Solved Examples: Numbers

We shall solve this question using two methods.
If you really do not know how to begin to solve this question, use the first method.
For faster solution because the ACT is a timed test, use the second method.

First Method - Arithmetic method (Test Numbers)

$ a\:\: and\:\: b\:\: are\:\: real\:\: numbers \\[3ex] a \gt 0, b \lt 0 \\[3ex] Let\:\: a = 1, b = -1 \\[3ex] |a| - |b| = |1| - |-1| = 1 - 1 = 0 \\[3ex] Test \\[3ex] |a - b| = |1 - (-1)| = |1 + 1| = |2| = 2 \:\: NO \\[3ex] |a + b| = |1 + (-1)| = |1 - 1| = |0| = 0 \:\: Maybe \\[3ex] |a| + |b| = |1| + |-1| = 1 + 1 = 2 \:\: NO \\[3ex] a - b = 1 - (-1) = 1 + 1 = 2 \:\: NO \\[3ex] a + b = 1 + (-1) = 1 - 1 = 0 \:\: Maybe \\[3ex] $ So, we are left with two options: $|a + b|$ and $a + b$
Specific to the question, $a$ and $b$ are real numbers.
Their absolute values are also real numbers.
The subtraction of their absolute values should give us real numbers, rather than the absolute value of real numbers.
In that sense, $a + b$ is a better answer.

Second Method - Algebraic method

$ a\:\: and\:\: b\:\: are\:\: real\:\: numbers \\[3ex] a \gt 0, b \lt 0 \\[3ex] $ Be definition
$|a|$ means that $a$ could be positive, zero, or negative.
However, from the question; $a \gt 0 \implies a$ is positive
$|b|$ means that $b$ could be positive, zero, or negative.
However, from the question; $b \lt 0 \implies b$ is negative
This means:
$ |a| - |b| \\[3ex] = (+a) - (-b) \\[3ex] = a - (-b) \\[3ex] = a + b $

We shall solve this question using the Arithmetic Method by testing numbers.
We shall use maximum numbers to test.

Student: Why do we have to use "maximum numbers" to test?
Teacher: We shall use "maximum numbers" to test because the question is asking for the greatest possible integer value
Keep in mind that $x$ and $y$ do not have to be integer values.
But, we are interested in the integer value of their sum

$ \underline{Arithmetic\:\:Method:\:\:Test\:\:Numbers} \\[3ex] y \lt 4 \\[3ex] Let\:\: y = 3.9 \\[3ex] x \lt y \\[3ex] Let\:\: x = 3.8 \\[3ex] x + y = 3.9 + 3.8 = 7.7 \\[3ex] $ Greatest possible integer value = $7$

Student: Why is the greatest possible integer value not equal to $8$?
$ Teacher:\:\: 7.7 \ne 8 \\[3ex] 7.7 \gt 7 \\[3ex] 7.7 \lt 8 $

$ 5,34\;million \\[3ex] = 5.34 * 1000000 \\[3ex] = 5340000 $

$ ||-2| - |-10|| \\[3ex] |-2| = 2 \\[3ex] |-10| = 10 \\[3ex] \rightarrow |2 - 10| \\[3ex] = |-8| \\[3ex] = 8 $

$xy = 0$ means that:
Either $x = 0$ OR $y = 0$ OR both $x = 0$ AND $y = 0$
This is known as the Zero Product Property

$\dfrac{x}{y} = 0$ means that $x = 0$
$0$ divided by anything is $0$
However, $y$ cannot be $0$.
Dividing anything by $0$ is undefined.

Combining the two conditions:
$x = 0 \:\:and\:\: y \ne 0$

$ \sqrt{188 + \sqrt{51 + \sqrt{169}}} \\[3ex] \sqrt{169} = 13 \\[3ex] 51 + 13 = 64 \\[3ex] \sqrt{64} = 8 \\[3ex] 188 + 8 = 196 \\[3ex] \sqrt{196} = 14 $

$ (\sqrt{0.7} + \sqrt{70})^2 \\[3ex] = (\sqrt{0.7} + \sqrt{70})(\sqrt{0.7} + \sqrt{70}) \\[3ex] \sqrt{0.7} * \sqrt{0.7} = (\sqrt{0.7})^2 = 0.7 \\[3ex] \sqrt{0.7} * \sqrt{70} = \sqrt{0.7 * 70} = \sqrt{0.7 * 7 * 10} = \sqrt{7 * 7} = \sqrt{49} = 7 \\[3ex] \sqrt{70} * \sqrt{0.7} = \sqrt{70 * 0.7} = \sqrt{7 * 10 * 0.7} = \sqrt{7 * 7} = \sqrt{49} = 7 \\[3ex] \sqrt{70} * \sqrt{70} = (\sqrt{70})^2 = 70 \\[3ex] = 0.7 + 7 + 7 + 70 \\[3ex] = 84.7 $

$ |9(-6) + 5(4)| \\[3ex] = |-54 + 20| \\[3ex] = |-34| \\[3ex] = 34 $

$ \dfrac{5 * 2 - 3^2}{[2^2 - (-7)]^2} \\[5ex] \underline{Numerator} \\[3ex] 5 * 2 - 3^2 \\[3ex] = 5 * 2 - 9 \\[3ex] = 10 - 9 \\[3ex] = 1 \\[3ex] \underline{Denominator} \\[3ex] [2^2 - (-7)]^2 \\[3ex] = [4 - (-7)]^2 \\[3ex] = [4 + 7]^2 \\[3ex] = 11^2 \\[3ex] = 121 \\[3ex] \therefore \dfrac{5 * 2 - 3^2}{[2^2 - (-7)]^2} = \dfrac{1}{121} $

$ 6^2 - 64 \div 4^2 * 7 - 5 \\[3ex] = 36 - 64 \div 16 * 7 - 5 \\[3ex] = 36 - 4 * 7 - 5 \\[3ex] = 36 - 28 - 5 \\[3ex] = 8 - 5 \\[3ex] = 3 $

$ 4 - 6[-4(5 - 7) - 7(5 - 4)] \\[3ex] = 4 - 6[-4(-2) - 7(1)] \\[3ex] = 4 - 6[8 - 7] \\[3ex] = 4 - 6(1) \\[3ex] = 4 - 6 \\[3ex] = -2 $

