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Basic Numeracy

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Question 161Question →

Let N=25×34×52N = 2^5 \times 3^4 \times 5^2. How many positive integer factors of NN are divisible by 1212 but not divisible by 1515?

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Answer: 16

Answer

16 positive factors
The number NN is given by 25×34×522^5 \times 3^4 \times 5^2. A factor 2a×3b×5c2^a \times 3^b \times 5^c is divisible by 12=22×3112 = 2^2 \times 3^1 if a∈{2,3,4,5}a \in \{2,3,4,5\} (4 choices), b∈{1,2,3,4}b \in \{1,2,3,4\} (4 choices), and c∈{0,1,2}c \in \{0,1,2\} (3 choices), giving 4×4×3=484 \times 4 \times 3 = 48 factors. Among these, those also divisible by 15=31×5115 = 3^1 \times 5^1 must be divisible by LCM(12,15)=60=22×31×51\text{LCM}(12,15) = 60 = 2^2 \times 3^1 \times 5^1, which restricts c∈{1,2}c \in \{1,2\} (2 choices), yielding 4×4×2=324 \times 4 \times 2 = 32 factors. Subtracting these gives 48−32=1648 - 32 = 16 factors.

Step-by-Step Solution

1
Express NN in prime factorized form and define the general structure of its factors.
Any positive factor of N=25×34×52N = 2^5 \times 3^4 \times 5^2 has the form k=2a×3b×5ck = 2^a \times 3^b \times 5^c, where 0≤a≤50 \le a \le 5, 0≤b≤40 \le b \le 4, and 0≤c≤20 \le c \le 2.
Prime factorization determines all possible divisors of a composite number.
2
Calculate the total number of factors divisible by 12=22×3112 = 2^2 \times 3^1.
For kk to be divisible by 12, we require a≥2a \ge 2 (a∈{2,3,4,5}a \in \{2,3,4,5\}, 4 choices), b≥1b \ge 1 (b∈{1,2,3,4}b \in \{1,2,3,4\}, 4 choices), and c≥0c \ge 0 (c∈{0,1,2}c \in \{0,1,2\}, 3 choices). Total factors = 4×4×3=484 \times 4 \times 3 = 48.
A factor contains another number as a divisor if all prime exponent lower bounds of the divisor are satisfied.
3
Calculate the number of factors divisible by both 1212 and 1515, which is equivalent to being divisible by LCM(12,15)=60=22×31×51\text{LCM}(12, 15) = 60 = 2^2 \times 3^1 \times 5^1.
For kk to be divisible by 60, we require a≥2a \ge 2 (4 choices), b≥1b \ge 1 (4 choices), and c≥1c \ge 1 (c∈{1,2}c \in \{1,2\}, 2 choices). Total factors = 4×4×2=324 \times 4 \times 2 = 32.
To exclude factors divisible by 15, we must find the intersection of multiples of 12 and multiples of 15.
4
Subtract the factors divisible by 60 from the factors divisible by 12.
Factors divisible by 12 but not by 15 = 48−32=1648 - 32 = 16.
Applying the principle of set difference yields the exact count satisfying both conditions.

Key Concept

Counting Factors with Prime Factorization Constraints
Estimated Time:2m 0s
Question 162Question →

For the composite numerical expression E=(412120+317201)−814102E = (412^{120} + 317^{201}) - 814^{102}, which of the following represents its unit digit?

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Answer: 7

Answer

7
Evaluating each term by cyclicity yields unit digits of 6 for 412120412^{120}, 7 for 317201317^{201}, and 6 for 814102814^{102}. Combining these according to the operations gives (6+7)−6=7(6 + 7) - 6 = 7, resulting in a unit digit of 7.

Step-by-Step Solution

1
Determine the unit digit of 412120412^{120}
Unit digit is 6
The unit digit of base 412 is 2, which has a cyclicity of 4 (2, 4, 8, 6). The exponent 120 is divisible by 4 (120(mod4)=0120 \pmod 4 = 0), so we take the 4th power in the cycle: 24=162^4 = 16, giving a unit digit of 6.
2
Determine the unit digit of 317201317^{201}
Unit digit is 7
The unit digit of base 317 is 7, which has a cyclicity of 4 (7, 9, 3, 1). The exponent remainder is 201(mod4)=1201 \pmod 4 = 1, corresponding to 71=77^1 = 7.
3
Determine the unit digit of 814102814^{102}
Unit digit is 6
The unit digit of base 814 is 4, which has a cyclicity of 2 (4 for odd powers, 6 for even powers). Since 102 is an even exponent, the unit digit is 6.
4
Combine the unit digits according to the expression E=(412120+317201)−814102E = (412^{120} + 317^{201}) - 814^{102}
Unit digit is 7
Adding the unit digits of the first two terms gives 6+7=136 + 7 = 13, which has a unit digit of 3. Subtracting 6 gives 3−6=−33 - 6 = -3, which in modulo 10 arithmetic yields 13−6=713 - 6 = 7.

Key Concept

Cyclicity of numbers and modular arithmetic for unit digit determination
Estimated Time:1m 30s
Question 163Question →
What is the simplified numerical value of the mathematical expression 25−[10+{8÷(6−4−2‾)}]25 - \left[ 10 + \left\{ 8 \div \left( 6 - \overline{4 - 2} \right) \right\} \right] when evaluated using the standard BODMAS rule?
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Answer: 1313

Answer

The correct simplified value of the given expression is 13.
Evaluating the expression step-by-step using VBODMAS rules:
1. Vinculum: 4−2‾=2\overline{4 - 2} = 2
2. Round brackets: 6−2=46 - 2 = 4
3. Curly brackets: 8÷4=28 \div 4 = 2
4. Square brackets: 10+2=1210 + 2 = 12
5. Final subtraction: 25−12=1325 - 12 = 13.
Thus, the correct value is 13.

