 = 18$. Takes under 10 seconds!

**Step-by-step solution:**

• Let $x = 0.3\overline{85}$.

• To convert a mixed recurring decimal to a fraction: - Numerator $= \text{Total number formed by digits} - \text{Non-repeating part} = 385 - 3 = 382$. - Denominator $=$ Write as many $9$s as there are repeating digits (two digits, $8$ and $5 \implies 99$) followed by as many $0$s as there are non-repeating digits after the decimal point (one digit, $3 \implies 0$). Thus, denominator $= 990$.

• Form the fraction: $x = \frac{382}{990}$.

• Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor ($2$): - Numerator: $\frac{382}{2} = 191$ - Denominator: $\frac{990}{2} = 495$ - Simplest form $= \frac{191}{495}$ (since $191$ is a prime number). **Exam Trick:**

• Denominator rule: The denominator must end in a $9$ or $0$. Since there are two repeating digits and one non-repeating, the denominator is $990$. The simplified fraction must have a denominator that is a factor of $990$. Only $495$ and $330$ are factors of $990$ in the options. Since $127/330 = 381/990 \implies 381 = 384-3 \implies 0.3\overline{84}$, we can eliminate it and confidently mark $\frac{191}{495}$ without doing full simplification.

**Step-by-step solution:**

• Unit digit of $(234)^{103}$: - The base ends in $4$. For base ending in $4$, the unit digit has cyclicity of $2$: $4^1 = 4$ (odd powers) and $4^2 = 16 \implies 6$ (even powers). - Since the power $103$ is odd, the unit digit is $4$.

• Unit digit of $(357)^{146}$: - The base ends in $7$. For base ending in $7$, the cyclicity is $4$ ($7, 9, 3, 1$). - Divide the power $146$ by $4$: $146 = 4 \times 36 + 2$. The remainder is $2$. - So, unit digit is the same as the unit digit of $7^2 = 49$, which is $9$.

• Unit digit of $(623)^{83}$: - The base ends in $3$. For base ending in $3$, the cyclicity is $4$ ($3, 9, 7, 1$). - Divide the power $83$ by $4$: $83 = 4 \times 20 + 3$. The remainder is $3$. - So, unit digit is the same as the unit digit of $3^3 = 27$, which is $7$.

• Calculate the final unit digit of the expression: $4 + 9 - 7 = 13 - 7 = 6$. **Exam Trick:**

• Quickly compute powers: power of $4$ is odd $\implies 4$; power of $7$ is $146 \equiv 2 \pmod 4 \implies 7^2 \implies 9$; power of $3$ is $83 \equiv 3 \pmod 4 \implies 3^3 \implies 7$. Put them together: $4 + 9 - 7 = 6$. Done in 15 seconds!

**Step-by-step solution:**

• Since $17$ is a prime number, we can apply Fermat's Little Theorem which states that $a^{p-1} \equiv 1 \pmod p$ for any integer $a$ not divisible by prime $p$. - Here, $a = 2$ and $p = 17$, so $2^{16} \equiv 1 \pmod{17}$.

• Express the power $2026$ in terms of $16$: - Divide $2026$ by $16$: $2026 = 16 \times 126 + 10$.

• Therefore, we can rewrite the expression as: $2^{2026} = (2^{16})^{126} \times 2^{10}$ $\equiv 1^{126} \times 2^{10} \pmod{17}$ $\equiv 2^{10} \pmod{17}$

• Calculate $2^{10} = 1024$. Now, divide $1024$ by $17$ to find the remainder: - $1024 = 17 \times 60 + 4$. - Hence, the remainder is $4$. **Exam Trick:**

• Negative Remainder Shortcut: We know that $2^4 = 16 \equiv -1 \pmod{17}$. Therefore, we can rewrite $2^{2026}$ using base $2^4$: $2^{2026} = (2^4)^{506} \times 2^2 \equiv (-1)^{506} \times 4 \pmod{17}$. Since $506$ is even, $(-1)^{506} = 1$. So, remainder $= 1 \times 4 = 4$. This negative remainder trick bypasses Fermat's theorem and large calculations entirely!

**Step-by-step solution:**

• Find the smallest such natural number $N$ using reverse tracking: - Let the final quotient after successive division by $3$ be $0$ (for the smallest natural number). - Third division: divisor $= 3$, remainder $= 1 \implies$ quotient before this division $= 3 \times 0 + 1 = 1$. - Second division: divisor $= 4$, remainder $= 2 \implies$ quotient before this division $= 4 \times 1 + 2 = 6$. - First division: divisor $= 5$, remainder $= 3 \implies$ original number $N = 5 \times 6 + 3 = 33$.

• Now, successively divide the smallest number $33$ by $3, 4,$ and $5$: - $33 \div 3 \implies$ Quotient $= 11$, Remainder $= 0$. - $11 \div 4 \implies$ Quotient $= 2$, Remainder $= 3$. - $2 \div 5 \implies$ Quotient $= 0$, Remainder $= 2$.

• Calculate the sum of the successive remainders: - Sum $= 0 + 3 + 2 = 5$. **Exam Trick:**

• Synthetic Division Ladder: Set up a quick mental calculation working from bottom to top: - Stage 3 (last division): $(0 \text{ quotient}) \times 3 + 1 = 1$ - Stage 2 (second division): $1 \times 4 + 2 = 6$ - Stage 1 (first division): $6 \times 5 + 3 = 33$ Once $33$ is found, division downwards yields: - $33 / 3 = 11$ with remainder $0$ - $11 / 4 = 2$ with remainder $3$ - $2 / 5 = 0$ with remainder $2$ Sum of remainders $= 0 + 3 + 2 = 5$. This tabular method avoids any complex algebraic equations and prevents errors completely.

**Step-by-step solution:**

• To find the number of trailing zeros in $n!$, we divide $n$ by $5$ successively and sum the quotients until the quotient becomes less than $5$.

• First division: $\frac{100}{5} = 20$

• Second division: $\frac{20}{5} = 4$

• Since the quotient $4$ is less than $5$, we stop here.

• Sum of quotients = $20 + 4 = 24$.

• Therefore, there are $24$ trailing zeros in $100!$. **Exam Trick:**

• Shortcut formula: $\lfloor \frac{n}{5} \rfloor + \lfloor \frac{n}{25} \rfloor + \lfloor \frac{n}{125} \rfloor + \dots$

• For $n=100$: $\lfloor \frac{100}{5} \rfloor + \lfloor \frac{100}{25} \rfloor = 20 + 4 = 24$. This calculation takes less than 5 seconds!

**Step-by-step solution:**

• To convert decimal to binary, successively divide the number by $2$ and record the remainders from bottom to top: - $109 \div 2 = 54$, remainder $1$ - $54 \div 2 = 27$, remainder $0$ - $27 \div 2 = 13$, remainder $1$ - $13 \div 2 = 6$, remainder $1$ - $6 \div 2 = 3$, remainder $0$ - $3 \div 2 = 1$, remainder $1$ - $1 \div 2 = 0$, remainder $1$

• Reading the remainders from bottom to top, we get $1101101_2$. **Exam Trick:**

• Binary positional weights method: Write down powers of $2$: $64, 32, 16, 8, 4, 2, 1$.

• Fit $109$ into these weights: $109 = 64 + 32 + 8 + 4 + 1$.

• Put $1$ for used weights and $0$ for unused weights: - $64 \implies 1$ - $32 \implies 1$ - $16 \implies 0$ - $8 \implies 1$ - $4 \implies 1$ - $2 \implies 0$ - $1 \implies 1$

• Combining them gives $1101101_2$. This positional sum trick avoids long divisions entirely!