$ (12.8)^2 - (30 \div 0.375) \\[3ex] = 163.84 - 80 \\[3ex] = 83.84 $

$ 1st\:\:7 \\[3ex] Preceed\:\:3 = prime\:\:number \\[3ex] Succeed\:\:4 = even\:\:number \\[3ex] NO \\[3ex] 2nd\:\:7 \\[3ex] Preceed\:\:5 = prime\:\:number \\[3ex] Succeed\:\:3 = odd\:\:number \\[3ex] NO \\[3ex] 3rd\:\:7 \\[3ex] Preceed\:\:2 = prime\:\:number \\[3ex] Succeed\:\:1 = odd\:\:number \\[3ex] NO \\[3ex] 4th\:\:7 \\[3ex] Preceed\:\:8 = not\:\:prime\:\:number \\[3ex] Succeed\:\:5 = odd\:\:number \\[3ex] NO \\[3ex] 5th\:\:7 \\[3ex] Preceed\:\:4 = not\:\:prime\:\:number \\[3ex] Succeed\:\:5 = odd\:\:number \\[3ex] NO \\[3ex] 6th\:\:7 \\[3ex] Preceed\:\:4 = not\:\:prime\:\:number \\[3ex] Succeed\:\:2 = even\:\:number \\[3ex] YES \\[3ex] There\:\:is\:\:only\:\:one\:\:7 $

$ Let\:\:the\:\:number = x \\[3ex] \sqrt{x} = 9.2371 \\[3ex] Square\:\:both\:\:sides \\[3ex] (\sqrt{x})^2 = 9.2371^2 \\[3ex] x \gt 9^2 \\[3ex] x \lt 10^2 \\[3ex] 9^2 \lt x \lt 10^2 \\[3ex] 81 \lt x \lt 100 $

Let us analyze each of the options.
$a$ is any nonzero value
This means that $a$ could be positive or negative

$b$ is any nonzero value
This means that $b$ could be positive or negative

Neither $a$ nor $b$ is zero

$ Option\:\:F \\[3ex] a - b \\[3ex] Assume\:\: a = 5, b = 2 \\[3ex] 5 - 2 = 3 \\[3ex] 3\:\:is\:\:not\:\:negative \\[3ex] Incorrect \\[3ex] Option\:\:G \\[3ex] -a - b \\[3ex] Assume\:\: a = -5, b = 2 \\[3ex] -(-5) - 2 = 5 - 2 = 3 \\[3ex] 3\:\:is\:\:not\:\:negative \\[3ex] Incorrect \\[3ex] Option\:\:H \\[3ex] |a| + |b| \\[3ex] Assume\:\: a = 5, b = 2 \\[3ex] |5| + |2| = 5 + 2 = 7 \\[3ex] 7\:\:is\:\:not\:\:negative \\[3ex] Incorrect \\[3ex] Option\:\:J \\[3ex] |a| - |b| \\[3ex] Assume\:\: a = 5, b = 2 \\[3ex] |5| - |2| = 5 - 2 = 3 \\[3ex] 3\:\:is\:\:not\:\:negative \\[3ex] Incorrect \\[3ex] Option\:\:K \\[3ex] -|a| - |b| \\[3ex] Assume\:\: a = 5, b = 2 \\[3ex] -|5| - |2| = -5 - 2 = -7...okay \\[3ex] Assume\:\: a = -5, b = 2 \\[3ex] -|-5| - |2| = -5 - 2 = -7...okay \\[3ex] Assume\:\: a = 5, b = -2 \\[3ex] -|5| - |-2| = -5 - 2 = -7...okay \\[3ex] Assume\:\: a = -5, b = -2 \\[3ex] -|-5| - |-2| = -5 - 2 = -7...okay \\[3ex] Correct! $

$ \sqrt{85} \approx 9.21954 \\[3ex] $ Next integer $\gt 9.21954$ is $10$

$ (343)^{\dfrac{1}{3}} * (0.14)^{-1} * (25)^{-\dfrac{1}{2}} \\[7ex] (343)^{\dfrac{1}{3}} = \sqrt[3]{343} = 7 \\[7ex] (0.14)^{-1} = \dfrac{1}{(0.14)^1} = \dfrac{1}{0.14} \\[5ex] (25)^{-\dfrac{1}{2}} = \dfrac{1}{(25)^{\dfrac{1}{2}}} \\[7ex] (25)^{\dfrac{1}{2}} = \sqrt{25} = 5 \\[3ex] \rightarrow (25)^{-\dfrac{1}{2}} = \dfrac{1}{5} \\[5ex] = 7 * \dfrac{1}{0.14} * \dfrac{1}{5} \\[5ex] = \dfrac{7 * 1 * 1}{0.14 * 5} \\[5ex] = \dfrac{7 * 100}{0.14 * 100 * 5} \\[5ex] = \dfrac{7 * 100}{14 * 5} \\[5ex] = \dfrac{1 * 20}{2 * 1} \\[5ex] = \dfrac{20}{2} \\[5ex] = 10 $

$ |5 - x| \\[3ex] x = 9 \\[3ex] = |5 - 9| \\[3ex] = |-4| \\[3ex] = 4 $

$ 3(-9)(2 - 10 - 2(10)) \\[3ex] = -27(2 - 10 - 20) \\[3ex] = -27(-8 - 20) \\[3ex] = -27(-28) \\[3ex] = 756 $

$ 5[2 + 2(3 * 9 - 19)] \\[3ex] = 5[2 + 2(27 - 19)] \\[3ex] = 5[2 + 2(8)] \\[3ex] = 5[2 + 16] \\[3ex] = 5[18] \\[3ex] = 90 $

Positive integer factors (exclude $6$) of $6 = 1, 2, 3$

$ 1 + 2 + 3 = 6 \\[3ex] 6 \ngtr 6 \\[3ex] $ Positive integer factors (exclude $8$) of $8 = 1, 2, 4$

$ 1 + 2 + 4 = 7 \\[3ex] 7 \ngtr 8 \\[3ex] $ Positive integer factors (exclude $10$) of $10 = 1, 2, 5$

$ 1 + 2 + 5 = 8 \\[3ex] 8 \ngtr 10 \\[3ex] $ Positive integer factors (exclude $12$) of $12 = 1, 2, 3, 4, 6$

$ 1 + 2 + 3 + 4 + 6 = 16 \\[3ex] 16 \gt 12 \\[3ex] $ Only one of those integers, $12$ is abundant.
The answer is Option $B$

$ 0.\bar{3} = 0.33333\bar{3} \\[3ex] 0.33333\bar{3} \gt 0.3333 \gt 0.333 \gt 0.33 \gt 0.3 \\[3ex] \therefore Greatest\:\:value = 0.\bar{3} $

