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Basic Numeracy

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Question 181Question →

If x=11+46−11−46x = \sqrt{11 + 4\sqrt{6}} - \sqrt{11 - 4\sqrt{6}} and (116)x2−15=64y+1\left(\frac{1}{16}\right)^{x^2 - 15} = 64^{y + 1}, what is the value of yy?

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Answer: 1

Answer

The value of yy is 11.
Simplifying 11±46\sqrt{11 \pm 4\sqrt{6}} as 8±3\sqrt{8} \pm \sqrt{3} yields x=23x = 2\sqrt{3} and x2=12x^2 = 12. Substituting x2x^2 gives (1/16)−3=163=212(1/16)^{-3} = 16^3 = 2^{12}. Matching exponents with 64y+1=26y+664^{y+1} = 2^{6y+6} yields 12=6y+612 = 6y + 6, giving y=1y = 1.

Step-by-Step Solution

1
Simplify the nested surds for xx
x=23x = 2\sqrt{3}
Express 11±4611 \pm 4\sqrt{6} as 11±224=(8±3)211 \pm 2\sqrt{24} = (\sqrt{8} \pm \sqrt{3})^2, giving 11+46−11−46=(22+3)−(22−3)=23\sqrt{11 + 4\sqrt{6}} - \sqrt{11 - 4\sqrt{6}} = (2\sqrt{2} + \sqrt{3}) - (2\sqrt{2} - \sqrt{3}) = 2\sqrt{3}.
2
Evaluate x2x^2
x2=12x^2 = 12
Squaring x=23x = 2\sqrt{3} yields x2=4×3=12x^2 = 4 \times 3 = 12.
3
Substitute x2=12x^2 = 12 into the exponent of the given equation
(116)−3=64y+1\left(\frac{1}{16}\right)^{-3} = 64^{y + 1}
The exponent x2−15x^2 - 15 evaluates to 12−15=−312 - 15 = -3.
4
Convert both sides to a common prime base (base 2)
212=26y+62^{12} = 2^{6y + 6}
(116)−3=163=(24)3=212\left(\frac{1}{16}\right)^{-3} = 16^3 = (2^4)^3 = 2^{12} and 64y+1=(26)y+1=26y+664^{y + 1} = (2^6)^{y + 1} = 2^{6y + 6}.
5
Equate powers and solve for yy
y=1y = 1
Since the bases are identical, 12=6y+6  ⟹  6y=6  ⟹  y=112 = 6y + 6 \implies 6y = 6 \implies y = 1.

Key Concept

Nested surd evaluation via perfect square conversion combined with solving exponential equations using index rules.
Question 182Question →
What is the simplified numerical value of the mathematical expression given below when evaluated strictly according to the VBODMAS rule?
15% of 200−[36÷{18−(2×7−4‾)}]×3.515\% \text{ of } 200 - \left[ 36 \div \left\{ 18 - \left( 2 \times \overline{7 - 4} \right) \right\} \right] \times 3.5
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Answer: 19.519.5

Answer

The simplified numerical value of the expression is 19.519.5.
Following the VBODMAS hierarchy, we first solve the bar term 7−4‾=3\overline{7-4} = 3, giving (2×3)=6(2 \times 3) = 6 inside the inner bracket. Next, the curly bracket simplifies to 18−6=1218 - 6 = 12. The square bracket simplifies to 36÷12=336 \div 12 = 3. Evaluating 15% of 200=3015\% \text{ of } 200 = 30 and the product 3×3.5=10.53 \times 3.5 = 10.5, we get 30−10.5=19.530 - 10.5 = 19.5.

Step-by-Step Solution

1
Evaluate the expression under the Vinculum (bar)
7−4‾=3\overline{7 - 4} = 3
According to VBODMAS, operations under the vinculum have the highest priority.
2
Evaluate the innermost round bracket ()( )
2×3=62 \times 3 = 6
Resolve the multiplication inside the round brackets.
3
Evaluate the curly bracket {}\{ \}
18−6=1218 - 6 = 12
Perform subtraction inside the curly brackets.
4
Evaluate the square bracket [][ ]
36÷12=336 \div 12 = 3
Perform division inside the square brackets.
5
Calculate the percentage portion
15% of 200=15100×200=3015\% \text{ of } 200 = \frac{15}{100} \times 200 = 30
'Of' represents multiplication in percentage evaluation.
6
Perform multiplication before final subtraction
3×3.5=10.53 \times 3.5 = 10.5
Multiplication takes precedence over final subtraction.
7
Perform final subtraction
30−10.5=19.530 - 10.5 = 19.5
Subtract the result of the bracketed expression term from the percentage term.

Key Concept

Order of Operations (VBODMAS Rule)
Estimated Time:1m 30s
Question 183Question →

What is the unit digit of the numerical expression M=3105×782+896M = 3^{105} \times 7^{82} + 8^{96}?

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Answer: 3

Answer

The unit digit of the given expression is 3.
The unit digit of an exponential term is determined by its base's cyclicity cycle and the exponent modulo 4. For 31053^{105}, 105(mod4)=1  ⟹  31=3105 \pmod 4 = 1 \implies 3^1 = 3. For 7827^{82}, 82(mod4)=2  ⟹  72=49  ⟹  982 \pmod 4 = 2 \implies 7^2 = 49 \implies 9. The product of these terms has a unit digit of (3×9)(mod10)=7(3 \times 9) \pmod{10} = 7. For 8968^{96}, 96(mod4)=096 \pmod 4 = 0, which indicates the 4th position in the cyclicity pattern of 8 (8, 4, 2, 6), yielding a unit digit of 6. Finally, adding the unit digits gives (7+6)(mod10)=3(7 + 6) \pmod{10} = 3.