Step-by-Step Solution

1
Evaluate the expression under the vinculum (bar)
4−2‾=2\overline{4 - 2} = 2
According to VBODMAS, operations under a vinculum (bar) have the highest precedence.
2
Simplify the innermost round brackets
6−2=46 - 2 = 4
Substitute the vinculum result into the round brackets (6−2)(6 - 2).
3
Evaluate the division inside the curly brackets
8÷4=28 \div 4 = 2
Division within the curly braces {8÷4}\{8 \div 4\} takes precedence.
4
Evaluate the addition inside the square brackets
10+2=1210 + 2 = 12
Add the terms inside the square brackets [10+2][10 + 2].
5
Perform the final subtraction
25−12=1325 - 12 = 13
Subtract the bracket result from 25.

Key Concept

Order of operations following the VBODMAS rule (Vinculum, Brackets, Orders, Division, Multiplication, Addition, Subtraction)
Question 164Question →

Consider the positive integer N=24×33×52N = 2^4 \times 3^3 \times 5^2. Which of the following statements regarding the properties of the factors and multiples of NN is/are correct?

Select all that apply

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Answer: The number of positive factors of NN that are perfect squares is equal to 12.; The sum of the reciprocals of all positive factors of NN is equal to σ(N)N\frac{\sigma(N)}{N}, where σ(N)\sigma(N) is the sum of all positive factors of NN.; The number of ordered pairs of positive integers (x,y)(x, y) such that lcm(x,y)=N\text{lcm}(x, y) = N is equal to 315.

Answer

The statements asserting that the number of perfect square factors is 12, the sum of factor reciprocals equals σ(N)N\frac{\sigma(N)}{N}, and the number of ordered pairs (x,y)(x, y) such that lcm(x,y)=N\text{lcm}(x, y) = N is 315 are all correct.
The statements asserting that NN has 12 perfect square factors, that the sum of reciprocals of all factors is σ(N)N\frac{\sigma(N)}{N}, and that there are 315 ordered pairs (x,y)(x, y) with lcm(x,y)=N\text{lcm}(x, y) = N are all mathematically correct applications of prime factorization principles.

Step-by-Step Solution

1
Determine the number of perfect square factors of N=24×33×52N = 2^4 \times 3^3 \times 5^2.
Any factor d=2a×3b×5cd = 2^a \times 3^b \times 5^c is a perfect square if a,b,ca, b, c are even. The allowed values are a∈{0,2,4}a \in \{0, 2, 4\} (3 options), b∈{0,2}b \in \{0, 2\} (2 options), and c∈{0,2}c \in \{0, 2\} (2 options). Total perfect square factors = 3×2×2=123 \times 2 \times 2 = 12.
Perfect square factors require non-negative even exponents for each prime factor.
2
Evaluate the sum of the reciprocals of all factors of NN.
As dd ranges over all divisors of NN, Nd\frac{N}{d} also ranges over all divisors of NN. Therefore, ∑d∣N1d=∑d∣NN/dN=1N∑d′∣Nd′=σ(N)N\sum_{d|N} \frac{1}{d} = \sum_{d|N} \frac{N/d}{N} = \frac{1}{N} \sum_{d'|N} d' = \frac{\sigma(N)}{N}.
Pairing each factor dd with its complementary factor N/dN/d converts the reciprocal sum into the total factor sum divided by NN.
3
Calculate the number of ordered pairs (x,y)(x, y) such that lcm(x,y)=N\text{lcm}(x, y) = N.
For each prime pieip_i^{e_i}, the maximum exponent in xx and yy must be eie_i. The number of choices for (ai,bi)(a_i, b_i) such that max⁡(ai,bi)=ei\max(a_i, b_i) = e_i is 2ei+12e_i + 1. For N=24×33×52N = 2^4 \times 3^3 \times 5^2, total ordered pairs = (2(4)+1)(2(3)+1)(2(2)+1)=9×7×5=315(2(4)+1)(2(3)+1)(2(2)+1) = 9 \times 7 \times 5 = 315.
The exponent of each prime factor in lcm(x,y)\text{lcm}(x, y) is determined independently by max⁡(ai,bi)=ei\max(a_i, b_i) = e_i.
4
Compute the product of all positive factors of NN.
Total number of factors D(N)=(4+1)(3+1)(2+1)=60D(N) = (4+1)(3+1)(2+1) = 60. The product of all factors is ND(N)/2=N60/2=N30N^{D(N)/2} = N^{60/2} = N^{30}.
Factors can be grouped into D(N)/2D(N)/2 pairs, each multiplying to NN.

Key Concept

Advanced properties of prime factorization including perfect square factor enumeration, reciprocal factor sum identity, LCM pair counting formula, and factor product relations.
Question 165Question →
Evaluate the expression given below by applying the standard order of operations (VBODMAS):
2.5×[18÷{4.5+12 of (7.8−3.4+1.4‾)}]+3.52.5 \times \left[ 18 \div \left\{ 4.5 + \frac{1}{2} \text{ of } \left( 7.8 - \overline{3.4 + 1.4} \right) \right\} \right] + 3.5
Which of the following represents the correct simplified result?
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Answer: 11

Answer

The correct simplified result is 11.
Evaluating step-by-step using VBODMAS order yields: Vinculum gives 4.8; Round bracket yields 7.8 - 4.8 = 3.0; 'of' operation gives 0.5 * 3.0 = 1.5; Curly brackets give 4.5 + 1.5 = 6.0; Square bracket gives 18 / 6.0 = 3; Final calculation gives 2.5 * 3 + 3.5 = 7.5 + 3.5 = 11.