$ \dfrac{|-3 - 2|^2 + (-1)^3}{16 \div 4 * 2 - 5} \\[5ex] \underline{Numerator} \\[3ex] |-3 - 2|^2 \\[3ex] = |-5|^2 \\[3ex] = 5^2 \\[3ex] = 25 \\[3ex] (-1)^3 \\[3ex] = (-1)(-1)(-1) \\[3ex] = -1 \\[3ex] \therefore |-3 - 2|^2 + (-1)^3 \\[3ex] = 25 + -1 \\[3ex] = 25 - 1 \\[3ex] = 24 \\[3ex] \underline{Denominator} \\[3ex] 16 \div 4 * 2 - 5 \\[3ex] = 4 * 2 - 5 \\[3ex] = 8 - 5 \\[3ex] = 3 \\[3ex] \underline{Entire\:\:Question} \\[3ex] = \dfrac{24}{3} \\[5ex] = 8 $

$ \dfrac{3}{5} - \dfrac{-1}{3} \\[5ex] = \dfrac{3}{5} - -\dfrac{1}{3} \\[5ex] = \dfrac{3}{5} + \dfrac{1}{3} \\[5ex] LCD = 15 \\[3ex] = \dfrac{9}{15} + \dfrac{5}{15} \\[5ex] = \dfrac{9 + 5}{15} \\[5ex] = \dfrac{14}{15} \\[5ex] = 0.933333333 \\[3ex] \dfrac{4}{5} = 0.8 \\[3ex] $ Looking at the options, $\dfrac{14}{15}$ is between $\dfrac{4}{5}$ and $1$

$ 21 = 3 * 7 \\[3ex] 22 = 2 * 11 \\[3ex] 23 = 23 \\[3ex] 24 = 2 * 2 * 2 * 3 \\[3ex] 25 = 5 * 5 \\[3ex] 26 = 2 * 13 \\[3ex] 27 = 3 * 3 * 3...three\:\:factors \\[3ex] 28 = 2 * 2 * 7 ...three\:\:factors \\[3ex] 29 = 29 \\[3ex] $ Two numbers, $27$ and $28$ have exactly three factors.

Positive integer divisors of $5 = 1, 5$
Number of positive integer divisors = $2$

Positive integer divisors of $6 = 1, 2, 3, 6$
Number of positive integer divisors = $4$

Positive integer divisors of $12 = 1, 2, 3, 4, 6, 12$
Number of positive integer divisors = $6$

Positive integer divisors of $16 = 1, 2, 4, 8, 16$
Number of positive integer divisors = $5$

Positive integer divisors of $18 = 1, 2, 3, 6, 9, 18$
Number of positive integer divisors = $6$

The correct option is $D.$

Ascending order means from least to greatest

$ \dfrac{9}{5}, 1.\overline{8}, 1.08, 1.\overline{08} \\[5ex] \dfrac{9}{5} = 1.8 \\[5ex] 1.\overline{8} = 1.888888888888... \\[3ex] 1.08 \\[3ex] 1.\overline{08} = 1.0808080808080808... \\[3ex] 1.88 \gt 1.8 \\[3ex] 1.8 \gt 1.0808 \\[3ex] 1.0808 \gt 1.08 \\[3ex] 1.08 \lt 1.0808 \lt 1.8 \lt 1.88 \\[3ex] \therefore 1.08 \lt 1.\overline{08} \lt \dfrac{9}{5} \lt 1.\overline{8} $

$2.5$ million is two million, five hundred thousand

$ 2.5\:\:million \\[3ex] = 2,500,000 \\[3ex] = 2.5 * 10^6 $

$ Real\:\:Numbers\:\:includes\:\:positive\:\:numbers,\:\:negative\:\:numbers,\:\:and\:\:zero \\[3ex] \underline{Arithmetic\:\:Method:\:\:Test\:\:Numbers} \\[3ex] If\:\:(a + b)^2 = a^2 + b^2 \\[3ex] Option\:A:\:\:either\:\:a\:\:or\:\:b\:\: = 0 \\[3ex] This\:\:means\:\:that\:\:a\:\:could\:\:be\:\:0\:\:OR\:\:b\:\:could\:\:be\:\:0\:\:OR\:\:both\:\:a\:\:and\:\:b\:are\:\:zeros \\[3ex] Test\:\:a = 0,\:\: b = -3 \\[3ex] LHS:\:\:(a + b)^2 \implies (0 + -3)^2 = (0 - 3)^2 = (-3)^2 = 9 \\[3ex] RHS:\:\:a^2 + b^2 \implies 0^2 + (-3)^2 = 0 + 9 = 9 \\[3ex] Test\:\:a = 3,\:\: b = 0 \\[3ex] LHS:\:\:(a + b)^2 \implies (3 + 0)^2 = (3)^2 = 9 \\[3ex] RHS:\:\:a^2 + b^2 \implies 3^2 + 0^2 = 9 + 0 = 9 \\[3ex] Test\:\:a = 0,\:\: b = 0 \\[3ex] LHS:\:\:(a + b)^2 \implies (0 + 0)^2 = 0^2 = 0 \\[3ex] RHS:\:\:a^2 + b^2 \implies 0^2 + 0^2 = 0 + 0 = 0 \\[3ex] Works...but\:\:let\:\:us\:\:analyze\:\:the\:\:other\:\:options \\[3ex] Option\:B:\:\:both\:\:a\:\:and\:\:b\:\: = 0 \\[3ex] This\:\:option\:\:is\:\:included\:\:in\:\:Option\:\:A \\[3ex] So,\:\:it\:\:cannot\:\:be\:\:the\:\:answer \\[3ex] Option\:C:\:\:both\:\:a\:\:and\:\:b\:\:are\:\:positive \\[3ex] Test\:\:a = 1,\:\: b = 3 \\[3ex] LHS:\:\:(a + b)^2 \implies (1 + 3)^2 = 4^2 = 16 \\[3ex] RHS:\:\:a^2 + b^2 \implies 1^2 + 3^2 = 1 + 9 = 10 \\[3ex] 16 \ne 10 \\[3ex] This\:\:does\:\:not\:\:work...NO \\[3ex] Option\:D:\:\:a\:\:is\:\:positive\:\:and\:\:b\:\:is\:\:negative \\[3ex] Test\:\:a = 1,\:\: b = -3 \\[3ex] LHS:\:\:(a + b)^2 \implies (1 + -3)^2 = (1 - 3)^2 = (-2)^2 = 4 \\[3ex] RHS:\:\:a^2 + b^2 \implies 1^2 + (-3)^2 = 1 + 9 = 10 \\[3ex] 4 \ne 10 \\[3ex] This\:\:does\:\:not\:\:work...NO \\[3ex] Option\:E:\:\:a\:\:is\:\:negative\:\:and\:\:b\:\:is\:\:positive \\[3ex] Test\:\:a = -1,\:\: b = 3 \\[3ex] LHS:\:\:(a + b)^2 \implies (-1 + 3)^2 = 2^2 = 4 \\[3ex] RHS:\:\:a^2 + b^2 \implies (-1)^2 + 3^2 = 1 + 9 = 10 \\[3ex] 4 \ne 10 \\[3ex] This\:\:does\:\:not\:\:work...NO \\[3ex] \therefore the\:\:correct\:\:option\:\:is\:\:A $