Step-by-Step Solution

1
Determine the cyclicity and remainder for 31053^{105}
Unit digit is 3
The unit digits of powers of 3 repeat in a cycle of 4 (3, 9, 7, 1). Dividing the exponent by 4 gives 105(mod4)=1105 \pmod 4 = 1, corresponding to 31=33^1 = 3.
2
Determine the cyclicity and remainder for 7827^{82}
Unit digit is 9
The unit digits of powers of 7 repeat in a cycle of 4 (7, 9, 3, 1). Dividing the exponent by 4 gives 82(mod4)=282 \pmod 4 = 2, corresponding to 72=497^2 = 49 (unit digit 9).
3
Calculate the unit digit of the product 3105×7823^{105} \times 7^{82}
Unit digit is 7
Multiply the unit digits of the two factors: (3×9)=27(3 \times 9) = 27, which has a unit digit of 7.
4
Determine the cyclicity and remainder for 8968^{96}
Unit digit is 6
The unit digits of powers of 8 repeat in a cycle of 4 (8, 4, 2, 6). The exponent 96 is completely divisible by 4 (96(mod4)=096 \pmod 4 = 0). When remainder is 0, we use the 4th power in the cycle (84=40968^4 = 4096), which ends in 6.
5
Sum the unit digits of the two main terms
Unit digit is 3
Adding the unit digits gives (7+6)=13(7 + 6) = 13, which has a unit digit of 3.

Key Concept

Determining the unit digit of composite exponential expressions using power cyclicity.
Question 184Question →

If p=12+12+12+…p = \sqrt{12 + \sqrt{12 + \sqrt{12 + \dots}}} and q=12−12−12−…q = \sqrt{12 - \sqrt{12 - \sqrt{12 - \dots}}}, such that 2p+q−z=1642^{p+q-z} = \frac{1}{64}, what is the value of zz?

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Answer: 1313

Answer

The value of zz is 1313.
Solving p=12+pp = \sqrt{12 + p} yields the quadratic equation p2−p−12=0p^2 - p - 12 = 0, giving p=4p = 4. Solving q=12−qq = \sqrt{12 - q} yields q2+q−12=0q^2 + q - 12 = 0, giving q=3q = 3. Adding pp and qq gives p+q=7p + q = 7. Substituting this into 2p+q−z=1642^{p+q-z} = \frac{1}{64} produces 27−z=2−62^{7-z} = 2^{-6}. Equating the indices gives 7−z=−67 - z = -6, which solves to z=13z = 13.

Step-by-Step Solution

1
Evaluate the first infinite nested surd pp
p=4p = 4
Let p=12+pp = \sqrt{12 + p}. Squaring both sides yields p2=12+p  ⟹  p2−p−12=0p^2 = 12 + p \implies p^2 - p - 12 = 0. Factoring gives (p−4)(p+3)=0(p-4)(p+3) = 0. Since p>0p > 0, p=4p = 4.
2
Evaluate the second infinite nested surd qq
q=3q = 3
Let q=12−qq = \sqrt{12 - q}. Squaring both sides yields q2=12−q  ⟹  q2+q−12=0q^2 = 12 - q \implies q^2 + q - 12 = 0. Factoring gives (q+4)(q−3)=0(q+4)(q-3) = 0. Since q>0q > 0, q=3q = 3.
3
Substitute pp and qq into the exponent expression
p+q=7p + q = 7, giving 27−z=1642^{7-z} = \frac{1}{64}
Summing pp and qq gives 4+3=74 + 3 = 7.
4
Solve the exponential equation for zz
z=13z = 13
Express 164\frac{1}{64} as a base 22 power: 164=2−6\frac{1}{64} = 2^{-6}. Equating exponents gives 7−z=−6  ⟹  z=7+6=137 - z = -6 \implies z = 7 + 6 = 13.

Key Concept

Evaluation of infinite nested surds and solving exponential equations using laws of indices
Estimated Time:2m 0s
Question 185Question →

What is the simplified value of the expression (64125)−23\left(\frac{64}{125}\right)^{-\frac{2}{3}}?

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Answer: 2516\frac{25}{16}

Answer

2516\frac{25}{16}
The expression (64125)−23\left(\frac{64}{125}\right)^{-\frac{2}{3}} is simplified by first converting the negative power to positive by inverting the fraction to (12564)23\left(\frac{125}{64}\right)^{\frac{2}{3}}. Taking the cube root yields 54\frac{5}{4}, and squaring that result gives 2516\frac{25}{16}.

Step-by-Step Solution

1
Apply the negative exponent rule a−n=1ana^{-n} = \frac{1}{a^n} to invert the fraction
(64125)−23=(12564)23\left(\frac{64}{125}\right)^{-\frac{2}{3}} = \left(\frac{125}{64}\right)^{\frac{2}{3}}
A negative exponent indicates taking the reciprocal of the base.
2
Express the base numbers as perfect cubes
125=53125 = 5^3 and 64=4364 = 4^3, so 12564=(54)3\frac{125}{64} = \left(\frac{5}{4}\right)^3
Rewriting bases into prime factors with power multiples simplifies fractional exponents.
3
Apply the power of a power rule (am)n=am×n(a^m)^n = a^{m \times n}
\left(\left(\frac{5}{4}\right)^3\right)^{\frac{2}{3}} = \left(\frac{5}{4}\right)^{3 \times \frac{2}{3}} = \left(\frac{5}{4}\right)^2
Multiplying the inner exponent 33 by the outer exponent 23\frac{2}{3} yields 22.
4
Square the fraction
(54)2=2516\left(\frac{5}{4}\right)^2 = \frac{25}{16}
Square both the numerator and the denominator to get the final numerical value.

