Often while working, an arithmetical answer is needed — exact is sometimes required, approximate is often acceptable — maybe the basis of a decision or a path forward is based on a few quick calculations, with more rigorous calculations to follow
Let’s illustrate with examples from the two groups
First, we’ll learn to multiply by 11 by just writing down the answer
Question: What is $\bold{6{,}587{,}236\cdot11}$ ?
Solution:
$= \bold{72{,}459{,}596}$
Try it by yourself this time
Try it by yourself this time
Remember the binomial theorem or Pascal’s Triangle. Let’s go back a few years and refresh.
\[(a+b)^2 = a^2+\bold{2}ab+b^2\] \[(a+b)^3 = a^3+\bold{3}a^2b+\bold{3}ab^2+b^3\] \[(a+b)^4 = a^4+\bold{4}a^3b+\bold{6}a^2b^2+\bold{4}ab^3+b^4\]
\[\bold{1}\] \[\bold{1}\ \ \ \bold{1}\] \[\bold{1}\ \ \ \bold{2}\ \ \ \bold{1}\] \[\bold{1}\ \ \ \bold{3}\ \ \ \bold{3}\ \ \ \bold{1}\] \[\bold{1}\ \ \ \bold{4}\ \ \ \bold{6}\ \ \ \bold{4}\ \ \ \bold{1}\] \[\bold{1}\ \ \ \bold{5}\ \ \ \bold{10}\ \ \bold{10}\ \ \ \bold{5}\ \ \ \bold{1}\] \[\bold{1}\ \ \ \bold{6}\ \ \ \bold{15}\ \ \bold{20}\ \ \bold{15}\ \ \ \bold{6}\ \ \ \bold{1}\]
Second, we’ll try a few exponent calculations
Question: What is $\bold{1.10^2}$ ?
Solution:
$= 1+2(0.10)+(0.10)^2$
$= 1+0.20+0.01 = \bold{1.21}$
$= 1+0.20+0.01 = \bold{1.21}$
And $\bold{1.10^3}$ ?
Solution:
$= 1+3(0.10)+3(0.10)^2+(0.10)^3$
$= 1+0.300+0.030+0.001$
$= \bold{1.331}$
$= 1+0.300+0.030+0.001$
$= \bold{1.331}$
Question: What is $\bold{1.05^4}$ ?
Solution:
$1+4(0.05)+6(0.05)^2+4(0.05)^3+(0.05)^4$
Now dropping the last two terms, we have
Now dropping the last two terms, we have
$\approx 1+0.200000+6(0.002500) … $
$\approx 1+0.200000+ (0.015000) … $
$\approx \bold{1.215}$
$\approx 1+0.200000+ (0.015000) … $
$\approx \bold{1.215}$
The exact answer is $1.21550625$
Therefore, by using only three terms we have an error of less than $0.05%$
Therefore, by using only three terms we have an error of less than $0.05%$
Now let’s consider a few squares.
Question: What is $\bold{73^2}$ ?
Recall that $(a+b)^2 = a^2+\bold{2}ab+b^2$
Solution: $(70)^2+\bold{2}(70)(3)+(3)^2$
$= 4900+6 \cdot 70+9$
$= 4900+420+9 = \bold{5329}$
$= 4900+420+9 = \bold{5329}$
Question: What is $\bold{88^2}$ ?
Recall that $(a-b)^2 = a^2-\bold{2}ab+b^2$
Solution: $(90)^2-\bold{2}(90)(2)+(2)^2$
$= 8100-4 \cdot 90+4$
$= 8100-360+4$
$= 8100-400+44 = \bold{7744}$
$= 8100-360+4$
$= 8100-400+44 = \bold{7744}$
Question: What is $\bold{36^2}$ ?
Again $(a-b)^2 = a^2-\bold{2}ab+b^2$
Solution: $(40)^2-\bold{2}(40)(4)+(4)^2$
$= 1600-8 \cdot 40+4$
$= 1600-320+16$
$= 1600-400+96 = \bold{1296}$
$= 1600-320+16$
$= 1600-400+96 = \bold{1296}$
Now let’s consider common temperature conversions.
Start with determining the Fahrenheit equivalent of a Centigrade temperature ?