$ \left(9^{\dfrac{1}{2}} + 16^{\dfrac{1}{2}}\right)^2 \\[5ex] = \left(\sqrt{9} + \sqrt{16}\right)^2 \\[3ex] = (3 + 4)^2 \\[3ex] = 7^2 \\[3ex] = 49 $

$ \underline{Prime\;\;Factorization\;\;Method} \\[3ex] 45 = 3 * 3 * 5 \\[3ex] 50 = 2 * 5 * 5 \\[3ex] 84 = 2 * 2 * 3 * 7 \\[3ex] GCF = 1 \\[3ex] $ This is because there is no other commom factor of the three numbers.
$1$ is a factor of everything because $1$ * any thing is that thing.
$1$ was not listed in that method because it is not a prime number.
However, $1$ is a factor of those three numbers.

$ 0.\overline{3178} \\[3ex] means\:\: 0.\overline{3178317831783178....} \\[3ex] The\:\:digits\:\: 3178 \:\:repeats \\[3ex] Every\:\:4th\:\:digit = 8 \\[3ex] Every\:\:8th\:\:digit = 8 \\[3ex] Every\:\:12th\:\:,16th\:\:,20th\:\:...digits = 8 \\[3ex] Every\:\:multiple\:\:of\:\:4\:\:-th\:\:digit = 8 \\[3ex] For\:\:358th\:\:digit, \\[3ex] \dfrac{358}{4} = 89.5 \\[3ex] Integer\:\:part = 89 \\[3ex] 89 * 4 = 356 \\[3ex] 356th\:\:digit = 8 \\[3ex] Back-to-repeating \\[3ex] 357th\:\:digit = 3 \\[3ex] 358th\:\:digit = 1 \\[3ex] OR \\[3ex] \dfrac{358}{4} = 89.5 \\[3ex] Decimal\:\:part = 0.5 \\[3ex] 0.5 * 4 = 2 \\[3ex] Count\:\:the\:\:2nd\:\:digit\:\:in\:\:the\:\:sequence \\[3ex] The\:\:2nd\:\:digit = 358th\:\:digit \\[3ex] 2nd\:\:digit = 1 \\[3ex] \therefore the\:\: 358th\:\:digit = 1 $

$ \underline{First\:\:Method:\:\:Arithmetically} \\[3ex] Try\:\:several\:\:numbers \\[3ex] Real\:\:Numbers\:\:includes\:\:positive\:\:numbers,\:\:negative\:\:numbers,\:\:and\:\:zero \\[3ex] Use\:\:both\:\:positive\:\:and\:\:negative\:\:real\:\:numbers\:\:as\:\:applicable \\[3ex] b \gt 0 \\[3ex] Let\:\: b = 1 \\[3ex] a \gt b \\[3ex] Let\:\: a = 2 \\[3ex] b \gt c \\[3ex] Let\:\: c = -1 \\[3ex] a \gt b \gt c \implies 2 \gt 1 \gt -1 \\[3ex] I.\:\: |a| \gt |b| \\[3ex] |2| \gt |1| ? \\[3ex] 2 \gt 1 ...correct...YES \\[3ex] I\:\:works \\[3ex] II.\:\: |a| \gt |c| \\[3ex] |2| \gt |-1| ? \\[3ex] 2 \gt 1 ...correct \\[3ex] But\:\:what\:\:if\:\:c = -2 \\[3ex] c\:\:can\:\:be\:\:-2\:\:because\:\: b \gt c \implies 1 \gt -2 \\[3ex] Try\:\: c = -2 \\[3ex] |2| \gt |-2| ? \\[3ex] 2 \gt 2 ...NO \\[3ex] II\:\:will\:\:not\:\:work \\[3ex] III.\:\: |b| \gt |c| \\[3ex] |1| \gt |-1| ? \\[3ex] 1 \gt -1...NO \\[3ex] III\:\:will\:\:not\:\:work \\[3ex] Correct\:\:Option = I\:\:only = A $

$ \dfrac{1}{4} = 0.25 \\[5ex] 0.03 \lt 0.2 \lt 0.25 \\[3ex] \therefore 0.03 \lt 0.2 \lt \dfrac{1}{4} $

We need a positive integer that can be divided by $3$, $7$, and $9$ without a remainder.
The least possible value of that positive integer is the least common multiple of $3$, $7$, and $9$

$ \underline{Prime\;\;Factorization\;\;Method} \\[3ex] 3 = \color{black}{3} \\[3ex] 7 = 7 \\[3ex] 9 = \color{black}{3} * 3 \\[3ex] LCM = \color{black}{3} * 7 * 3 = 63 \\[3ex] x = 63 $

$ (a.) \\[3ex] \sqrt{\dfrac{49}{196}} \\[5ex] \dfrac{\sqrt{49}}{\sqrt{196}} \\[5ex] \dfrac{7}{14} \\[5ex] \dfrac{1}{2} \\[5ex] (b.) \\[3ex] -\sqrt{\dfrac{196}{225}} \\[5ex] -1 * \sqrt{\dfrac{196}{225}} \\[5ex] -1 * \dfrac{\sqrt{196}}{\sqrt{225}} \\[5ex] -1 * \dfrac{14}{15} \\[5ex] -\dfrac{14}{15} $

We shall solve this question by testing with numbers
I am going to solve by fractions because I would like for you to be comfortable with fractions.
However, you may choose to convert to decimals if you want to.