Key Concept

Negative and Fractional Indices Rules
Question 186Question →

What is the unit digit of the numerical value of the expression N=528144+36389×24465−17951N = 528^{144} + 363^{89} \times 244^{65} - 179^{51}?

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Answer: 9

Answer

9
Evaluating each term's unit digit yields 6 for the first term, 3 for the second term, 4 for the third term, and 9 for the fourth term. Applying BODMAS rules: multiplication of 3 and 4 gives a unit digit of 2. Then adding 6 gives 8. Subtracting 9 from 8 gives a unit digit of 9 (since 18 - 9 = 9). Thus, the correct unit digit is 9.

Step-by-Step Solution

1
Find the unit digit of 528144528^{144}
The unit digit of the base is 8. The cyclicity of 8 is 4. Dividing the exponent 144 by 4 gives a remainder of 0. When remainder is 0, we take the 4th power: 84=40968^4 = 4096, which has a unit digit of 6.
Exponents divisible by the cyclicity length correspond to the 4th power in the cycle.
2
Find the unit digit of 36389363^{89}
The unit digit of the base is 3. The cyclicity of 3 is 4. Dividing 89 by 4 gives a remainder of 1. Thus, the unit digit is 31=33^1 = 3.
The unit digit pattern for powers of 3 repeats every 4 powers.
3
Find the unit digit of 24465244^{65}
The unit digit of the base is 4. The cyclicity of 4 is 2 (powers alternate: 41=4,42=64^1 = 4, 4^2 = 6). Since the exponent 65 is odd, the unit digit is 4.
Powers of 4 with odd exponents always end in 4.
4
Find the unit digit of 17951179^{51}
The unit digit of the base is 9. The cyclicity of 9 is 2 (powers alternate: 91=9,92=19^1 = 9, 9^2 = 1). Since 51 is odd, the unit digit is 9.
Powers of 9 with odd exponents always end in 9.
5
Combine the unit digits using BODMAS rules
Unit digit of expression = [6+(3×4)−9](mod10)=(6+2−9)(mod10)=(8−9)(mod10)=9[6 + (3 \times 4) - 9] \pmod{10} = (6 + 2 - 9) \pmod{10} = (8 - 9) \pmod{10} = 9.
Multiplication is performed before addition/subtraction, and a negative intermediate unit digit is adjusted by adding 10.

Key Concept

Unit digit calculation using cyclicity of digits and order of operations
Estimated Time:1m 30s
Question 187Question →

Consider the positive integer N=60N = 60. Which of the following statements regarding the factors and prime factorization of NN is/are correct?

Select all that apply

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Answer: The total number of positive factors of 6060 is 1212.; The prime factorization of 6060 is 22×31×512^2 \times 3^1 \times 5^1.

Answer

The total number of positive factors of 60 is 12, and the prime factorization of 60 is 22×31×512^2 \times 3^1 \times 5^1.
The statement declaring that the total number of positive factors is 1212 is correct because (2+1)(1+1)(1+1)=12(2+1)(1+1)(1+1) = 12. Furthermore, the statement giving prime factorization as 22×31×512^2 \times 3^1 \times 5^1 is correct as it accurately breaks down 6060 into prime bases.

Step-by-Step Solution

1
Find the prime factorization of 6060
60=22×31×5160 = 2^2 \times 3^1 \times 5^1
Decomposing 6060 into its prime components provides the exponents needed for factor analysis.
2
Calculate the total number of positive factors
Total factors = (2+1)(1+1)(1+1)=3×2×2=12(2+1)(1+1)(1+1) = 3 \times 2 \times 2 = 12
Adding 11 to each prime exponent and multiplying gives the total count of positive factors.
3
Evaluate the distinct prime factors and their sum
Distinct prime factors are 2,3,52, 3, 5, and their sum is 2+3+5=102 + 3 + 5 = 10
Verifying distinct prime sum shows that claiming a sum of 1111 is incorrect.
4
Calculate the number of odd factors
Odd factors = (1+1)(1+1)=4(1+1)(1+1) = 4
Odd factors depend only on odd prime bases (31×513^1 \times 5^1), yielding 44 odd factors (1,3,5,151, 3, 5, 15).

Key Concept

Prime Factorization and Factor Counting Formula
Question 188Question →

If x=10+221−10−221x = \sqrt{10 + 2\sqrt{21}} - \sqrt{10 - 2\sqrt{21}} and 9x2−k=(1243)k−69^{x^2 - k} = \left(\frac{1}{243}\right)^{k - 6}, find the value of kk.

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Answer: 2

Answer

The value of kk is 2.
Simplifying xx yields x=(7+3)−(7−3)=23x = (\sqrt{7} + \sqrt{3}) - (\sqrt{7} - \sqrt{3}) = 2\sqrt{3}, which gives x2=12x^2 = 12. Substituting x2=12x^2 = 12 into the exponential equation gives 912−k=(1/243)k−69^{12 - k} = (1/243)^{k - 6}. Expressing both sides with prime base 3 gives 32(12−k)=3−5(k−6)3^{2(12 - k)} = 3^{-5(k - 6)}, which simplifies to 324−2k=330−5k3^{24 - 2k} = 3^{30 - 5k}. Equating the exponents gives 24−2k=30−5k24 - 2k = 30 - 5k, which solves to k=2k = 2.

