QuanticaAlgebra IOpen in the courseYou can rewrite an expression in a different order, or group its numbers differently. Sometimes the value stays exactly the same, so the rewrite is free and just makes the arithmetic easier. Sometimes it changes, and you land on a different number. This lesson is about telling the two apart.
Problem
A running total to knock out in your head. Find \(38+65+62+35\). Straight down the line this is a grind, but the four numbers were chosen so that a little rearranging pairs them into round hundreds. What is the sum?
Show a hint
- Two of these four addends already sit close to a hundred. Can you find a pairing that lands on round numbers?
Show the full solution
Reorder to \(38+62=100\) and \(65+35=100\), so the sum is \(100+100=\boxed{200}\). Reordering addends never changes a sum, so it is always worth scanning a list for pairs that make round numbers.
Problem
Now a product built the same way. Find the value of \(8\cdot 17\cdot 125\) by choosing which two factors to multiply first.
Show a hint
- You may multiply the factors of a product in any order. Which two of these three make a power of ten?
Show the full solution
Pair \(8\cdot 125=1000\) first, then \(1000\cdot 17=\boxed{17000}\). Starting with \(8\cdot 17=136\) lands in the same place, but the arithmetic is much uglier.
Problem
A classmate evaluates \(40-12-6\). Instead of working left to right, they group the last two numbers, compute \(12-6=6\) first, and report \(40-6=34\). That is not what the expression says. Worked correctly, what is \(40-12-6\)?
Show a hint
- Subtraction is done left to right. Grouping a different pair first is a change, not a shortcut. Does it give the same number?
Show the full solution
Left to right, \(40-12=28\) and \(28-6=\boxed{22}\). The classmate's regrouping computes \(40-(12-6)=34\), a different value, because subtraction is not associative.
Problem
Rewrite every subtraction in \(58-24+42-8\) as adding a negative, then reorder to pair friendly numbers. What is the value?
Show a hint
- Turn the chain into \(58+(-24)+42+(-8)\). Now it is one long sum, so reorder freely.
- Pair the two that reach a round number, then add the two negatives.
Show the full solution
Rewritten, \(58+(-24)+42+(-8)\) is a single sum with each minus stuck to its number, so reorder to \(58+42=100\) and \((-24)+(-8)=-32\). Then \(100+(-32)=\boxed{68}\).
Problem
A student wants to reorder \(30-18+5\) and writes \(30-5+18\), swapping the \(18\) and the \(5\) but leaving the signs in place. That gives \(43\), which is wrong, because the minus belongs to the \(18\). Evaluated correctly, what does \(30-18+5\) equal?
Show a hint
- Read it as \(30+(-18)+5\). When you move a number, its sign has to move with it.
Show the full solution
Write it as \(30+(-18)+5\). Left to right, \(30-18=12\) and \(12+5=\boxed{17}\). The minus has to travel with the \(18\). Writing \(30-5+18\) flips both signs, which is how it lands on \(43\).
Problem
Rewrite the division in \(34\cdot 19\div 34\) as multiplying by a reciprocal, then reorder the factors so a pair cancels. What is the value?
Show a hint
- Read it as \(34\cdot 19\cdot\frac{1}{34}\). Which two factors undo each other?
Show the full solution
As \(34\cdot 19\cdot\frac{1}{34}\), the factors reorder to \(\left(34\cdot\frac{1}{34}\right)\cdot 19=1\cdot 19\). The \(34\) and its reciprocal cancel, leaving \(\boxed{19}\).
Problem
Two clever pieces, one minus sign between them. Find the value of \(25\cdot 63\cdot 4-(88+47+12+53)\). Reorder inside the product and inside the sum, but do not move a piece across the minus sign.
Show a hint
- Inside the product, \(25\cdot 4\) is a round number. Inside the parentheses, look for two pairs that each make \(100\).
- The freedom to reorder lives within each level. You may not pull a factor out of the product and add it to the sum.