Step-by-Step Solution

1
Evaluate the expression under the vinculum (bar line).
\overline{3.4 + 1.4} = 4.8
According to VBODMAS, the vinculum takes precedence over all standard brackets.
2
Simplify the innermost round brackets.
7.8 - 4.8 = 3.0
Resolve terms within parenthetical brackets next.
3
Execute the 'of' operation inside the curly brackets.
\frac{1}{2} \text{ of } 3.0 = 1.5
The 'of' operator represents multiplication with higher priority than division and standard addition.
4
Complete the addition within the curly brackets.
4.5 + 1.5 = 6.0
Finish evaluating all operations contained within the curly brackets.
5
Perform division inside the square brackets.
18÷6.0=318 \div 6.0 = 3
Evaluate the remaining square bracket operation.
6
Perform final multiplication and addition operations from left to right according to operator precedence.
2.5 \times 3 + 3.5 = 7.5 + 3.5 = 11
Multiplication takes precedence over final addition.

Key Concept

VBODMAS Rule (Vinculum, Brackets, Of, Division, Multiplication, Addition, Subtraction)
Estimated Time:1m 15s
Question 166Question →

What is the unit digit of the expression P=(678120+35945)×24337−71482P = (678^{120} + 359^{45}) \times 243^{37} - 714^{82}?

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Answer: 9

Answer

The unit digit of the expression is 9.
Evaluating each term using cyclicity rules gives 678120→6678^{120} \rightarrow 6, 35945→9359^{45} \rightarrow 9, 24337→3243^{37} \rightarrow 3, and 71482→6714^{82} \rightarrow 6. Combining these gives (6+9)×3−6≡15×3−6≡5×3−6≡15−6≡9(mod10)(6 + 9) \times 3 - 6 \equiv 15 \times 3 - 6 \equiv 5 \times 3 - 6 \equiv 15 - 6 \equiv 9 \pmod{10}.

Step-by-Step Solution

1
Calculate the unit digit of 678120678^{120}
Unit digit is 6
The cyclicity of 8 is 4 (8, 4, 2, 6). Exponent 120 mod 4 = 0, which corresponds to the 4th term in the cycle.
2
Calculate the unit digit of 35945359^{45}
Unit digit is 9
The cyclicity of 9 is 2 (9 for odd powers, 1 for even powers). Exponent 45 is odd.
3
Find the unit digit of (678120+35945)(678^{120} + 359^{45})
Unit digit is 5
Sum of unit digits is 6 + 9 = 15, which has unit digit 5.
4
Calculate the unit digit of 24337243^{37}
Unit digit is 3
The cyclicity of 3 is 4 (3, 9, 7, 1). Exponent 37 mod 4 = 1, so the unit digit is 3^1 = 3.
5
Multiply the sum by 24337243^{37}
Unit digit is 5
Product of unit digits is 5 * 3 = 15, which has unit digit 5.
6
Calculate the unit digit of 71482714^{82}
Unit digit is 6
The cyclicity of 4 is 2 (4 for odd powers, 6 for even powers). Exponent 82 is even.
7
Compute the final unit digit of the overall expression
Unit digit is 9
Subtracting unit digits yields 5 - 6. Adding 10 for borrowing gives 15 - 6 = 9.

Key Concept

Unit Digit and Cyclicity Rules for Exponential Expressions
Estimated Time:1m 30s
Question 167Question →

Arrange the following real numbers in ascending order (from smallest to largest):

Drag items to arrange them in the correct order

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Answer

The correct ascending order is 2340<0.57‾<0.57‾<712\frac{23}{40} < 0.\overline{57} < 0.5\overline{7} < \frac{7}{12}.
Converting all numbers to decimal expansions gives: 2340=0.5750\frac{23}{40} = 0.5750, 0.57‾=0.575757...0.\overline{57} = 0.575757..., 0.57‾=0.577777...0.5\overline{7} = 0.577777..., and 712=0.583333...\frac{7}{12} = 0.583333.... Comparing these digit by digit confirms the correct ascending sequence is 2340<0.57‾<0.57‾<712\frac{23}{40} < 0.\overline{57} < 0.5\overline{7} < \frac{7}{12}.

Step-by-Step Solution

1
Convert each fraction and recurring decimal into its expanded decimal representation.
2340=0.57500...\frac{23}{40} = 0.57500..., 0.57‾=0.575757...0.\overline{57} = 0.575757..., 0.57‾=0.577777...0.5\overline{7} = 0.577777..., and 712=0.583333...\frac{7}{12} = 0.583333...
Converting all terms into a common decimal format allows direct place-value comparison.
2
Compare the values digit by digit from left to right starting at the tenths place.
All terms start with 0.570.57, except 712\frac{7}{12} which is 0.58...0.58... (making 712\frac{7}{12} the largest). For the remaining three terms, compare the thousandths digit: 2340\frac{23}{40} has 00 in the ten-thousandths place (0.57500.5750), 0.57‾0.\overline{57} has 77 (0.57570.5757), and 0.57‾0.5\overline{7} has 77 in the thousandths place (0.57770.5777).
Determining place-value differences establishes the relative magnitudes clearly.
3
Sequence the original numbers according to their evaluated decimal order.
0.5750<0.5757...<0.5777...<0.5833...  ⟹  2340<0.57‾<0.57‾<7120.5750 < 0.5757... < 0.5777... < 0.5833... \implies \frac{23}{40} < 0.\overline{57} < 0.5\overline{7} < \frac{7}{12}
Re-substituting the original expressions yields the required ascending order sequence.

Key Concept

Ordering and comparison of rational numbers, terminating fractions, pure recurring decimals, and mixed recurring decimals.
Question 168Question →

What is the simplified numerical value of the expression 15+{20−(8÷6−2‾)}15 + \left\{ 20 - \left( 8 \div \overline{6 - 2} \right) \right\} when evaluated using the standard BODMAS rule?