Step-by-Step Solution

1
Simplify the nested surds for xx
10+221=7+3\sqrt{10 + 2\sqrt{21}} = \sqrt{7} + \sqrt{3} and 10−221=7−3\sqrt{10 - 2\sqrt{21}} = \sqrt{7} - \sqrt{3}
Nested surds of the form a±2b\sqrt{a \pm 2\sqrt{b}} simplify to m±n\sqrt{m} \pm \sqrt{n} where m+n=am + n = a and m⋅n=bm \cdot n = b.
2
Evaluate xx and calculate x2x^2
x=23x = 2\sqrt{3} and x2=12x^2 = 12
Subtracting the two surds gives (7+3)−(7−3)=23(\sqrt{7} + \sqrt{3}) - (\sqrt{7} - \sqrt{3}) = 2\sqrt{3}. Squaring 232\sqrt{3} gives (2)2×3=12(2)^2 \times 3 = 12.
3
Express both sides of the exponential equation with base 3
324−2k=330−5k3^{24 - 2k} = 3^{30 - 5k}
Since 9=329 = 3^2 and 243=35243 = 3^5, 912−k=(32)12−k=324−2k9^{12 - k} = (3^2)^{12 - k} = 3^{24 - 2k} and (1/243)k−6=(3−5)k−6=330−5k(1/243)^{k - 6} = (3^{-5})^{k - 6} = 3^{30 - 5k}.
4
Equate the powers and solve for kk
k=2k = 2
Since bases are equal (3>0,3≠13 > 0, 3 \neq 1), setting exponents equal gives 24−2k=30−5k  ⟹  3k=6  ⟹  k=224 - 2k = 30 - 5k \implies 3k = 6 \implies k = 2.

Key Concept

Simplification of nested square root surds a±2b\sqrt{a \pm 2\sqrt{b}} combined with equating exponents using prime base factorization.
Question 189Question →

What is the numerical value of the mathematical expression given below when evaluated strictly according to the VBODMAS rule?

18% of 250+[34−{12×2.5−(14÷2+8−3‾)}]18\% \text{ of } 250 + \left[ 34 - \left\{ 12 \times 2.5 - \left( 14 \div 2 + \overline{8 - 3} \right) \right\} \right]
Show answer & explanation

Answer: 61

Answer

61
Following the VBODMAS priority rule, we first clear the vinculum 8−3‾=5\overline{8-3}=5. Next, within the round brackets, division precedes addition: 14÷2+5=7+5=1214 \div 2 + 5 = 7 + 5 = 12. Inside the curly brackets, multiplication precedes subtraction: 12×2.5−12=30−12=1812 \times 2.5 - 12 = 30 - 12 = 18. The square bracket reduces to 34−18=1634 - 18 = 16. Evaluating 18%18\% of 250250 gives 4545. Adding 45+1645 + 16 yields the correct final result of 61.

Step-by-Step Solution

1
Evaluate the vinculum bar term
\overline{8 - 3} = 5
According to VBODMAS, operations under a vinculum (bar) must be simplified first.
2
Simplify the innermost round brackets
(14 \div 2 + 5) = 7 + 5 = 12
Division takes precedence over addition inside the parentheses.
3
Simplify the curly brackets
\{ 12 \times 2.5 - 12 \} = 30 - 12 = 18
Multiplication is performed before subtraction inside the brace group.
4
Simplify the square brackets
[ 34 - 18 ] = 16
Complete the bracket hierarchy by evaluating the outer square brackets.
5
Evaluate the 'of' percentage term
18\% \text{ of } 250 = \frac{18}{100} \times 250 = 45
The percentage 'of' operator is calculated as multiplication.
6
Perform the final addition
45 + 16 = 61
Combine the percentage term with the simplified bracket result.

Key Concept

VBODMAS Rule (Vinculum, Brackets, Of, Division, Multiplication, Addition, Subtraction)
Question 190Question →

Find the unit digit of the numerical expression N=(1!+2!+3!+⋯+50!)25+888N = (1! + 2! + 3! + \dots + 50!)^{25} + 8^{88}.

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Answer: 9

Answer

The unit digit of the given expression is 9.
All factorials from 5!5! to 50!50! end in 0, so the sum (1!+2!+3!+⋯+50!)(1! + 2! + 3! + \dots + 50!) ends in 1+2+6+24=331 + 2 + 6 + 24 = 33, which has a unit digit of 3. Raising 3 to the power 25 yields a unit digit of 3 because 25(mod4)=125 \pmod 4 = 1. For 8888^{88}, since 88 is divisible by 4 (88(mod4)=088 \pmod 4 = 0), we take the 4th term of the cyclicity of 8, which is 6. Summing the unit digits yields 3+6=93 + 6 = 9.