$ a, b, c, d \;\;are\;\; positive \;\; integers \\[3ex] a \lt b \lt c \lt d \\[3ex] Let: \\[3ex] a = 1 \\[3ex] b = 2 \\[3ex] c = 3 \\[3ex] d = 4 \\[3ex] Option\;F \\[3ex] \dfrac{a}{b} = \dfrac{1}{2} \\[5ex] Option\;G \\[3ex] \dfrac{b}{c} = \dfrac{2}{3} \\[5ex] Option\;H \\[3ex] \dfrac{c}{d} = \dfrac{3}{4} \\[5ex] Option\;J \\[3ex] \dfrac{a + b}{b + c} = \dfrac{1 + 2}{2 + 3} = \dfrac{3}{5} \\[5ex] Option\;K \\[3ex] \dfrac{b + c}{c + d} = \dfrac{2 + 3}{3 + 4} = \dfrac{5}{7} \\[5ex] LCM \;\; 2, 3, 4, 5, 7 = 4 * 3 * 5 * 7 = 420 \\[3ex] Option\;F:\;\; \dfrac{1}{2} = \dfrac{210}{420} \\[5ex] Option\;G:\;\; \dfrac{2}{3} = \dfrac{280}{420} \\[5ex] Option\;H:\;\; \dfrac{3}{4} = \dfrac{315}{420} \\[5ex] Option\;J:\;\; \dfrac{3}{5} = \dfrac{252}{420} \\[5ex] Option\;K:\;\; \dfrac{5}{7} = \dfrac{300}{420} \\[5ex] Same\;\;Denominator = 420 \\[3ex] Greatest\;\;Numerator = 315 \\[3ex] \therefore Greatest\;\;Value = Option\;H:\;\; \dfrac{c}{d} $

We need a positive integer that can be divided by $3$, $7$, and $9$ without a remainder.
The least possible value of that positive integer is the least common multiple of $3$, $7$, and $9$

$ \underline{Prime\;\;Factorization\;\;Method} \\[3ex] 3 = \color{black}{3} \\[3ex] 7 = 7 \\[3ex] 9 = \color{black}{3} * 3 \\[3ex] LCM = \color{black}{3} * 7 * 3 = 63 \\[3ex] x = 63 $

We shall solve this question by testing with numbers
I am going to solve by fractions because I would like for you to be comfortable with fractions.
However, you may choose to convert to decimals if you want to.

$ a, b, c, d \;\;are\;\; positive \;\; integers \\[3ex] a \lt b \lt c \lt d \\[3ex] Let: \\[3ex] a = 1 \\[3ex] b = 2 \\[3ex] c = 3 \\[3ex] d = 4 \\[3ex] Option\;F \\[3ex] \dfrac{a}{b} = \dfrac{1}{2} \\[5ex] Option\;G \\[3ex] \dfrac{b}{c} = \dfrac{2}{3} \\[5ex] Option\;H \\[3ex] \dfrac{c}{d} = \dfrac{3}{4} \\[5ex] Option\;J \\[3ex] \dfrac{a + b}{b + c} = \dfrac{1 + 2}{2 + 3} = \dfrac{3}{5} \\[5ex] Option\;K \\[3ex] \dfrac{b + c}{c + d} = \dfrac{2 + 3}{3 + 4} = \dfrac{5}{7} \\[5ex] LCM \;\; 2, 3, 4, 5, 7 = 4 * 3 * 5 * 7 = 420 \\[3ex] Option\;F:\;\; \dfrac{1}{2} = \dfrac{210}{420} \\[5ex] Option\;G:\;\; \dfrac{2}{3} = \dfrac{280}{420} \\[5ex] Option\;H:\;\; \dfrac{3}{4} = \dfrac{315}{420} \\[5ex] Option\;J:\;\; \dfrac{3}{5} = \dfrac{252}{420} \\[5ex] Option\;K:\;\; \dfrac{5}{7} = \dfrac{300}{420} \\[5ex] Same\;\;Denominator = 420 \\[3ex] Greatest\;\;Numerator = 315 \\[3ex] \therefore Greatest\;\;Value = Option\;H:\;\; \dfrac{c}{d} $

For this question, we have to find a contradiction to each statement.
In other words, we have to try to prove each statement incorrect.
If we cannot, then that is the correct option.
Try to write several examples of $4$ consecutive positive integers
Let us begin with $I.$
$I.$ At least $1$ of the $4$ integers is prime.
Test Case: $32, 33, 34, 35$...these are $4$ consecutive positive integers
As we can see, all $4$ integers are not prime numbers.
They are all composite numbers.
Hence $I.$ is incorrect.
This eliminates three options: $F.$, $H.$ and $K.$

Next is $II.$
$II.$ At least $2$ of the $4$ integers have a common prime factor.
This is a true statement because $4$ consecutive positive integers must include $2$ even integers
$2$ is a prime factor of every even integer
Therefore, two of the four integers must have a common prime factor, and that common prime factor is $2$
At least two means two or more
So, this statement is true

Then, for $III.$
Using the same example we used for $I.$
Test Case: $32, 33, 34, 35$...these are $4$ consecutive positive integers
$32$ is not a factor of $33$, $34$, or $35$
Neither is any of those integers a factor of any of the other integers.
Hence, this is a false statement.

The only true statement is $II.$
The correct option is $G.$

$ 9 - 3 \div \dfrac{1}{3} + 1 \\[5ex] = 9 - \left(3 \div \dfrac{1}{3}\right) + 1 \\[5ex] = 9 - \left(3 * \dfrac{3}{1}\right) + 1 \\[5ex] = 9 - 9 + 1 \\[3ex] = 0 + 1 \\[3ex] = 1 $

$ 8 \div 2(2 + 2) \\[3ex] = 8 \div 2 * (2 + 2) \\[3ex] = 8 \div 2 * 4 \\[3ex] = 4 * 4 \\[3ex] = 16 \\[3ex] \underline{Common\;\;Mistake} \\[3ex] 8 \div 2(2 + 2) \\[3ex] = 8 \div 2(4) \\[3ex] = 8 \div 8 \\[3ex] = 1 \\[3ex] $ Please note:
In PEMDAS:
Multiplication and Division occurs from left to right in the order it was written (whichever is first)

$ \dfrac{3}{4} * \sqrt{9 673} - 0.5(5.9352 + 2.16937) \\[5ex] = 0.75 * 98.35141077 - 0.5(8.10457) \\[3ex] = 73.76355808 - 4.052285 \\[3ex] = 69.71127308 $

$ x : y = 2 : 3 \\[3ex] \implies \\[3ex] x = 2 \\[3ex] y = 3 \\[3ex] \underline{Numerator} \\[3ex] x^2 - y^2 \\[3ex] = 2^2 - 3^2 \\[3ex] = 4 - 9 \\[3ex] = -5 \\[3ex] \underline{Denominator} \\[3ex] y^2 + x^2 \\[3ex] = 3^2 + 2^2 \\[3ex] = 9 + 4 \\[3ex] = 13 \\[3ex] Answer = -\dfrac{5}{13} $