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Answer: 33

Answer

The simplified numerical value of the expression is 33.
Applying VBODMAS order (Vinculum →\rightarrow Brackets →\rightarrow Division →\rightarrow Subtraction →\rightarrow Addition) yields: 6−2‾=4\overline{6-2} = 4, 8÷4=28 \div 4 = 2, 20−2=1820 - 2 = 18, and 15+18=3315 + 18 = 33.

Step-by-Step Solution

1
Evaluate the expression under the vinculum (bar)
6−2‾=4\overline{6 - 2} = 4
The vinculum (bar line) takes precedence over standard operators.
2
Perform division inside the round brackets
8÷4=28 \div 4 = 2
Operations inside the innermost round brackets are evaluated next.
3
Perform subtraction inside the curly brackets
20−2=1820 - 2 = 18
Evaluate the expression enclosed in the curly brackets.
4
Perform final addition outside brackets
15+18=3315 + 18 = 33
Complete the evaluation by adding the remaining outer terms.

Key Concept

BODMAS / VBODMAS Rule (Order of Operations with Vinculum and Brackets)
Question 169Question →

What is the simplified numerical value of the following mathematical expression when evaluated strictly according to the BODMAS rule?

25% of 120−[4×{15−(18÷7−4‾)}]÷1.525\% \text{ of } 120 - \left[ 4 \times \left\{ 15 - \left( 18 \div \overline{7 - 4} \right) \right\} \right] \div 1.5
Show answer & explanation

Answer: 6

Answer

6
Evaluating the vinculum 7−4‾\overline{7 - 4} yields 33. The round bracket becomes 18÷3=618 \div 3 = 6. The curly bracket yields 15−6=915 - 6 = 9. The square bracket evaluates to 4×9=364 \times 9 = 36. Concurrently, 25%25\% of 120120 equals 3030. Following operator hierarchy, division is executed next: 36÷1.5=2436 \div 1.5 = 24. Finally, performing subtraction gives 30−24=630 - 24 = 6.

Step-by-Step Solution

1
Evaluate the expression under the bar (vinculum)
7−4‾=3\overline{7 - 4} = 3
According to BODMAS, vinculum (bar bracket) has the highest priority among all grouping symbols.
2
Evaluate the innermost round bracket
18÷3=618 \div 3 = 6
Perform division inside the round bracket after resolving the vinculum.
3
Evaluate the curly bracket
15−6=915 - 6 = 9
Subtract the result of the round bracket from 15 inside the curly bracket.
4
Evaluate the square bracket and the percentage term
4×9=364 \times 9 = 36 and 25% of 120=25100×120=3025\% \text{ of } 120 = \frac{25}{100} \times 120 = 30
Multiply 4 by the curly bracket result, and independently calculate the percentage term.
5
Perform division before final subtraction
36÷1.5=2436 \div 1.5 = 24
BODMAS mandates performing division before subtraction.
6
Perform the final subtraction
30−24=630 - 24 = 6
Subtract the quotient from the percentage value.

Key Concept

BODMAS Order of Operations with Vinculum and Mixed Numbers
Question 170Question →

Three quantities are given as 0.6‾0.\overline{6}, 1225\frac{12}{25}, and 1.41.4. What is the exact ratio of the Least Common Multiple (LCM) of these three quantities to their Highest Common Factor (HCF)?

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Answer: 6300

Answer

6300
Converting the given values yields the reduced fractions 23\frac{2}{3}, 1225\frac{12}{25}, and 75\frac{7}{5}. Applying standard fraction formulas, LCM=LCM(2,12,7)HCF(3,25,5)=841=84\text{LCM} = \frac{\text{LCM}(2,12,7)}{\text{HCF}(3,25,5)} = \frac{84}{1} = 84 and HCF=HCF(2,12,7)LCM(3,25,5)=175\text{HCF} = \frac{\text{HCF}(2,12,7)}{\text{LCM}(3,25,5)} = \frac{1}{75}. Taking the ratio LCMHCF\frac{\text{LCM}}{\text{HCF}} gives 84÷175=630084 \div \frac{1}{75} = 6300.

Step-by-Step Solution

1
Convert all terms to irreducible fractions
0.6‾=69=230.\overline{6} = \frac{6}{9} = \frac{2}{3}, 1225=1225\frac{12}{25} = \frac{12}{25}, and 1.4=1410=751.4 = \frac{14}{10} = \frac{7}{5}
All numbers must be expressed as simplified fractions ab\frac{a}{b} in lowest terms before applying fraction HCF and LCM formulas.
2
Calculate the Least Common Multiple (LCM) of the fractions
LCM(23,1225,75)=LCM(2,12,7)HCF(3,25,5)=841=84\text{LCM}\left(\frac{2}{3}, \frac{12}{25}, \frac{7}{5}\right) = \frac{\text{LCM}(2, 12, 7)}{\text{HCF}(3, 25, 5)} = \frac{84}{1} = 84
The LCM of a set of fractions is given by LCM of numeratorsHCF of denominators\frac{\text{LCM of numerators}}{\text{HCF of denominators}}.
3
Calculate the Highest Common Factor (HCF) of the fractions
HCF(23,1225,75)=HCF(2,12,7)LCM(3,25,5)=175\text{HCF}\left(\frac{2}{3}, \frac{12}{25}, \frac{7}{5}\right) = \frac{\text{HCF}(2, 12, 7)}{\text{LCM}(3, 25, 5)} = \frac{1}{75}
The HCF of a set of fractions is given by HCF of numeratorsLCM of denominators\frac{\text{HCF of numerators}}{\text{LCM of denominators}}.
4
Find the ratio of the LCM to the HCF
Ratio=84175=84×75=6300\text{Ratio} = \frac{84}{\frac{1}{75}} = 84 \times 75 = 6300
Dividing the computed LCM by the computed HCF yields the required ratio.