Step-by-Step Solution

1
Determine the unit digit of the base sum (1!+2!+3!+⋯+50!)(1! + 2! + 3! + \dots + 50!).
Unit digit of base is 3.
Factorials 5!5! and above all end in 0 (5!=1205! = 120), so only 1!+2!+3!+4!=331! + 2! + 3! + 4! = 33 contributes to the unit digit.
2
Calculate the unit digit of 3253^{25}.
Unit digit is 3.
The cyclicity pattern of 3 has period 4 (3, 9, 7, 1). Exponent 25(mod4)=125 \pmod 4 = 1, giving 31=33^1 = 3.
3
Calculate the unit digit of 8888^{88}.
Unit digit is 6.
The cyclicity pattern of 8 has period 4 (8, 4, 2, 6). Exponent 88(mod4)=088 \pmod 4 = 0, so we take the 4th position in the cycle, which is 6.
4
Add the unit digits of both terms.
Final unit digit is 9.
3+6=93 + 6 = 9, which gives a unit digit of 9.

Key Concept

Cyclicity of numbers and factorial unit digit properties
Question 191Question →

A logistics company assigns identification codes to container shipments based on a positive integer NN. The number NN has exactly 2424 positive factors, and its prime factorization contains only the prime factors 22, 33, and 77. If NN is a multiple of 1414 and the number of odd positive factors of NN is 66, what is the smallest possible value of NN?

Show answer & explanation

Answer: 504504

Answer

The smallest possible value of NN is 504504.
The number 504504 has prime factorization 23×32×712^3 \times 3^2 \times 7^1. Total positive factors = (3+1)(2+1)(1+1)=24(3+1)(2+1)(1+1) = 24, and odd positive factors = (2+1)(1+1)=6(2+1)(1+1) = 6. It is divisible by 1414 (504=14×36504 = 14 \times 36) and is the smaller of the two valid numbers (504504 and 11761176).

Step-by-Step Solution

1
Express NN in terms of its prime factors.
N=2a×3b×7cN = 2^a \times 3^b \times 7^c, where a≥1,b≥1,c≥1a \ge 1, b \ge 1, c \ge 1 as NN contains prime factors 2,3,2, 3, and 77, and is a multiple of 14=2×714 = 2 \times 7.
The question specifies that NN is divisible by 1414 and contains prime factors 22, 33, and 77.
2
Formulate equations for total factors and odd factors.
Total positive factors T(N)=(a+1)(b+1)(c+1)=24T(N) = (a + 1)(b + 1)(c + 1) = 24. Odd positive factors depend only on odd prime powers (3b×7c3^b \times 7^c), so (b+1)(c+1)=6(b + 1)(c + 1) = 6.
Odd factors correspond to choosing 20=12^0 = 1 for the even prime factor.
3
Solve for exponent aa.
(a+1)×6=24  ⟹  a+1=4  ⟹  a=3(a + 1) \times 6 = 24 \implies a + 1 = 4 \implies a = 3.
Dividing total factors by odd factors isolates the term (a+1)(a + 1).
4
Find possible values for bb and cc.
Since (b+1)(c+1)=6(b + 1)(c + 1) = 6 with b,c≥1b, c \ge 1, the possible factor pairs for (b+1,c+1)(b+1, c+1) are (2,3)(2, 3) or (3,2)(3, 2). Case 1: b=1,c=2  ⟹  N=23×31×72=1176b=1, c=2 \implies N = 2^3 \times 3^1 \times 7^2 = 1176. Case 2: b=2,c=1  ⟹  N=23×32×71=504b=2, c=1 \implies N = 2^3 \times 3^2 \times 7^1 = 504.
Integer factorizations of 66 give two distinct exponent assignments.
5
Determine the minimum value.
Comparing 504504 and 11761176, the minimum value is 504504.
Assigning the larger exponent to the smaller prime base (33 instead of 77) minimizes the product.

Key Concept

Factors and Prime Factorization Constraints
Estimated Time:2m 0s
Question 192Question →

What is the total number of positive even factors of the integer 180180?

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Answer: 12

Answer

The total number of positive even factors of 180180 is 1212.
The prime factorization of 180180 is 22×32×512^2 \times 3^2 \times 5^1. An even factor must be divisible by 22, so the power of 22 must be either 11 or 22 (22 choices). The power of 33 can be 0,1,0, 1, or 22 (33 choices), and the power of 55 can be 00 or 11 (22 choices). Total even factors = 2×3×2=122 \times 3 \times 2 = 12.

Step-by-Step Solution

1
Express 180180 in terms of its prime factors.
180=22×32×51180 = 2^2 \times 3^2 \times 5^1
Prime factorization allows systematic calculation of factor counts using exponent combinations.
2
Determine the number of valid choices for each prime factor's exponent to form an even factor.
Choices for exponent of 22: 11 or 22 (22 choices). Choices for exponent of 33: 0,10, 1, or 22 (33 choices). Choices for exponent of 55: 00 or 11 (22 choices).
An even factor must contain at least one factor of 22, so the exponent of 22 cannot be 00.
3
Multiply the counts of choices together.
2×3×2=122 \times 3 \times 2 = 12
By the fundamental principle of counting, multiplying the available choices for each prime factor gives the total number of even factors.

Key Concept

Counting Even Factors using Prime Factorization
Question 193Question →

What is the simplified numerical value of the mathematical expression given below when evaluated strictly according to the VBODMAS rule?

16.5+{4.5×[12−(3.2+4.8−1.6‾÷0.8)]}16.5 + \left\{ 4.5 \times \left[ 12 - \left( 3.2 + \overline{4.8 - 1.6} \div 0.8 \right) \right] \right\}
Show answer & explanation

Answer: 38.138.1

Answer

The simplified numerical value of the expression is 38.138.1.
Following the VBODMAS priority sequence: resolving the vinculum gives 3.23.2; evaluating division inside the round bracket yields 3.2÷0.8=43.2 \div 0.8 = 4; finishing the round bracket yields 3.2+4=7.23.2 + 4 = 7.2; evaluating the square bracket gives 12−7.2=4.812 - 7.2 = 4.8; evaluating the curly bracket gives 4.5×4.8=21.64.5 \times 4.8 = 21.6; finally, adding 16.5+21.616.5 + 21.6 yields 38.138.1.

