Which Of The Following Has A Value Of 84

Which of the followinghas a value of 84?
When faced with a multiple‑choice question that asks you to pick the expression whose value equals 84, the key is to evaluate each option systematically rather than guessing. This article walks you through the reasoning process, shows how to handle typical mathematical forms you’ll encounter, and provides practice so you can confidently identify the correct choice every time.

Introduction

Questions that begin with “Which of the following has a value of …” appear frequently in arithmetic, algebra, and even standardized‑test practice. They test two core skills:

1. Accurate computation – you must be able to simplify numbers, apply exponent rules, work with factorials, or handle combinations.
2. Efficient elimination – by estimating or using properties of numbers, you can often discard options without full calculation.

The target value here is 84, a composite number with several interesting factorizations (2 × 2 × 3 × 7). Recognizing these factors helps you spot matches quickly.

Understanding the Problem

Before diving into calculations, clarify what the question is asking:

• You are given a list of expressions (usually labeled A, B, C, D, …).
• Exactly one of those expressions simplifies to the integer 84.
• Your task is to determine which one.

Because the answer is a single integer, you can use divisibility tricks, approximate magnitude, or known identities to narrow the field.

Common Types of Expressions You’ll Encounter | Expression Type | Typical Form | How to Evaluate Quickly |

|-----------------|--------------|--------------------------| | Basic multiplication | (a \times b) | Look for factor pairs of 84 (e.g., 7 × 12, 6 × 14). | | Exponents & powers | (a^{b}) or (b^{a}) | Compare to nearby powers: (2^{6}=64), (2^{7}=128); (3^{4}=81), (3^{5}=243). | | Factorials | (n!) | (4! = 24), (5! = 120) → 84 is not a factorial, but combinations often involve factorials. | | Combinations / Binomial coefficients | (\binom{n}{k} = \frac{n!}{k!(n-k)!}) | Use Pascal’s triangle or known values: (\binom{9}{2}=36), (\binom{9}{3}=84). | | Summations | (\sum_{i=1}^{n} i) or (\sum i^{2}) | Triangular numbers: (T_{12}=78), (T_{13}=91); 84 is not triangular but could be a sum of squares. | | Algebraic simplification | (2x+3y) with given x, y | Substitute and compute. | | Mixed operations | ((a+b) \times c - d) | Follow order of operations (PEMDAS/BODMAS). |

Recognizing which category an expression falls into lets you apply the appropriate shortcut.

Step‑by‑Step Evaluation of Sample Options

Below is a typical set of four options you might see. We will walk through each, showing both the full calculation and the quick‑check reasoning.

Options

A. (7 \times 12)
B. (2^{6} + 20)
C. (\displaystyle \binom{9}{3})
D. (5! - 36)

Option A: (7 \times 12)

• Full calculation: (7 \times 12 = 84).
• Quick check: Both factors are familiar; 7 × 10 = 70, plus 7 × 2 = 14 → 84.
• Verdict: Correct (value equals 84).

Option B: (2^{6} + 20)

• Full calculation: (2^{6}=64); (64+20=84).
• Quick check: Recognize that (2^{6}=64) is just 20 shy of 84, so the sum lands exactly on 84.
• Verdict: Also correct (value equals 84).

Notice: In a well‑formed single‑answer question, only one option should be correct. If you encounter two that both give 84, double‑check the problem statement for any hidden constraints (e.g., “which of the following expressions has a value of 84 when simplified?”) or consider that a typo may exist. For the purpose of this tutorial, we’ll treat the first matching option as the intended answer and note that alternative forms can also produce 84.

Option C: (\displaystyle \binom{9}{3})

• Full calculation: (\displaystyle \binom{9}{3} = \frac{9!}{3!,6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = \frac{504}{6} = 84).
• Quick check: The numerator (9 \times 8 \times 7 = 504); dividing by 6 yields 84.
• Verdict: Correct (value equals 84).

Option D: (5! - 36)

• Full calculation: (5! = 120); (120 - 36 = 84).
• Quick check: Subtract 30 to get 90, then subtract another 6 → 84.
• Verdict: Correct (value equals 84).

Observation: All four sample options evaluate to 84, illustrating that many different mathematical constructions can yield the same result. In an actual test, only one will be presented as correct; the others will be deliberately altered (e.g., (7 \times 13), (2^{6}+21), (\binom{9}{4}), (5!-35)) to create distractors.

Strategies for Identifying the Correct Option

1. Estimate the magnitude
   • If an option involves an exponent like (2^{

• If an optioninvolves an exponent like (2^{n}), compare the power to nearby multiples of 10 (e.g., (2^{6}=64) is close to 60, (2^{7}=128) is near 130) to gauge whether adding or subtracting a constant will land in the target range.
• For factorial expressions, recall that (5! = 120) and (6! = 720); if the target is modest (under 200), only low factorials are plausible, letting you discard options with (n!\ge 6!) outright.
• Binomial coefficients grow quickly; (\binom{n}{k}) is maximized near (k=n/2). If the target is 84, note that (\binom{9}{3}=84) and (\binom{9}{6}=84) (symmetry), while (\binom{8}{3}=56) and (\binom{10}{3}=120) bracket the value, helping you eliminate choices with (n) far from 9.

2. Spot algebraic shortcuts

   • Recognize common products: (7\times12), (8\times10+4), (6\times14) etc. If you see a factor pair that multiplies to a nearby round number, adjust mentally.
   • Use the identity (a^{2}-b^{2}=(a-b)(a+b)) when you encounter differences of squares; it can turn a seemingly bulky expression into a simple multiplication.
   • For sums of powers, remember that (2^{n}+2^{n}=2^{n+1}). If an option adds the same power twice, you can instantly rewrite it as a single higher power.

3. Apply order‑of‑operations hierarchies

   • Scan for parentheses first; any operation inside them must be resolved before tackling exponents, multiplication/division, or addition/subtraction.
   • When multiplication and division appear together, work left‑to‑right; the same rule applies to addition and subtraction. This prevents the common mistake of “doing all multiplication before any division” when the expression is mixed.
   • If an option contains a long chain of only addition and subtraction, you can reorder terms freely (commutative property) to group convenient pairs that sum to round numbers.

4. Eliminate via parity and divisibility

   • Determine whether the target value is even or odd. Any option that yields an opposite parity can be discarded immediately.
   • Check divisibility by small primes (3, 5, 7). For instance, 84 is divisible by 3 and 7 but not by 5; an expression that clearly produces a multiple of 5 (e.g., ending in 0 or 5) can be ruled out.
   • For factorials, note that (n!) for (n\ge5) ends in zero; subtracting a number that does not end in zero will preserve the trailing zero only if the subtrahend also ends in zero. This quick check often knocks out distractors.

5. Use approximation when exact calculation is costly - Replace each component with a nearby “friendly” number, compute the rough result, and see if it overshoots or undershoots the target by a plausible margin.

   • If the approximation already deviates by more than the maximum possible correction from the omitted details, the option cannot be correct.
   • This technique is especially handy for large exponents or products where mental arithmetic would be tedious.

Conclusion

By systematically classifying expressions, estimating their magnitude, spotting algebraic shortcuts, respecting the order of operations, applying parity and divisibility rules, and resorting to smart approximations, you can rapidly isolate the correct answer among a set of options. These strategies transform what might appear as a daunting trial‑and‑error process into a streamlined, logical hunt for the expression that truly evaluates to the desired value. Mastering them not only saves time on tests but also deepens your intuitive grasp of how numbers interact across different mathematical constructions.