Key Concept

HCF and LCM of Fractions and Recurring Decimals
Question 171Question →

A positive integer NN has exactly 1212 positive factors. The sum of all distinct prime factors of NN is 1010. If NN is divisible by 1212 but not divisible by 99, what is the total number of positive factors of 5N5N?

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Answer: 18

Answer

The total number of positive factors of 5N5N is 1818.
The distinct prime factors of NN that sum to 1010 and satisfy divisibility by 1212 are 2,3,2, 3, and 55. Because NN has 1212 factors, its exponents must be 2,1,12, 1, 1. Since NN is divisible by 1212 and not divisible by 99, N=22×31×51=60N = 2^2 \times 3^1 \times 5^1 = 60. Consequently, 5N=22×31×525N = 2^2 \times 3^1 \times 5^2, which possesses (2+1)(1+1)(2+1)=18(2+1)(1+1)(2+1) = 18 factors.

Step-by-Step Solution

1
Determine the set of distinct prime factors of NN
The distinct prime factors of NN are 22, 33, and 55.
The sum of distinct prime factors is 1010. Possible sets of distinct primes summing to 1010 are {3,7}\{3, 7\} and {2,3,5}\{2, 3, 5\}. Because NN is divisible by 1212, its prime factorization must include 22 and 33, forcing the set of prime factors to be {2,3,5}\{2, 3, 5\}.
2
Determine the exponent structure of NN
The prime factorization of NN takes the form p12×p21×p31p_1^2 \times p_2^1 \times p_3^1 for distinct primes p1,p2,p3p_1, p_2, p_3.
The number of positive factors is given by (a1+1)(a2+1)(a3+1)=12(a_1+1)(a_2+1)(a_3+1) = 12. The only integer factorization of 1212 into three factors greater than 11 is 3×2×23 \times 2 \times 2, requiring exponents 2,1,12, 1, 1 in some arrangement.
3
Assign exponents to prime factors 22, 33, and 55
N=22×31×51=60N = 2^2 \times 3^1 \times 5^1 = 60.
Since NN is divisible by 12=22×3112 = 2^2 \times 3^1, the exponent of 22 must be at least 22 and the exponent of 33 at least 11. Since NN is not divisible by 9=329 = 3^2, the exponent of 33 cannot be 22 or greater, so it must be exactly 11. Thus, the prime 22 takes exponent 22, leaving prime 55 with exponent 11.
4
Calculate the prime factorization and factor count of 5N5N
5N=22×31×525N = 2^2 \times 3^1 \times 5^2, which has 1818 positive factors.
Multiplying NN by 55 increments the exponent of 55 from 11 to 22. The factor count formula yields (2+1)×(1+1)×(2+1)=3×2×3=18(2+1) \times (1+1) \times (2+1) = 3 \times 2 \times 3 = 18.

Key Concept

Factors, Multiples, and Prime Factorization
Question 172Question →

What is the unit digit of the value obtained from the numerical expression N=(512160+737203)×434116−618140N = (512^{160} + 737^{203}) \times 434^{116} - 618^{140}?

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Answer: 8

Answer

8
Evaluating each term using cyclicity rules gives 512160→6512^{160} \rightarrow 6, 737203→3737^{203} \rightarrow 3, 434116→6434^{116} \rightarrow 6, and 618140→6618^{140} \rightarrow 6. The composite expression evaluates to (6+3)×6−6=54−6(6 + 3) \times 6 - 6 = 54 - 6. Performing the unit digit subtraction 4−64 - 6 with borrowing yields 14−6=814 - 6 = 8.

Step-by-Step Solution

1
Determine the unit digits of the terms inside the parentheses: 512160512^{160} and 737203737^{203}
Unit digit of 512160512^{160} is 66; unit digit of 737203737^{203} is 33.
For 512512, the base unit digit is 22 (cyclicity pattern 2, 4, 8, 6). Exponent 160÷4160 \div 4 leaves remainder 00, indicating the 4th4^{\text{th}} power in cyclicity: 24=16→62^4 = 16 \rightarrow 6. For 737737, the base unit digit is 77 (cyclicity pattern 7, 9, 3, 1). Exponent 203÷4203 \div 4 leaves remainder 33, indicating 73=343→37^3 = 343 \rightarrow 3.
2
Add the unit digits inside the parentheses and multiply by the unit digit of 434116434^{116}
Unit digit of the product term is 44.
Sum inside parentheses: 6+3=96 + 3 = 9. For 434116434^{116}, the base unit digit is 44 (cyclicity pattern 4, 6). An even exponent 116116 gives 42=16→64^2 = 16 \rightarrow 6. The product is 9×6=54→49 \times 6 = 54 \rightarrow 4.
3
Subtract the unit digit of 618140618^{140} from the product's unit digit
The final unit digit is 88.
For 618140618^{140}, base unit digit is 88 (cyclicity pattern 8, 4, 2, 6). Exponent 140÷4140 \div 4 leaves remainder 00, giving 84=4096→68^4 = 4096 \rightarrow 6. Subtracting gives 4−64 - 6. Since 4<64 < 6, add 1010 to borrow from the tens place: 14−6=814 - 6 = 8.

Key Concept

Cyclicity of Unit Digits and Modular Subtraction
Estimated Time:1m 30s
Question 173Question →

Let N=2a×3b×7cN = 2^a \times 3^b \times 7^c be a positive integer, where aa, bb, and cc are positive integers. The number of positive factors of NN that are divisible by 66 is 3636, the number of positive factors of NN that are divisible by 1414 is 3232, and the number of positive factors of NN that are divisible by 2121 is 3030. What is the total number of positive factors of NN?