Step-by-Step Solution

1
Evaluate the expression under the vinculum (bar)
4.8−1.6‾=3.2\overline{4.8 - 1.6} = 3.2
According to the VBODMAS rule, operations under a vinculum take highest priority.
2
Evaluate the division inside the innermost round bracket
3.2÷0.8=43.2 \div 0.8 = 4
Division takes precedence over addition within parentheses.
3
Complete the evaluation inside the round bracket
3.2+4=7.23.2 + 4 = 7.2
Add the remaining terms within the round bracket.
4
Evaluate the expression inside the square bracket
12−7.2=4.812 - 7.2 = 4.8
Subtract the result of the round bracket from 1212.
5
Evaluate the multiplication inside the curly bracket
4.5×4.8=21.64.5 \times 4.8 = 21.6
Multiply the term outside the square bracket by the simplified square bracket value.
6
Perform the final addition
16.5+21.6=38.116.5 + 21.6 = 38.1
Add the initial constant to the simplified curly bracket term.

Key Concept

Order of Operations (VBODMAS Rule)
Estimated Time:1m 30s
Question 194Question →

A teacher wants to divide 8484 students into equal groups such that the number of students in each group is a prime factor of 8484. What is the sum of all distinct prime numbers that can represent the size of these groups?

Show answer & explanation

Answer: 1212

Answer

The sum of all distinct prime group sizes is 1212.
The prime factorization of 8484 is 22×3×72^2 \times 3 \times 7. The distinct prime factors are 22, 33, and 77. Summing these distinct prime numbers yields 2+3+7=122 + 3 + 7 = 12.

Step-by-Step Solution

1
Find the prime factorization of 8484.
84=22×31×7184 = 2^2 \times 3^1 \times 7^1.
Decomposing 8484 into prime factors reveals all prime numbers that divide 8484 exactly.
2
Identify the distinct prime factors of 8484.
The distinct prime factors are 22, 33, and 77.
The question specifies distinct prime factors, so each prime base is listed once regardless of its exponent.
3
Calculate the sum of these distinct prime factors.
2+3+7=122 + 3 + 7 = 12.
Adding the distinct prime factors gives the sum of all possible prime group sizes.

Key Concept

Prime Factorization and Distinct Prime Factors
Question 195Question →

Let NN be a positive integer with the prime factorization N=24×32×53N = 2^4 \times 3^2 \times 5^3. Which of the following statements regarding the positive factors of NN are correct?

Select all that apply

Show answer & explanation

Answer: The total number of positive factors of NN that are perfect squares is 1212.; The number of positive factors of NN that are divisible by 1515 is 3030.

Answer

The correct statements are that the total number of positive factors of NN that are perfect squares is 1212, and the number of positive factors of NN that are divisible by 1515 is 3030.
The statement asserting that NN has 1212 factors that are perfect squares is correct because choosing even powers (0,2,40, 2, 4 for 22; 0,20, 2 for 33; and 0,20, 2 for 55) yields 3×2×2=123 \times 2 \times 2 = 12 factors. The statement asserting that 3030 factors are divisible by 1515 is also correct because forcing exponents of 33 and 55 to be at least 11 yields 5×2×3=305 \times 2 \times 3 = 30 factors.

Step-by-Step Solution

1
Analyze the prime factorization N=24×32×53N = 2^4 \times 3^2 \times 5^3
Any positive factor of NN is of the form 2a×3b×5c2^a \times 3^b \times 5^c, where 0≤a≤40 \le a \le 4, 0≤b≤20 \le b \le 2, and 0≤c≤30 \le c \le 3.
By the Fundamental Theorem of Arithmetic, factors inherit prime bases with constrained exponents.
2
Count the number of factors that are perfect squares
For a perfect square, a∈{0,2,4}a \in \{0, 2, 4\} (3 choices), b∈{0,2}b \in \{0, 2\} (2 choices), and c∈{0,2}c \in \{0, 2\} (2 choices). Total square factors =3×2×2=12= 3 \times 2 \times 2 = 12.
All prime exponents in a perfect square must be even non-negative integers.
3
Count the number of factors divisible by 1515
Since 15=31×5115 = 3^1 \times 5^1, a∈{0,1,2,3,4}a \in \{0, 1, 2, 3, 4\} (5 choices), b∈{1,2}b \in \{1, 2\} (2 choices), and c∈{1,2,3}c \in \{1, 2, 3\} (3 choices). Total factors divisible by 15=5×2×3=3015 = 5 \times 2 \times 3 = 30.
Divisibility by 1515 mandates at least one factor of 33 and at least one factor of 55.
4
Verify even and odd factor counts to evaluate the remaining claims
Even factors require a≥1a \ge 1, giving 4×3×4=484 \times 3 \times 4 = 48 even factors. Odd factors require a=0a = 0, giving 1×3×4=121 \times 3 \times 4 = 12 odd factors.
Evaluating a≥1a \ge 1 vs a=0a = 0 confirms that claims specifying 5050 even factors and 1515 odd factors are incorrect.