Recall $\bold{F} = \frac{9}{5} \cdot \bold{C}+32$
Or $\bold{F} = [1.80] \cdot \bold{C}+32$
Easier if approached using $\bold{F} = [0.90 \cdot 2] \cdot \bold{C}+32$
Which is simply $[2 \cdot \bold{C}]$ less $10%$ plus $32$
Or $\bold{F} = [1.80] \cdot \bold{C}+32$
Easier if approached using $\bold{F} = [0.90 \cdot 2] \cdot \bold{C}+32$
Which is simply $[2 \cdot \bold{C}]$ less $10%$ plus $32$
Question: What is the Fahrenheit equivalent of $350^\circ\text{C}$ ?
Say $[2 \cdot \bold{350}]$ less $10%$ plus $32$
Say $700-70$ plus $32$
Say $630$ plus $32$
Therefore, $\bold{662}^\circ\text{F}$
Say $700-70$ plus $32$
Say $630$ plus $32$
Therefore, $\bold{662}^\circ\text{F}$
Recall $\bold{C} = \frac{5}{9} \cdot (\bold{F}-32)$
Or $\bold{C} = [0.55\bar5] \cdot [\bold{F}-32]$
Easier if approached using $\bold{C} = [0.50+(0.1) \cdot 0.50+(0.01) \cdot 0.50] \cdot [\bold{F}-32]$
So say $[\bold{F}-32]$ times half plus $[\bold{F}-32]$ times a tenth of a half and so forth
Or $\bold{C} = [0.55\bar5] \cdot [\bold{F}-32]$
Easier if approached using $\bold{C} = [0.50+(0.1) \cdot 0.50+(0.01) \cdot 0.50] \cdot [\bold{F}-32]$
So say $[\bold{F}-32]$ times half plus $[\bold{F}-32]$ times a tenth of a half and so forth
Question: What is the Centigrade equivalent of $350^\circ\text{F}$ ?
Say $[350-32] = 318$
Say half of $318 = 159$
Say $159+15.9 \text{ etc}$
$= 159+16-0.1+1.6+0.16$
$ = 174.9+1.6+0.16 = 176.66$
Therefore, $\bold{176.66}^\circ\text{C}\approx \bold{177}^\circ\text{C}$.
Say half of $318 = 159$
Say $159+15.9 \text{ etc}$
$= 159+16-0.1+1.6+0.16$
$ = 174.9+1.6+0.16 = 176.66$
Therefore, $\bold{176.66}^\circ\text{C}\approx \bold{177}^\circ\text{C}$.
Question: What is the Centigrade equivalent of $351^\circ\text{F}$ ?
Say $[351-32] = 319 \text{ (odd number)}$
Say half of $319 = 159$ with a remainder of $1$
Say $159+15.9 \text{ etc}$
$= 159+16-0.1+1.6+0.16+[0.6]$
$= 176.66+[0.6]$
Say half of $319 = 159$ with a remainder of $1$
Say $159+15.9 \text{ etc}$
$= 159+16-0.1+1.6+0.16+[0.6]$
$= 176.66+[0.6]$
The extra $0.6$ is to account for the remainder of 1
Therefore, $\bold{177.26}^\circ\text{C}\approx \bold{177}^\circ\text{C}$.
Question: What is the Centigrade equivalent of $1650^\circ\text{F}$ ?
Say $[1650-32] = 1618}$
Say half of $1618 = 809$
Say $159+15.9 \text{ etc}$
$= 809+80.9+8.1+0.8$
$= 898.8$
Therefore, $\bold{898.8}^\circ\text{C}\approx \bold{899}^\circ\text{C}\approx \bold{900}^\circ\text{C}$.
Say half of $1618 = 809$
Say $159+15.9 \text{ etc}$
$= 809+80.9+8.1+0.8$
$= 898.8$
Therefore, $\bold{898.8}^\circ\text{C}\approx \bold{899}^\circ\text{C}\approx \bold{900}^\circ\text{C}$.