$ \dfrac{(6 - 7)^2 - 3|4 - 9|}{128 - 2 * 6^2} \\[5ex] \underline{Numerator} \\[3ex] (6 - 7)^2 - 3|4 - 9| \\[3ex] = (-1)^2 - 3|-5| \\[3ex] = (-1)(-1) - 3(5) \\[3ex] = 1 - 15 \\[3ex] = -14 \\[3ex] \underline{Denominator} \\[3ex] 128 - 2 * 6^2 \\[3ex] = 128 - 2 * 36 \\[3ex] = 128 - 72 \\[3ex] = 56 \\[3ex] \therefore \dfrac{(6 - 7)^2 - 3|4 - 9|}{128 - 2 * 6^2} = \dfrac{-14}{56} \\[5ex] = -\dfrac{1}{4} $

$ x : y = 3 : 5 \\[3ex] \implies \\[3ex] x = 3 \\[3ex] y = 5 \\[3ex] \underline{Numerator} \\[3ex] 2x^2 - y^2 \\[3ex] = 2 * 3^2 - 5^2 \\[3ex] = 2(9) - 25 \\[3ex] = 18 - 25 \\[3ex] = -7 \\[3ex] \underline{Denominator} \\[3ex] y^2 - x^2 \\[3ex] = 5^2 - 3^2 \\[3ex] = 25 - 9 \\[3ex] = 16 \\[3ex] Answer = -\dfrac{7}{16} $

$ w, z \gt 0 ...positive\;\;integers \\[3ex] w + z = 5 \\[3ex] \underline{Possibilities} \\[3ex] 1 + 4 = 5...eqn.(1) \\[3ex] 4 + 1 = 5...eqn.(2) \\[3ex] 2 + 3 = 5...eqn.(3) \\[3ex] 3 + 2 = 5...eqn.(4) \\[3ex] But: \\[3ex] x = 2^w \;\;\;and\;\;\; x^z = 64 \\[5ex] For:\;\; x^z = 64 \\[3ex] Possibilities\;\;are: \\[3ex] 2^6 = 64 \;\;\;or\;\;\; 4^3 = 64\;\;\;or\;\; 8^2 = 64 \\[3ex] \implies \\[3ex] z = 6 \;\;\;or\;\;\; z = 3 \;\;\;or\;\;\;\; z = 2 \\[3ex] z\;\;cannot\;\;be\;\;equal\;\;to\;\; 6...Possibilities \\[3ex] \therefore z = 3 \;\;\;or\;\;\;\; z = 2 \\[5ex] \implies w = 2 \;\;\;or\;\;\; w = 3...eqn.(3)\;\;\;and\;\;eqn.(4) \\[3ex] $ There are two values of w that satisfies oth $2^w = x$ and $x^z = 64$

You may ask students to find the values of x

The colors besides red indicate the common factors that should be counted only one time.
Begin with them in the multiplication for the LCM.
Then, include the rest.

$ Numbers = 50, 30, 70 \\[3ex] 50 = \color{black}{2} * \color{darkblue}{5} * 5 \\[3ex] 30 = \color{black}{2} * 3 * \color{darkblue}{5} \\[3ex] 70 = \color{black}{2} * \color{darkblue}{5} * 7 \\[5ex] LCM = \color{black}{2} * \color{darkblue}{5} * 5 * 3 * 7 \\[3ex] LCM = 1,050 $

The colors besides red indicate the common factors that should be counted only one time.
They are the only ones to be included in the calculation of the GCF.

$ Numbers = 60, 84, 126 \\[3ex] 60 = \color{black}{2} * 2 * \color{darkblue}{3} * 5 \\[3ex] 84 = \color{black}{2} * 2 * \color{darkblue}{3} * 7 \\[3ex] 126 = \color{black}{2} * \color{darkblue}{3} * 3 * 7 \\[5ex] GCF = \color{black}{2} * \color{darkblue}{3} \\[3ex] GCF = 6 $

The colors besides red indicate the common factors that should be counted only one time.
Begin with them in the multiplication for the LCM.
Then, include the rest.

$ Numbers = 70, 60, 50 \\[3ex] 70 = \color{black}{2} * \color{darkblue}{5} * 7 \\[3ex] 60 = \color{black}{2} * 2 * 3 * \color{darkblue}{5} \\[3ex] 50 = \color{black}{2} * \color{darkblue}{5} * 5 \\[5ex] LCM = \color{black}{2} * \color{darkblue}{5} * 7 * 2 * 3 * 5 \\[3ex] LCM = 2,100 $

$ 3d + 17 = 13 \\[3ex] 3d = 13 - 17 \\[3ex] 3d = -4 \\[3ex] d = -\dfrac{4}{3} = 1.33\overline{3} \\[5ex] $ d is a rational number because it is a repeating decimal
d is not irrational because it is rational
d is negative
d is not positive because it is negative
d is not an integer because it is a fraction that can be expressed as a repeating decimal

B. I and IV only

Test each option
Option F. 4 and 9
6 is not a factor of either 4 or 9...NEXT

Test each option
Option G. 9 and 12
6 is not a factor of 9...NEXT

Test each option
Option H. 12 and 15
6 is not a factor of 15...NEXT

Test each option
Option J. 12 and 18
6 is a factor of both
Find the LCM

The colors besides red indicate the common factors that should be counted only one time.
Begin with them in the multiplication for the LCM.
Then, include the rest.

$ Numbers = 12, 18 \\[3ex] 12 = \color{black}{2} * 2 * \color{darkblue}{3} \\[3ex] 18 = \color{black}{2} * \color{darkblue}{3} * 3 \\[3ex] LCM = \color{black}{2} * \color{darkblue}{3} * 2 * 3 \\[3ex] LCM = 36...STOP \\[3ex] $

First, let us find the GCF of $x^2y^2$ and $xy^3$
The colors besides red indicate the common factors that should be counted only one time.
They are the only ones to be included in the calculation of the GCF.

$ x^2y^2 = \color{black}{x} * x * \color{darkblue}{y} * \color{purple}{y} \\[3ex] xy^3 = \color{black}{x} * y * \color{darkblue}{y} * \color{purple}{y} \\[5ex] GCF = \color{black}{x} * \color{darkblue}{y} * \color{purple}{y} \\[3ex] GCF = xy^2 \\[3ex] GCF = 45 \\[3ex] \implies \\[3ex] xy^2 = 45 \\[3ex] $ Note: $y$ must be squared in order to compute the GCF
Let us analyze each option
A. 45
45 is the GCF
This option is incorrect

B. 15
This is also incorrect because the square of 15 is greater than 45

C. 9
This is incorrect because the square of 9 is greater than 45

D. 5
This is incorrect because even if we square 5, which gives 25; there is no positive integer whose product with 25 gives 45

$ Assume\;\; y = 5 \\[3ex] y^2 = 5^2 = 25 \\[3ex] x * 25 = 45 \\[3ex] $ $x$ should be a positive integer, but it is not. It is a decimal. So, this option is incorrect.