Key Concept

Counting factors with constrained exponents in prime factorizations
Question 196Question →
What is the final numerical value of the following expression when evaluated strictly according to the VBODMAS rule?
12.5×4−[16+{25÷(3.5+2.8−1.3‾)}]12.5 \times 4 - \left[ 16 + \left\{ 25 \div \left( 3.5 + \overline{2.8 - 1.3} \right) \right\} \right]
Show answer & explanation

Answer: 29

Answer

The simplified numerical value of the expression is 29.
Applying VBODMAS step-by-step resolves the vinculum first (2.8 - 1.3 = 1.5), then the round brackets (3.5 + 1.5 = 5), curly brackets (25 / 5 = 5), square brackets (16 + 5 = 21), and finally the multiplication and subtraction (50 - 21 = 29).

Step-by-Step Solution

1
Evaluate the vinculum (bar) operation first
\overline{2.8 - 1.3} = 1.5
According to VBODMAS, the vinculum takes top priority over all other operations.
2
Simplify the expression inside the innermost round brackets
3.5 + 1.5 = 5
Next in hierarchy are the round brackets (parentheses).
3
Simplify the expression inside the curly brackets
25÷5=525 \div 5 = 5
Evaluate operations inside curly braces next.
4
Simplify the expression inside the square brackets
16 + 5 = 21
Complete the evaluation of the outer square brackets.
5
Perform the multiplication and final subtraction outside the brackets
12.5 \times 4 = 50 \text{ and then } 50 - 21 = 29
Perform multiplication before final subtraction following standard operator precedence.

Key Concept

VBODMAS Rule (Vinculum, Brackets, Orders, Division, Multiplication, Addition, Subtraction)
Estimated Time:1m 30s
Question 197Question →

A positive integer NN has the prime factorization N=2a×3b×5cN = 2^a \times 3^b \times 5^c, where aa, bb, and cc are positive integers. It is known that NN has exactly 4848 positive factors, N/2N/2 has 3636 positive factors, and N/5N/5 has 4040 positive factors. What is the total number of positive factors of N/3N/3?

Show answer & explanation

Answer: 24

Answer

The total number of positive factors of N/3N/3 is 24.
By writing the factor formula (a+1)(b+1)(c+1)=48(a+1)(b+1)(c+1) = 48 and comparing it with d(N/2)=a(b+1)(c+1)=36d(N/2) = a(b+1)(c+1) = 36 and d(N/5)=(a+1)(b+1)c=40d(N/5) = (a+1)(b+1)c = 40, we uniquely determine the exponents a=3a = 3, b=1b = 1, and c=5c = 5. Dividing NN by 33 changes the prime exponent of 33 to 00. Applying the factor count formula to 23×30×552^3 \times 3^0 \times 5^5 yields (3+1)(0+1)(5+1)=24(3+1)(0+1)(5+1) = 24.

Step-by-Step Solution

1
Express the total number of factors of NN using its prime factorization
d(N)=(a+1)(b+1)(c+1)=48d(N) = (a+1)(b+1)(c+1) = 48
The number of positive factors of a number p1e1p2e2…pkekp_1^{e_1} p_2^{e_2} \dots p_k^{e_k} is given by (e1+1)(e2+1)…(ek+1)(e_1+1)(e_2+1)\dots(e_k+1).
2
Use the factor counts of N/2N/2 and N/5N/5 to determine the exponents aa and cc
a=3a = 3 and c=5c = 5
Since d(N/2)=a(b+1)(c+1)=36d(N/2) = a(b+1)(c+1) = 36, taking the ratio d(N/2)/d(N)d(N/2)/d(N) yields a/(a+1)=36/48=3/4a/(a+1) = 36/48 = 3/4, which gives a=3a = 3. Similarly, d(N/5)=(a+1)(b+1)c=40d(N/5) = (a+1)(b+1)c = 40, so d(N/5)/d(N)=c/(c+1)=40/48=5/6d(N/5)/d(N) = c/(c+1) = 40/48 = 5/6, giving c=5c = 5.
3
Solve for the remaining exponent bb
b=1b = 1
Substituting a=3a = 3 and c=5c = 5 into d(N)=48d(N) = 48 gives (3+1)(b+1)(5+1)=48  ⟹  24(b+1)=48  ⟹  b+1=2  ⟹  b=1(3+1)(b+1)(5+1) = 48 \implies 24(b+1) = 48 \implies b+1 = 2 \implies b = 1.
4
Calculate the number of positive factors of N/3N/3
d(N/3)=(3+1)(0+1)(5+1)=4×1×6=24d(N/3) = (3+1)(0+1)(5+1) = 4 \times 1 \times 6 = 24
Dividing NN by 33 reduces the exponent of 33 from b=1b = 1 to b−1=0b-1 = 0, so N/3=23×30×55N/3 = 2^3 \times 3^0 \times 5^5.

Key Concept

Prime Factorization and Number of Factors
Estimated Time:2m 0s
Question 198Question →

Consider the positive integer 9090. Which of the following statements regarding the factors and prime factorization of 9090 are correct?

Select all that apply

Show answer & explanation

Answer: The prime factorization of 9090 is 21×32×512^1 \times 3^2 \times 5^1.; The total number of positive factors of 9090 is 1212.

Answer

The prime factorization of 90 is 2^1 × 3^2 × 5^1, and the total number of positive factors of 90 is 12.
The prime factorization of 90 is indeed 2^1 × 3^2 × 5^1. The total number of positive factors is calculated as (1+1)(2+1)(1+1) = 12, making both statement options correct.