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Answer: 60

Answer

The total number of positive factors of NN is 60.
By setting up the system of equations based on exponent constraints for each divisor (6,14,216, 14, 21), we find a=4,b=3,c=2a=4, b=3, c=2. Substituting these into the total factor formula (a+1)(b+1)(c+1)(a+1)(b+1)(c+1) yields 5×4×3=605 \times 4 \times 3 = 60.

Step-by-Step Solution

1
Formulate algebraic expressions for the constrained factor counts
ab(c+1)=36ab(c+1) = 36, ac(b+1)=32ac(b+1) = 32, and bc(a+1)=30bc(a+1) = 30
A factor 2x×3y×7z2^x \times 3^y \times 7^z is divisible by 6=21×316 = 2^1 \times 3^1 if x≥1x \ge 1, y≥1y \ge 1, z≥0z \ge 0, yielding a⋅b⋅(c+1)a \cdot b \cdot (c+1) factors. Similarly, divisibility by 1414 requires x≥1,z≥1x \ge 1, z \ge 1, giving a⋅c⋅(b+1)a \cdot c \cdot (b+1), and divisibility by 2121 requires y≥1,z≥1y \ge 1, z \ge 1, giving b⋅c⋅(a+1)b \cdot c \cdot (a+1).
2
Expand equations and substitute P=abcP = abc
ab=36−Pab = 36 - P, ac=32−Pac = 32 - P, bc=30−Pbc = 30 - P
Expanding ab(c+1)=abc+ab=36ab(c+1) = abc + ab = 36 gives ab=36−Pab = 36 - P. Applying the same expansion to the other two equations gives expressions for acac and bcbc in terms of PP.
3
Multiply the three equations together and solve for PP
(ab)(ac)(bc)=(abc)2=P2⇒(36−P)(32−P)(30−P)=P2⇒P=24(ab)(ac)(bc) = (abc)^2 = P^2 \Rightarrow (36-P)(32-P)(30-P) = P^2 \Rightarrow P = 24
Testing positive integer values of PP that divide into products reveals that for P=24P = 24, (36−24)(32−24)(30−24)=12×8×6=576=242(36-24)(32-24)(30-24) = 12 \times 8 \times 6 = 576 = 24^2.
4
Calculate the values of abab, acac, bcbc, and solve for aa, bb, cc
ab=12ab = 12, ac=8ac = 8, bc=6bc = 6, giving a=4a = 4, b=3b = 3, c=2c = 2
a2=(ab)(ac)bc=12×86=16⇒a=4a^2 = \frac{(ab)(ac)}{bc} = \frac{12 \times 8}{6} = 16 \Rightarrow a = 4. Similarly, b2=12×68=9⇒b=3b^2 = \frac{12 \times 6}{8} = 9 \Rightarrow b = 3, and c2=8×612=4⇒c=2c^2 = \frac{8 \times 6}{12} = 4 \Rightarrow c = 2.
5
Compute the total number of positive factors of NN
(4+1)(3+1)(2+1)=5×4×3=60(4+1)(3+1)(2+1) = 5 \times 4 \times 3 = 60
The total number of factors of N=2a×3b×7cN = 2^a \times 3^b \times 7^c is (a+1)(b+1)(c+1)(a+1)(b+1)(c+1).

Key Concept

Constrained Factor Counting and Prime Factorization Systems
Question 174Question →

If 3x+2=813^{x + 2} = 81, what is the value of xx?

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Answer: 2

Answer

The value of xx is 2.
Rewriting 81 as 343^4 gives 3x+2=343^{x+2} = 3^4. Since the bases are equal, setting the exponents equal gives x+2=4x + 2 = 4, which solves to x=2x = 2.

Step-by-Step Solution

1
Express the right-hand side with a base of 3
81=3481 = 3^4
Both sides must have the same base to apply the laws of indices.
2
Equate the exponents
x+2=4x + 2 = 4
If am=ana^m = a^n (where a>0a > 0 and a≠1a \neq 1), then m=nm = n.
3
Solve for xx
x=2x = 2
Subtracting 2 from both sides isolates xx.

Key Concept

Equating exponents of equal bases
Question 175Question →

What is the unit digit of the expression K=(377123+428164)×533200−61986K = (377^{123} + 428^{164}) \times 533^{200} - 619^{86}?

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Answer: 8

Answer

The unit digit of the expression is 8.
Using cyclicity principles for each power term, 377123377^{123} yields unit digit 3 (remainder 3), 428164428^{164} yields unit digit 6 (remainder 0 →\rightarrow 4th power), 533200533^{200} yields unit digit 1 (remainder 0 →\rightarrow 4th power), and 61986619^{86} yields unit digit 1 (even power of 9). Combining these gives (3+6)×1−1=8(3 + 6) \times 1 - 1 = 8.

Step-by-Step Solution

1
Find the unit digit of 377123377^{123}
Unit digit is 3
The unit digit of 377 is 7, which follows a cyclicity pattern of 4 (7, 9, 3, 1). Dividing the exponent 123 by 4 leaves a remainder of 3. Thus, the unit digit is given by 73=343→37^3 = 343 \rightarrow 3.
2
Find the unit digit of 428164428^{164}
Unit digit is 6
The unit digit of 428 is 8, which has a cyclicity of 4 (8, 4, 2, 6). The exponent 164 is divisible by 4 (remainder 0), which corresponds to the 4th position in the cyclicity cycle (84=4096→68^4 = 4096 \rightarrow 6).
3
Evaluate the sum inside the parentheses (377123+428164)(377^{123} + 428^{164})
Unit digit is 9
Adding the individual unit digits gives 3+6=93 + 6 = 9.
4
Find the unit digit of 533200533^{200}
Unit digit is 1
The unit digit of 533 is 3, which has a cyclicity of 4 (3, 9, 7, 1). Exponent 200 is divisible by 4 (remainder 0), corresponding to the 4th power in the cycle (34=81→13^4 = 81 \rightarrow 1).
5
Multiply the bracketed sum by 533200533^{200}
Unit digit is 9
Multiplying the unit digits yields 9×1=99 \times 1 = 9.
6
Find the unit digit of 61986619^{86}
Unit digit is 1
The unit digit of 619 is 9, which has a cyclicity of 2 (9 for odd powers, 1 for even powers). Since 86 is even, the unit digit is 1.
7
Subtract 61986619^{86} from the product
Unit digit is 8
Subtracting the unit digits gives 9−1=89 - 1 = 8.

