/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 In how many different ways can t... [FREE SOLUTION] | 91Ó°ÊÓ

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In how many different ways can the missing digits in this short multiplication be completed? $$ \begin{array}{r} \quad \square 6 \\ \times \quad \square \\ \hline \square 28 \\ \hline \end{array} $$ One would like students not only to master the direct operation of multiplying digits effectively, but also to notice that the inverse procedure of "identifying the multiples of a given integer that give rise to a specified output" depends on the HCF of the multiplier and the base (10) of the numeral system. \- Multiplying by \(1,3,7,\) or 9 induces a one-to-one mapping on the set of ten digits \(0-9\); so an inverse problem such as \(" 7 \times \square\) ends in 6 " has just one digit-solution. \- Multiplying by \(2,4,6,\) or 8 induces a two-to-one mapping onto the set of even digits (multiples of 2); so an inverse problem such as " \(6 \times \square\) ends in 4" has two digit-solutions, and an inverse problem such as " \(6 \times \square\) ends in 3 " has no digit-solutions. \- Multiplying by 5 induces a five-to-one mapping onto the multiples (0 and 5 ) of 5 , so an inverse problem such as " \(5 \times \square\) ends in 0 " has five digit-solutions and an inverse problem such as " \(5 \times \square\) ends in 3 " has no digit-solutions at all. \- Multiplying by 0 induces a ten-to-one mapping onto the multiples of 0 (namely 0 ); so an inverse problem such as " \(0 \times \square\) ends in 0 " has ten digit-solutions and an inverse problem such as " \(0 \times \square\) ends in 3 (or any digit other than 0 )" has no digit-solutions at all. The next problem shows - in a very simple setting - how elusive inverse problems can be. Here, instead of being asked to perform a direct calculation, the rules and the answer are given, and we are simply asked to invent a calculation that gives the specified output.

Short Answer

Expert verified
There are 2 solutions: using 0 or 5 for the missing digit in \( \square 6 \).

Step by step solution

01

Interpret the Problem

You need to fill in the missing digits in the multiplication problem. The multiplication setup is given, where the multiplicand is \( \square 6 \), the multiplier is \( \square \), and the product ends in \( \square 28 \). The task is to identify possible digits for each \( \square \).
02

Analyze the Final Digit of the Product

The final digit of the product is 8, which provides a clue for the inverse problem. Consider the last digit multiplication: "If \( \square \times 6 \) ends in an 8, what could \( \square \) be?" Since \( 6 \times 3 = 18 \), ending in 8, \( \square \) must be 3.
03

Determine the Possibilities for the Multiplicand's Leading Digit

Now, fill in the digit of the multiplicand \( \square 6 \), knowing \( 6 \times 3 \equiv 8 \pmod{10} \). Set up the multiplication: \( (a \times 3 \times 10) + (3 \times 6) = c28 \). Simplify it as \( 30a + 18 = c28 \). For this equation to be true, \( a \) must make \( 30a + 18 \equiv 28 \pmod{100} \).
04

Solve the Modulo Equation for \( a \)

We solve the equation \( 30a + 18 \equiv 28 \pmod{100} \). Rearrange to \( 30a \equiv 10 \pmod{100} \), and simplify to find \( a \equiv 1/3 \pmod{10} \). Solve this directly: \( a = 0, 5 \) meet the condition when thoroughly checked within the integer range of 0 to 9.
05

Count Different Solutions

For each value of \( a \) found in step 4, there is a unique solution for the multiplication configuration. Since \( a = 0 \) and \( a = 5 \) are valid, check \( 06 \times 3 = 18 \) gives \( 028 \), and \( 56 \times 3 = 168 \) gives \( 168 \). Thus, there are two possible solutions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inverse Operations
Inverse operations refer to methods that reverse the action of a given operation. In the context of multiplication, the inverse operation is division. For elementary problem solving, inverse operations are vital for deducing the missing components of a given problem like our exercise. When you are presented with the outcome of a multiplication and are asked to find the missing number, you use the concept of inverse operations to 'backtrack'. This involves thinking of what numbers multiply together to give the result you see, especially focusing on clues from the output like the digits.

Consider the exercise where you need to determine the digits such that a multiplication ends in a specific number. Knowing the product ends in 8, you have to figure out which digits, when multiplied with 6, end in an 8. Using inverse operations here means considering the reverse logic of how multiplication works.
Modulo Arithmetic
Modulo arithmetic, often referred to as "clock arithmetic," deals with numbers wrapped around upon reaching a particular value, called the modulus. In our case, we apply this concept by focusing on the last digits of numbers, crucially understanding how multiplication affects them. The original problem uses modulo 10 to focus on the units digit of numbers.

When solving these types of problems, it helps to think in terms of equivalence classes under addition and multiplication modulo 10. For example, when you see that the product ends in 28, you are specifically interested in the congruence property: numbers before 28 that are equivalent modulo 10. It’s about understanding which multipliers, under this system, will lead us back to the desired end digit.
  • Find the possible multipliers for a specific ending digit.
  • Use simple multiplication rules, like with small numbers, to discover patterns.
  • In our problem: figuring out, via modulo, what times 6 leads to a number ending in 8.
Multiplication Strategies
Effective multiplication strategies simplify the process, especially when dealing with known numbers. In elementary problems, it's about more than just times tables—it's understanding patterns. Strategies might include breaking down problems into smaller parts, using repeated addition, or combining known results for simplicity.