E. 3

$ Assume\;\; y = 3 \\[3ex] y^2 = 3^2 = 9 \\[3ex] x * 9 = 45 \\[3ex] $ This option is correct.

$\dfrac{b + c}{d}$ is undefined implies that $d = 0$

Because division by zero gives undefined

$ abc = d \\[3ex] abc = 0 \\[3ex] ac * b = d \\[3ex] ac = 1 \\[3ex] \implies \\[3ex] 1 * b = 0 \\[3ex] b = 0 \\[3ex] $ The value of b must be zero

$ (a.) \\[3ex] ((-16 - (-2 + 1)) * 2) \div 5 \\[3ex] PEMDAS:\;\;Inner\;\;Parenthesis \\[3ex] ((-16 - (-1)) * 2) \div 5 \\[3ex] Two\;\;negatives\;\;make\;\;a\;\;positive \\[3ex] ((-16 + 1) * 2) \div 5 \\[3ex] PEMDAS:\;\;Inner\;\;Parenthesis \\[3ex] (-15 * 2) \div 5 \\[3ex] PEMDAS \\[3ex] -30 \div 5 \\[3ex] -6 \\[5ex] (b.) \\[3ex] (-11 - 6 - -5 + 1 + 3 * 2) \div -5 \\[3ex] Two\;\;negatives\;\;make\;\;a\;\;positive \\[3ex] (-11 - 6 + 5 + 1 + 3 * 2) \div -5 \\[3ex] PEMDAS \\[3ex] (-11 - 6 + 5 + 1 + 6) \div -5 \\[3ex] (-17 + 5 + 1 + 6) \div -5 \\[3ex] (-12 + 1 + 6) \div -5 \\[3ex] (-11 + 6) \div -5 \\[3ex] -5 \div -5 \\[3ex] 1 \\[5ex] (c.) \\[3ex] 2 - 8 \div -2 - 3 - -12 \div -6 * -2 \\[3ex] Two\;\;negatives\;\;make\;\;a\;\;positive \\[3ex] 2 - 8 \div -2 - 3 + 12 \div -6 * -2 \\[3ex] PEMDAS \\[3ex] Left\;\;to\;\;Right:\;\; Division \\[3ex] 2 - (8 \div -2) - 3 + (12 \div -6) * -2 \\[3ex] 2 - (-4) - 3 + (-2) * -2 \\[3ex] Two\;\;negatives\;\;make\;\;a\;\;positive \\[3ex] A\;\;positive\;\;and\;\;a\;\;negative\;\;make\;\;a\;\;negative \\[3ex] 2 + 4 - 3 - 2 * -2 \\[3ex] PEMDAS \\[3ex] Multiplication \\[3ex] 2 + 4 - 3 - (2 * -2) \\[3ex] 2 + 4 - 3 - (-4) \\[3ex] Two\;\;negatives\;\;make\;\;a\;\;positive \\[3ex] 2 + 4 - 3 + 4 \\[3ex] Left\;\;to\;\;Right:\;\; Addition\;\;and\;\;Subtraction \\[3ex] 6 - 3 + 4 \\[3ex] 3 + 4 \\[3ex] 7 $

$ PEMDAS \\[3ex] (43 * 8) - (234 \div 6) \\[3ex] = 344 - 39 \\[3ex] = 305 $

The colors besides red indicate the common factors that should be counted only one time.
Begin with them in the multiplication for the LCM.
Then, include the rest.

$ Numbers = 60, 70, 90 \\[3ex] 60 = \color{black}{2} * 2 * \color{purple}{3} * \color{darkblue}{5} \\[3ex] 70 = \color{black}{2} * \color{darkblue}{5} * 7 \\[3ex] 90 = \color{black}{2} * 3 * \color{purple}{3} * \color{darkblue}{5} \\[5ex] LCM = \color{black}{2} * \color{purple}{3} * \color{darkblue}{5} * 2 * 7 * 3 \\[3ex] LCM = 1,260 $

Let us analyze each option

Option A. $\sqrt{2} = 1.414213562$
Non-repeating and non-terminanting decimal
This is an irrational number

Option B. $\sqrt{\pi} = 1.772453851$
Non-repeating and non-terminanting decimal
This is an irrational number

Option C. $\sqrt{7} = 2.645751311$
Non-repeating and non-terminanting decimal
This is an irrational number

Option D. $\sqrt{\dfrac{5}{25}} = \dfrac{\sqrt{5}}{\sqrt{25}} = \dfrac{2.236067977}{5} = 0.4472135955$

Non-repeating and non-terminanting decimal
This is an irrational number

Option E. $\sqrt{\dfrac{64}{49}} = \dfrac{\sqrt{64}}{\sqrt{49}} = \dfrac{8}{7}$

This is expressed as an exact ratio of two integers
It is a rational number.

The colors besides red indicate the common factors that should be counted only one time.
Begin with them in the multiplication for the LCM.
Then, include the rest.

$ Numbers = 1, 2, 3, 4, 5, 6 \\[3ex] 1 = 1 \\[3ex] 2 = \color{black}{2} \\[3ex] 3 = \color{darkblue}{3} \\[3ex] 4 = \color{black}{2} * 2 \\[3ex] 5 = 5 \\[3ex] 6 = \color{black}{2} * \color{darkblue}{3} \\[5ex] LCM = \color{black}{2} * \color{darkblue}{3} * 1 * 2 * 5 \\[3ex] LCM = 60 $

Because the ACT is a timed test, my advice in solving this question is to begin with the highest number
Why? Because the question is asking for the greatest value.
Begin with the highest number and if it does not work, try the next higher number, and keep going that way through the options.
The highest number = 72
So, let us find the LCM of 108 and 72

The colors besides red indicate the common factors that should be counted only one time.
Begin with them in the multiplication for the LCM.
Then, include the rest.