Key Concept

Unit Digit and Cyclicity
Question 176Question →

A district officer allocates a development fund among four sectors: Health, Education, Infrastructure, and Agriculture. Health receives a fraction equal to 0.16‾0.1\overline{6} of the total fund. Education receives 38\frac{3}{8} of the remaining fund after the Health allocation. Infrastructure is directly allocated 0.43‾0.4\overline{3} of the original total fund. The remaining amount is given to Agriculture. If the amount allocated to Infrastructure exceeds the amount allocated to Education by ₹ 87,00087,000, what is the total development fund in Rupees?

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Answer: 720000

Answer

The total development fund is ₹ 720,000.
Converting the recurring decimals gives Health's share as 16\frac{1}{6} and Infrastructure's share as 1330\frac{13}{30}. The remaining fund after Health allocation is 56\frac{5}{6}, making Education's share equal to 38×56=516\frac{3}{8} \times \frac{5}{6} = \frac{5}{16} of the total. The difference between Infrastructure's and Education's shares is 1330−516=29240\frac{13}{30} - \frac{5}{16} = \frac{29}{240}. Equating 29240\frac{29}{240} of the total fund to ₹ 87,000 yields the total fund as ₹ 720,000.

Step-by-Step Solution

1
Convert the recurring decimals into simplified vulgar fractions.
0.16‾=16−190=1590=160.1\overline{6} = \frac{16-1}{90} = \frac{15}{90} = \frac{1}{6} and 0.43‾=43−490=3990=13300.4\overline{3} = \frac{43-4}{90} = \frac{39}{90} = \frac{13}{30}.
Converting mixed recurring decimals to fractions allows exact algebraic calculations.
2
Calculate the fraction of the total fund received by Education.
Remaining fund after Health = 1−16=561 - \frac{1}{6} = \frac{5}{6}. Education's fraction = 38×56=516\frac{3}{8} \times \frac{5}{6} = \frac{5}{16}.
Education receives a fraction of the remaining fund, not the original total fund.
3
Find the difference between Infrastructure's fraction and Education's fraction.
1330−516=104−75240=29240\frac{13}{30} - \frac{5}{16} = \frac{104 - 75}{240} = \frac{29}{240}.
The LCM of denominators 30 and 16 is 240.
4
Set up the linear equation with the given monetary difference to find the total fund FF.
29240×F=87,000  ⟹  F=87,000×24029=3,000×240=720,000\frac{29}{240} \times F = 87,000 \implies F = \frac{87,000 \times 240}{29} = 3,000 \times 240 = 720,000.
Solving for FF yields the original total fund allocation.

Key Concept

Converting mixed recurring decimals to vulgar fractions and evaluating multi-step compound fractional parts of a total quantity.
Question 177Question →

If x=6+25−6−25x = \sqrt{6 + 2\sqrt{5}} - \sqrt{6 - 2\sqrt{5}} and (127)x−5=9y+1\left(\frac{1}{27}\right)^{x - 5} = 9^{y + 1}, what is the value of x+yx + y?

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Answer: 112\frac{11}{2}

Answer

The value of x+yx + y is 112\frac{11}{2}.
Simplifying 6+25\sqrt{6 + 2\sqrt{5}} yields 5+1\sqrt{5} + 1 and 6−25\sqrt{6 - 2\sqrt{5}} yields 5−1\sqrt{5} - 1. Taking their difference gives x=2x = 2. Substituting x=2x = 2 into the exponential equation gives (1/27)−3=273=(33)3=39(1/27)^{-3} = 27^3 = (3^3)^3 = 3^9. The right side is 9y+1=(32)y+1=32y+29^{y+1} = (3^2)^{y+1} = 3^{2y+2}. Equating exponents gives 2y+2=92y + 2 = 9, which leads to y=7/2y = 7/2. Adding xx and yy gives 2+7/2=11/22 + 7/2 = 11/2.

Step-by-Step Solution

1
Simplify the nested surds for xx
6+25=5+1\sqrt{6 + 2\sqrt{5}} = \sqrt{5} + 1 and 6−25=5−1\sqrt{6 - 2\sqrt{5}} = \sqrt{5} - 1
For a nested surd a±2b\sqrt{a \pm 2\sqrt{b}}, express aa as m+nm+n and bb as m⋅nm \cdot n. Here 5+1=65+1=6 and 5⋅1=55 \cdot 1 = 5.
2
Compute the value of xx
x=(5+1)−(5−1)=2x = (\sqrt{5} + 1) - (\sqrt{5} - 1) = 2
Subtracting the second simplified surd from the first cancels out 5\sqrt{5} and adds the integer terms.
3
Express the exponential equation using a common prime base 33
(3−3)2−5=(32)y+1  ⟹  (3−3)−3=32y+2  ⟹  39=32y+2(3^{-3})^{2 - 5} = (3^2)^{y + 1} \implies (3^{-3})^{-3} = 3^{2y + 2} \implies 3^9 = 3^{2y + 2}
Rewriting 127=3−3\frac{1}{27} = 3^{-3} and 9=329 = 3^2 allows direct comparison of exponents.
4
Solve for yy and evaluate x+yx + y
2y+2=9  ⟹  y=722y + 2 = 9 \implies y = \frac{7}{2}, so x+y=2+72=112x + y = 2 + \frac{7}{2} = \frac{11}{2}
Equating exponents with identical bases and adding the resolved values of xx and yy.