For this problem, once you know the units digit (from inverse operations), you can use that to deduce other parts of the problem quickly. The process involves both logic and direct multiplication, like identifying what underlies simple multiplication results.
  • Knowing tables helps visualize results quickly.
  • Checking smaller divisions and common factors ensures quick verification.
  • Trying different known tables can guide evaluating further unknown digits.
Digit Identification
Digit identification is the technique of determining unknown digits in number problems, an essential skill in arithmetic and mathematical reasoning. When some digits in a number are hidden, like in the exercise, you have to use logical deduction to unravel them.

Here, you find parts: one resulting from the product must end in a known number. This identification process involves checking feasible options and cross-confirming with other digits.
  • Refers to using strategic guessing and checking.
  • Understanding the properties of single-digit multiplication aids in solving directly.
  • Focuses on the effectiveness of matching digits in problems.
By solving simpler parts first, it becomes easier to piece together the correct multiplication combination.

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Most popular questions from this chapter

[This problem requires a mixture of serious thought and written proof.] (a) I choose six integers between 10 and 19 (inclusive). (i) Prove that some pair of integers among my chosen six must be relatively prime. (ii) Is it also true that some pair must have a common factor? (b) I choose six integers in the nineties (from \(90-99\) inclusive). (i) Prove that some pair among my chosen integers must be relatively prime. (ii) Is it also true that some pair must have a common factor? (c) I choose \(n+1\) integers from a run of \(2 n\) consecutive integers. (i) Prove that some pair among the chosen integers must be relatively prime. (ii) Is it also true that some pair must have a common factor?

(a) In the "24 game" you are given four numbers. Your job is to use each number once, and to combine the four numbers using any three of the four basic arithmetical operations - using the same operation more than once if you wish, and as many brackets as you like (but never concatenating different numbers, such as "3" and "4" to make " \(34 "\) ). If the given numbers are \(3,3,4,4,\) then one immediately sees \(3 \times 4+3 \times 4=24 .\) With 3,3,5 , 5 it may take a little longer, but is still fairly straightforward. However, you may find it more challenging to make 24 in this way: (i) using the four numbers 3,3,6,6 (ii) using the four numbers 3,3,7,7 (iii) using the four numbers 3,3,8,8 . (b) Suppose we restrict the numbers to be used each time to "four \(4 \mathrm{~s}\) " \((4,4,4,4),\) and change the goal from "make \(24 ",\) to "make each answer from \(0-10\) using exactly four \(4 \mathrm{~s} "\). (i) Which of the numbers \(0-10\) cannot be made? (ii) What if one is allowed to use squaring and square roots as well as the four basic operations? What is the first inaccessible integer? Calculating by turning the handle deterministically (as with addition, or multiplication, or multiplying out brackets, or differentiating) is a valuable skill. But such direct procedures are usually only the beginning. Using mathematics and solving problems generally depend on the corresponding inverse procedures \(-\) where a certain amount of juggling and insight is needed in order to work backwards (as with subtraction, or division, or factorisation, or integration). For example, in applications of calculus, the main challenge is to solve differential equations (an inverse problem) rather than to differentiate known functions. Problem 14 captures the spirit of this idea in the simplest possible context of arithmetic: the required answer is given, and we have to find how (or whether) that answer can be generated. We will meet more interesting examples of this kind throughout the rest of the collection.

(a)(i) Explain why any integer that is a factor (or a divisor) of both \(m\) and \(n\) must also be a factor of their difference \(m-n,\) and of their sum \(m+n\). (ii) Prove that $$ H C F(m, n)=H C F(m-n, n) $$ (iii) Use this to calculate in your head \(H C F(1001,91)\) without factorising either number. (b)(i) Prove that: \(H C F(m, m+1)=1\). (ii) Find \(H C F(m, 2 m+1)\). (iii) Find \(H C F\left(m^{2}+1, m-1\right)\).

The three integers \(a=3, b=4, c=5\) in the Pythagorean triple (3,4,5) form an arithmetic progression: that is, \(c-b=b-a\). Find all Pythagorean triples \((a, b, c)\) which form an arithmetic progression \(-\) that is, for which \(c-b=b-a\)

(a) In a sale which offers "15\% discount on all marked prices" I buy three articles: a pair of trainers priced at \(£ 57.74,\) a T-shirt priced at \(£ 17.28\), and a yo-yo priced at \(£ 4.98 .\) Using only mental arithmetic, work out how much I should expect to pay altogether. (b) Some retailers display prices without adding VAT - or "sales tax" - at \(20 \%\) (because their main customers need to know the pre-VAT price). Suppose the prices in part (a) are the prices before adding VAT. Each price then needs to be adjusted in two ways - adding VAT and subtracting the discount. Should I add the VAT first and then work out the discount? Or should I apply the discount first and then add the VAT? (c) Suppose the discount in part (b) is no longer \(15 \%\). What level of discount would exactly cancel out the addition of VAT at \(20 \% ?\)

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