$ Numbers = 108, 72 \\[3ex] 108 = \color{black}{2} * \color{darkblue}{2} * 3 * \color{purple}{3} * \color{green}{3} \\[3ex] 72 = \color{black}{2} * \color{darkblue}{2} * 2 * \color{purple}{3} * \color{green}{3} \\[5ex] LCM = \color{black}{2} * \color{darkblue}{2} * \color{purple}{3} * \color{green}{3} * 3 * 2 \\[3ex] LCM = 216 \\[3ex] $ This works.
The other number is 72

The colors besides red indicate the common factors that should be counted only one time.
Begin with them in the multiplication for the LCM.
Then, include the rest.

$ Numbers = 30, 20, 70 \\[3ex] 30 = \color{black}{2} * 3 * \color{darkblue}{5} \\[3ex] 20 = \color{black}{2} * 2 * \color{darkblue}{5} \\[3ex] 70 = \color{black}{2} * \color{darkblue}{5} * 7 \\[5ex] LCM = \color{black}{2} * \color{darkblue}{5} * 3 * 2 * 7 \\[3ex] LCM = 420 $

A good approach to solve this question is to assume a number
Say, the number that Tom wanted to divide by 2; is 6
6 divided by 2 is 3
But accidentally; Tom did: 6 * 2 = 12

So, the question is asking us:
What could Tom do to the result on the calculator screen to obtain the result he originally wanted?
He originally wanted to divide 6 by 2 to get 3
This means that he originally wanted to get 3
So, in other words, what could Tom do to 12 in order to get 3?

12 divided by 4 is 3
Tom should divide by 4

Let us analyze each option.
Let us use some examples. We may change these examples if needed.
Let the two rational numbers be: $5$ and $\dfrac{3}{5}$

Let the two irrational numbers be: $\sqrt{2}$ and $\sqrt{2}$

$ \underline{Option\;F} \\[3ex] \sqrt{2} * \sqrt{2} \\[3ex] = (\sqrt{2})^2 \\[3ex] = 2 \\[3ex] is\;\; rational \\[3ex] This\;\;option\;\;is\;\;Incorrect \\[5ex] \underline{Option\;G} \\[3ex] \dfrac{\sqrt{2}}{\sqrt{2}} = 1 \\[5ex] Seems\;\;correct\;\;but\;\;let\;\;us\;\;try\;\;other\;\;values \\[3ex] Let\;\;the\;\;two\;\;irrational\;\;numbers = \sqrt{2} \;\;and\;\; \sqrt{3} \\[3ex] \dfrac{\sqrt{2}}{\sqrt{3}} \\[5ex] = \dfrac{\sqrt{2}}{\sqrt{3}} * \dfrac{\sqrt{3}}{\sqrt{3}} \\[5ex] = \dfrac{\sqrt{2 * 3}}{\sqrt{3 * 3}} \\[5ex] = \dfrac{\sqrt{6}}{\sqrt{9}} \\[5ex] = \dfrac{\sqrt{6}}{3} \\[5ex] is\;\; irrational \\[3ex] This\;\;option\;\;is\;\;Incorrect \\[5ex] \underline{Option\;H} \\[3ex] 5 * \dfrac{3}{5} \\[5ex] = 3 \\[3ex] is\;\;rational \\[3ex] The\;\;option\;\;is\;\;Incorrect \\[5ex] \underline{Option\;J} \\[3ex] 5 \div \dfrac{3}{5} \\[5ex] = 5 * \dfrac{5}{3} \\[5ex] = \dfrac{25}{3} \\[5ex] = 8.3\bar{3} \\[3ex] is\;\;rational \\[3ex] This\;\;option\;\;is\;\;Incorrect \\[5ex] \underline{Option\;K} \\[3ex] 5 + \dfrac{3}{5} \\[5ex] = \dfrac{25}{5} + \dfrac{3}{5} \\[5ex] = \dfrac{28}{5} \\[5ex] = 5.6 \\[3ex] is\;\;rational \\[3ex] This\;\;is\;\;the\;\;correct\;\;option $

$ 1st:\;\; c \lt d \rightarrow d \gt c \\[3ex] 2nd:\;\; e \lt c \rightarrow c \gt e \\[3ex] \implies d \gt c \gt e \\[3ex] 3rd:\;\; e \gt b \\[3ex] 4th:\;\; b \gt a \\[3ex] \implies \\[3ex] d \gt c \gt e \gt b \gt a \\[3ex] $ Therefore, the greatest of the numbers is d

Let us analyze each option.
We shall use several exmaples
NOTE: n is a positive integer
Let us test each option with two values of n: 2 and 3
If any of the values does not give an odd integer, move on to the next option.

$ \underline{Option\;F} \\[3ex] n = 2 \\[3ex] 3^n = 3^2 = 9 \\[3ex] n = 3 \\[3ex] 3^n = 3^3 = 27 \\[3ex] 9\;\;and\;\;27\;\;are\;\;odd\;\;integers \\[3ex] OKAY \\[5ex] $ You may stop here because this is the correct option.
However, for teaching purposes; let us test the remaining options.

$ \underline{Option\;G} \\[3ex] n = 2 \\[3ex] 2^n = 2^3 = 8 \\[3ex] 8\;\;is\;\;not\;\;an\;\;odd\;\;integer \\[3ex] NO...NEXT \\[3ex] \underline{Option\;H} \\[3ex] n = 2 \\[3ex] 3n = 3(2) = 6 \\[3ex] 8\;\;is\;\;not\;\;an\;\;odd\;\;integer \\[3ex] NO...NEXT \\[3ex] \underline{Option\;J} \\[3ex] n = 2 \\[3ex] \dfrac{n}{3} = \dfrac{2}{3} \\[5ex] \dfrac{2}{3}\;\;is\;\;not\;\;an\;\;odd\;\;integer \\[5ex] NO...NEXT \\[3ex] \underline{Option\;K} \\[3ex] n = 2 \\[3ex] 3 + n = 3 + 2 = 5 \\[3ex] n = 3 \\[3ex] 3 + n = 3 + 3 = 6 \\[3ex] 6\;\;is\;\;not\;\;an\;\;odd\;\;integer \\[3ex] NO $

The colors besides red indicate the common factors that should be counted only one time.
Begin with them in the multiplication for the LCM.
Then, include the rest.

$ Numbers = 80, 70, 30 \\[3ex] 80 = \color{black}{2} * 2 * 2 * 2 * \color{darkblue}{5} \\[3ex] 70 = \color{black}{2} * \color{darkblue}{5} * 7 \\[3ex] 30 = \color{black}{2} * 3 * \color{darkblue}{5} \\[5ex] LCM = \color{black}{2} * \color{darkblue}{5} * 2 * 2 * 2 * 7 * 3 \\[3ex] LCM = 1680 $