Key Concept

Simplification of nested surds a±2b\sqrt{a \pm 2\sqrt{b}} combined with solving exponential equations using common prime bases.
Estimated Time:2m 0s
Question 178Question →

What is the simplified value of the expression (256)0.16×(256)0.09(256)^{0.16} \times (256)^{0.09}?

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Answer: 44

Answer

The simplified value of (256)0.16×(256)0.09(256)^{0.16} \times (256)^{0.09} is 44.
According to the basic laws of indices, multiplying terms with equal bases requires adding their exponents: (256)0.16×(256)0.09=(256)0.16+0.09=(256)0.25(256)^{0.16} \times (256)^{0.09} = (256)^{0.16 + 0.09} = (256)^{0.25}. Expressing 0.250.25 as 14\frac{1}{4} converts the expression to the fourth root of 256256, which equals 44 because 44=2564^4 = 256.

Step-by-Step Solution

1
Apply the law of indices am×an=am+na^m \times a^n = a^{m+n}
(256)0.16+0.09=(256)0.25(256)^{0.16 + 0.09} = (256)^{0.25}
When multiplying exponential terms with identical bases, add their powers.
2
Convert the decimal exponent to a simplified fraction
0.25=25100=140.25 = \frac{25}{100} = \frac{1}{4}
Converting to a fraction simplifies root evaluation.
3
Express 256256 as a base with power 44
256=44256 = 4^4
Writing the base as 444^4 allows straightforward simplification with the exponent 14\frac{1}{4}.
4
Evaluate (44)1/4(4^4)^{1/4}
44×14=41=44^{4 \times \frac{1}{4}} = 4^1 = 4
Apply the power rule (am)n=am×n(a^m)^n = a^{m \times n} to arrive at the final simplified value.

Key Concept

Laws of Indices: Product Rule (am×an=am+na^m \times a^n = a^{m+n}) and Power of a Power Rule ((am)n=amn(a^m)^n = a^{mn})
Question 179Question →

Determine the exact numerical value of the following mathematical expression when evaluated strictly according to the VBODMAS rule:

45−[15+{24÷(3+12 of (16−8−2‾))}]45 - \left[ 15 + \left\{ 24 \div \left( 3 + \frac{1}{2} \text{ of } \left( 16 - \overline{8 - 2} \right) \right) \right\} \right]
Show answer & explanation

Answer: 27

Answer

The simplified numerical value of the given expression is 27.
Evaluating step-by-step using VBODMAS order of precedence (Vinculum -> Brackets -> Of -> Division -> Multiplication -> Addition -> Subtraction) yields the simplified result 27.

Step-by-Step Solution

1
Evaluate the vinculum (bar) operation
\overline{8 - 2} = 6
According to VBODMAS, operations under a vinculum take precedence over standard bracket operations.
2
Simplify the innermost subtraction inside the round brackets
16 - 6 = 10
Subtracted the value obtained from the vinculum within the innermost parentheses.
3
Evaluate the 'of' operator
\frac{1}{2} \text{ of } 10 = 5
The 'of' operator represents multiplication with priority immediately following brackets.
4
Complete the addition in the round brackets
3 + 5 = 8
Evaluated the remaining addition within the round brackets.
5
Perform division inside the curly brackets
24÷8=324 \div 8 = 3
Divided 24 by the simplified result of the round brackets.
6
Perform addition inside the square brackets
15 + 3 = 18
Added 15 to the result from the curly brackets.
7
Perform final subtraction
45 - 18 = 27
Subtracted the square bracket result from 45 to obtain the final simplified value.

Key Concept

Order of Operations (VBODMAS Rule)
Estimated Time:1m 30s
Question 180Question →

What is the unit digit of the expression S=284+497×354−945S = 2^{84} + 4^{97} \times 3^{54} - 9^{45}?

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Answer: 3

Answer

The unit digit of the expression is 3.
The unit digit of 2842^{84} is 6 (since 84 is a multiple of 4, corresponding to the 4th position in cyclicity). The product 497×3544^{97} \times 3^{54} has a unit digit of (4×9)(mod10)=6(4 \times 9) \pmod{10} = 6. The unit digit of 9459^{45} is 9. Substituting these into the original expression yields 6+6−9=36 + 6 - 9 = 3.

Step-by-Step Solution

1
Find the unit digit of 2842^{84}
Unit digit is 6
The cyclicity of 22 is 44 (2,4,8,62, 4, 8, 6). Since 84÷4=2184 \div 4 = 21 with remainder 00, we take the 4th term in the cyclicity pattern, which is 66.
2
Find the unit digit of 4974^{97} and 3543^{54}
Unit digit of 4974^{97} is 4; Unit digit of 3543^{54} is 9
The cyclicity of 44 is 22 (44 for odd powers, 66 for even powers). Since 9797 is odd, 4974^{97} ends in 44. The cyclicity of 33 is 44 (3,9,7,13, 9, 7, 1). Since 54=4×13+254 = 4 \times 13 + 2, remainder is 22, giving 32=93^2 = 9.
3
Find the unit digit of the product 497×3544^{97} \times 3^{54}
Unit digit of product is 6
Multiplying the unit digits: 4×9=364 \times 9 = 36, so the unit digit is 66.
4
Find the unit digit of 9459^{45}
Unit digit is 9
The cyclicity of 99 is 22 (99 for odd powers, 11 for even powers). Since 4545 is odd, 9459^{45} ends in 99.
5
Combine the unit digits following the algebraic operations
Unit digit is 3
Evaluating 6+6−9=12−9=36 + 6 - 9 = 12 - 9 = 3.

Key Concept

Unit Digit and Cyclicity of Exponential Numbers
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