Step-by-Step Solution

1
Find the prime factorization of 90.
90 = 2 × 45 = 2 × 3^2 × 5 = 2^1 × 3^2 × 5^1.
Decomposing 90 into prime base powers establishes the prime factorization.
2
Calculate the total number of positive factors using exponents.
Total factors = (1 + 1)(2 + 1)(1 + 1) = 2 × 3 × 2 = 12.
Adding 1 to each prime exponent and multiplying gives the total factor count.
3
Verify even factors and prime factor sum.
Even factors = 1 × (2 + 1)(1 + 1) = 6. Distinct prime sum = 2 + 3 + 5 = 10.
Confirming additional properties disproves the incorrect statements.

Key Concept

Prime factorization and formulas for determining total and specific factor counts of an integer
Question 199Question →

A rectangular courtyard measuring 105 cm105\text{ cm} in length and 135 cm135\text{ cm} in width is to be completely paved with identical square tiles without cutting any tile. What is the maximum possible side length of each square tile?

Show answer & explanation

Answer: 15 cm15\text{ cm}

Answer

The maximum possible side length of each square tile is 15 cm15\text{ cm}.
The maximum side length of a square tile that fits exact integer counts along both length and width is given by HCF(105,135)\text{HCF}(105, 135). By prime factorization, 105=3×5×7105 = 3 \times 5 \times 7 and 135=33×5135 = 3^3 \times 5. The common prime factors are 33 and 55, each with lowest power 11. Therefore, HCF=3×5=15 cm\text{HCF} = 3 \times 5 = 15\text{ cm}.

Step-by-Step Solution

1
Identify the required mathematical operation
To find the maximum side length of identical square tiles that fit whole dimensions of 105 cm105\text{ cm} and 135 cm135\text{ cm}, find the Highest Common Factor (HCF) of 105105 and 135135.
The side of each square tile must divide both the length and the width completely, and to maximize tile size, we need the largest such divisor.
2
Perform prime factorization of both numbers
105=3×5×7105 = 3 \times 5 \times 7
135=33×5=3×3×3×5135 = 3^3 \times 5 = 3 \times 3 \times 3 \times 5
Expressing numbers as products of prime factors allows easy identification of common factors.
3
Calculate the HCF using prime powers
HCF(105,135)=31×51=15\text{HCF}(105, 135) = 3^1 \times 5^1 = 15
HCF is obtained by taking the product of the smallest powers of all common prime factors.

Key Concept

Application of HCF and Prime Factorization in Spatial Tiling Problems
Estimated Time:45s
Question 200Question →

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 7272?

Show answer & explanation

Answer: 21

Answer

21
The correct answer is 21. By analyzing the exponent bounds of 2a×3b×5c2^a \times 3^b \times 5^c, there are 48 factors of NN divisible by 1212 (a≥2,b≥1a \ge 2, b \ge 1). Among these, 27 factors are also divisible by 7272 (a≥3,b≥2a \ge 3, b \ge 2). Subtracting the two counts gives 48−27=2148 - 27 = 21 factors divisible by 12 but not by 72.

Step-by-Step Solution

1
Represent the general form of a factor of NN
Every positive factor of N=25×34×52N = 2^5 \times 3^4 \times 5^2 can be uniquely written as f=2a×3b×5cf = 2^a \times 3^b \times 5^c, where a∈{0,1,2,3,4,5}a \in \{0, 1, 2, 3, 4, 5\}, b∈{0,1,2,3,4}b \in \{0, 1, 2, 3, 4\}, and c∈{0,1,2}c \in \{0, 1, 2\}.
By the Fundamental Theorem of Arithmetic, any factor of a number in prime-factored form takes exponents bounded by the original number's exponents.
2
Identify exponent constraints for divisibility by 12
Since 12=22×3112 = 2^2 \times 3^1, for ff to be divisible by 1212, we must have a≥2a \ge 2 and b≥1b \ge 1.
Divisibility requires prime factor exponents in the factor to be at least as large as those in the divisor.
3
Calculate the total number of factors divisible by 12
Number of choices for aa: 5−2+1=45 - 2 + 1 = 4 (values 2,3,4,52, 3, 4, 5). Number of choices for bb: 4−1+1=44 - 1 + 1 = 4 (values 1,2,3,41, 2, 3, 4). Number of choices for cc: 2−0+1=32 - 0 + 1 = 3 (values 0,1,20, 1, 2). Total count = 4×4×3=484 \times 4 \times 3 = 48.
By the multiplication principle of counting, independent choices for each exponent are multiplied.
4
Identify exponent constraints and count factors divisible by 72
Since 72=23×3272 = 2^3 \times 3^2, a factor is divisible by 7272 if a≥3a \ge 3 and b≥2b \ge 2. Number of choices: a∈{3,4,5}a \in \{3, 4, 5\} (3 choices), b∈{2,3,4}b \in \{2, 3, 4\} (3 choices), c∈{0,1,2}c \in \{0, 1, 2\} (3 choices). Total count = 3×3×3=273 \times 3 \times 3 = 27.
Every factor of NN that is divisible by 72 is inherently also a factor divisible by 12, forming a strict subset.
5
Apply the principle of subtraction to find the required count
Required count = (Factors divisible by 12) - (Factors divisible by 72) = 48−27=2148 - 27 = 21.
The set of factors divisible by 12 but not by 72 is the set difference between factors divisible by 12 and factors divisible by 72.

Key Concept

Counting Factors with Divisibility Constraints using Prime Factorization
Estimated Time:2m 0s
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