Show the full solution
In the product, \(25\cdot 4=100\), so \(25\cdot 63\cdot 4=100\cdot 63=6300\). In the sum, \(88+12=100\) and \(47+53=100\), so the parentheses hold \(200\). Then \(6300-200=\boxed{6100}\). The shuffling stays inside each piece, since the factors and the addends sit at different levels.
So order and grouping never change a sum or a product, while both can change a difference or a quotient. That changes once you rewrite the subtraction as adding a negative or the division as multiplying by a reciprocal, since the expression becomes a single chain of additions or multiplications that you can reorder freely. The next lesson is about turning word descriptions into expressions.
Practice these ideas
Practice
Add \(46+29+54+21\) by rearranging the four numbers into round pairs. What is the sum?
Show the solution
Reorder to \(46+54=100\) and \(29+21=50\), so \(100+50=\boxed{150}\).
Practice
Find the value of \(2\cdot 39\cdot 50\) by choosing which two factors to multiply first.
Show the solution
Pair \(2\cdot 50=100\), then \(100\cdot 39=\boxed{3900}\).
Practice
Rewrite each subtraction as adding a negative in \(73-26+27-14\), then pair friendly numbers. What is the value?
Show the solution
As one sum, \(73+27=100\) and \((-26)+(-14)=-40\), so \(100+(-40)=\boxed{60}\).
Practice
Rewrite the division as multiplying by a reciprocal, then find the value of \(61\cdot 7\div 61\).
Show the solution
As \(61\cdot\frac{1}{61}\cdot 7=1\cdot 7\), the \(61\) cancels its reciprocal, leaving \(\boxed{7}\).
Practice
A classmate reads \(70-35-8\) by grouping the last two numbers, computing \(35-8=27\) first, and reporting \(70-27=43\). Evaluated correctly, left to right, what does \(70-35-8\) equal?
Show the solution
Left to right, \(70-35=35\) and \(35-8=\boxed{27}\). The classmate computed \(70-(35-8)=43\), a different value, because subtraction is not associative.
Practice
Among the four operations \(+\), \(-\), \(\cdot\), and \(\div\), exactly two always let you swap the two numbers with no change in value. One of them is addition. Name the other one.
Show the solution
The other one is \(\boxed{\text{multiplication}}\), since \(a\cdot b=b\cdot a\). Subtraction and division fail the swap, because \(7-2\) is not \(2-7\) and \(20\div 4\) is not \(4\div 20\).
Practice
Add \(71+48+29+22\) by rearranging into round pairs. What is the sum?
Show the solution
Reorder to \(71+29=100\) and \(48+22=70\), so \(100+70=\boxed{170}\).
Practice
Division is not associative, so the grouping in \(96\div 8\div 2\) matters. Reading it correctly, left to right, what is its value? (Grouping as \(96\div(8\div 2)\) would give the different answer \(24\).)
Show the solution
Left to right, \(96\div 8=12\) and \(12\div 2=\boxed{6}\). Grouping \(8\div 2=4\) first gives \(96\div 4=24\), which is why division needs a fixed reading order.
Practice
Rewrite the divisions as multiplying by reciprocals, then reorder to cancel. Find the value of \(15\cdot 46\div 15\div 2\).
Show the solution
As \(15\cdot\frac{1}{15}\cdot 46\cdot\frac{1}{2}\), the \(15\) cancels, leaving \(46\cdot\frac{1}{2}=\boxed{23}\).
Practice
Insert exactly one pair of parentheses into \(24-8-6-2\) to make the result as large as possible. Enter that largest value.
Show the solution
Group as \(24-(8-6-2)\). The inside is \(8-6-2=0\), so the result is \(24-0=\boxed{24}\). Every other placement does worse, since \(24-(8-6)-2=20\), \(24-8-(6-2)=12\), and the ungrouped chain is only \(8\).