And last, Aliquot Parts
2, 5
$5 \cdot 2=10$
$2 \cdot 5=10$
To multiply by 2, we can divide by 5 if it’s easier that way
2, 50
$50 \cdot 2=100$
$2 \cdot 50=100$
4
$25 \cdot 4=100$
$4 \cdot 25=100$
To multiply by 25, we can divide by 4 if it’s easier that way
To divide by 25, we can multiply by 4 if it’s easier that way
8
$12.5 \cdot 8=100$
$8 \cdot 12.5=100$
3
$33.33 \cdot 3=100$
$3 \cdot 33.33=100$
6
$16.67 \cdot 6=16\frac{2}{3} \cdot 6=100$
$6 \cdot 16.67=6 \cdot 16\frac{2}{3}=100$
7
$14.29 \cdot 7=14\frac{2}{7} \cdot 7=100$
$7 \cdot 14.29=7 \cdot 14\frac{2}{7}=100$
9
$11.11 \cdot 9=11\frac{1}{9} \cdot 9=100$
$9 \cdot 11.11=9 \cdot 11\frac{1}{9}=100$
12
$8.33 \cdot 12=8\frac{1}{3} \cdot 12=100$
$12 \cdot 8.33=12 \cdot 8\frac{1}{3}=100$
15
$6.67 \cdot 15=6\frac{2}{3} \cdot 15=100$
$15 \cdot 6.67=15 \cdot 6\frac{2}{3}=100$
How to use the divisors of 1, of 10, and of 100
2, 5
$5 \cdot 2=10$
$2 \cdot 5=10$
To multiply by 5, we can divide by 2 and vice versa, whichever is easier!
What is $5 \cdot 184$ ?
Just say 184 over 2 times 10; therefore, $\bold{920}$.
What is $5 \cdot 121$ ?
Just say 121 times 5; therefore, $\bold{605}$.
What is $3243 / 5$ ?
Just say 3243 times 2 over 10; therefore, $\bold{648.6}$.
What is $2 \cdot 4585$ ?
Just say 4585 over 5 times 10; therefore, $\bold{9170}$.
What is $2 \cdot 1236$ ?
Just say 1236 times 2; therefore, $\bold{2472}$.
2, 50
$50 \cdot 2=100$
$2 \cdot 50=100$
5, 20
$20 \cdot 5=100$
$5 \cdot 20=100$
4
$25 \cdot 4=100$
$4 \cdot 25=100$
To multiply by 25, we can divide by 4 if it’s easier that way
What is $25 \cdot 160$ ?
Just say 160 over 4 times 100;
therefore, $\bold{4000}$.
therefore, $\bold{4000}$.
What is $600 / 25$ ?
Just say 6 times 4;
therefore, $\bold{24}$.
therefore, $\bold{24}$.
8
$12.5 \cdot 8=100$
$8 \cdot 12.5=100$
3
$33.33 \cdot 3=100$
$3 \cdot 33.33=100$
6
$16.67 \cdot 6=16\frac{2}{3} \cdot 6=100$
$6 \cdot 16.67=6 \cdot 16\frac{2}{3}=100$
7
$14.29 \cdot 7=14\frac{2}{7} \cdot 7=100$
$7 \cdot 14.29=7 \cdot 14\frac{2}{7}=100$
9
$11.11 \cdot 9=11\frac{1}{9} \cdot 9=100$
$9 \cdot 11.11=9 \cdot 11\frac{1}{9}=100$
12
$8.33 \cdot 12=8\frac{1}{3} \cdot 12=100$
$12 \cdot 8.33=12 \cdot 8\frac{1}{3}=100$
What is \(8.25%\) tax on \( \$47.00\) ?
Just say 47 over 12; therefore, ~\(\bold{\$4.00}\)
Could say 47 over 12 less say ~\(\$0.10\);
therefore, ~\(\bold{\$3.90}\).
The exact value is \(\bold{\$3.88}\)Could say 47 over 12 less say ~\(\$0.10\);
therefore, ~\(\bold{\$3.90}\).
15
$6.67 \cdot 15=6\frac{2}{3} \cdot 15=100$
$15 \cdot 6.67=15 \cdot 6\frac{2}{3}=100$
$\frac{2}{3} \cdot \frac{3}{2} = 1$
$\frac{2}{3} \cdot 1.5 = 1$
What is $\frac{2}{3}$ of $630$ ?
Just say 630 over 15 times 10; therefore, $42$ times $10$; therefore, $\bold{420}$.
