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Addition calculations to be carried out by hand, full worked solutions in feedback.

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

Addition of numbers of any size.

Calculate the following without using a calculator.

", "advice": "

Addition of numbers

Part a)

To add numbers we need to take into account their place value, so line numbers up placing the hundreds, tens and units in vertical columns.

\\[\\begin{align} \\var{a}&\\\\\\underline{+\\hspace{0.5cm}\\var{b}}&\\\\=\\hspace{1cm}\\end{align}\\]

Addition is started in the right hand column, the units. $\\var{au}+\\var{bu}=\\var{au+bu}$. Any tens in the answer to this are then carried across to be added to the tens total. In this case the answer has $\\var{((au+bu)-mod(au+bu,10))/10}$ in the tens column to carry across and $\\var{mod((au+bu),10)}$ in the units column which can go straight in for the answer.

\\[\\begin{align} \\var{a}&\\\\\\underline{+\\hspace{0.5cm}\\var{b}}&\\\\=\\hspace{0.5cm}\\ _\\var{((au+bu)-mod(au+bu,10))/10} \\var{mod(au+bu,10)}&\\\\\\end{align}\\]

The next column is the tens column, $\\var{at}+\\var{bt}=\\var{at+bt}$ Add to this anything that was carried over from the units, $\\var{((au+bu)-(mod(au+bu,10)))/10}$, makes it $\\var{at+bt+((au+bu)-mod(au+bu,10))/10}$. As we are adding the tens column, the $\\var{mod(at+bt+((au+bu)-mod(au+bu,10))/10,10)}$ is placed in the answer in this position and the $\\var{(at+bt+((au+bu)-mod(au+bu,10))/10-(mod(at+bt+((au+bu)-mod(au+bu,10))/10,10)))/10}$ is carried across to the hundreds column.

\\[\\begin{align} \\var{a}&\\\\\\underline{+\\hspace{0.5cm}\\var{b}}&\\\\=\\hspace{0.5cm}\\ _\\var{(at+bt-mod(at+bt,10))/10} \\var{mod(at+bt+((au+bu)-mod(au+bu,10))/10,10)}\\var{mod(au+bu,10)}&\\\\\\end{align}\\]

Then finally the hundreds column. There is only one value in this column so all that is needed is for any digits carried over from the total of the tens column need to be added to this value. So this is $\\var{bh}+\\var{(at+bt-mod(at+bt,10))/10}=\\var{bh+(at+bt-mod(at+bt,10))/10}$. This is then placed in the answer in the appropriate position, in the hundreds column, if the answer has two digits then the first of these is carried across and placed in the thousands position.

\\[\\begin{align} \\var{a}&\\\\\\underline{+\\hspace{0.5cm}\\var{b}}&\\\\=\\hspace{0.5cm}\\var{a+b}\\end{align}\\]

If there were more values then the process would continue until you had added each column in turn.

Part b)

\\[\\begin{align} \\var{2*a}&\\\\\\underline{+\\hspace{0.5cm}\\var{b}}&\\\\=\\hspace{1cm}\\end{align}\\]

Starting with the units $\\var{mod(2*a,10)}+\\var{bu}=\\var{mod(2*a,10)+bu}$.

So $\\var{((mod(2*a,10)+bu)-mod(mod(2*a,10)+bu,10))/10}$ is carried across to the tens column and $\\var{mod((mod(2*a,10)+bu),10)}$ in the units column answer.

\\[\\begin{align} \\var{2*a}&\\\\\\underline{+\\hspace{0.5cm}\\var{b}}&\\\\=\\hspace{0.5cm}\\ _\\var{((mod(2*a,10)+bu)-mod(mod(2*a,10)+bu,10))/10} \\var{mod(mod(2*a,10)+bu,10)}&\\\\\\end{align}\\]

The next column, $\\var{(mod(2*a,100)-mod(2*a,10))/10}+\\var{bt}=\\var{(mod(2*a,100)-mod(2*a,10))/10+bt}$

Add to this anything that was carried over from the units, $\\var{(mod(2*a,100)-mod(2*a,10))/10+bt}+\\var{((mod(2*a,10)+bu)-(mod(mod(2*a,10)+bu,10)))/10}=\\var{(mod(2*a,100)-mod(2*a,10))/10+bt+((mod(2*a,10)+bu)-mod(mod(2*a,10)+bu,10))/10}$.

So $\\var{mod((mod(2*a,100)-mod(2*a,10))/10+bt+((mod(2*a,10)+bu)-mod(mod(2*a,10)+bu,10))/10,10)}$ is placed in the answer and $\\var{((mod(2*a,100)-mod(2*a,10))/10+bt-mod((mod(2*a,100)-mod(2*a,10))/10+bt,10))/10}$ is carried across to the hundreds column.

\\[\\begin{align} \\var{2*a}&\\\\\\underline{+\\hspace{0.5cm}\\var{b}}&\\\\=\\hspace{0.5cm}\\ _\\var{((mod(2*a,100)-mod(2*a,10))/10+bt-mod((mod(2*a,100)-mod(2*a,10))/10+bt,10))/10} \\var{mod((mod(2*a,100)-mod(2*a,10))/10+bt+((mod(2*a,10)+bu)-mod(mod(2*a,10)+bu,10))/10,10)}\\var{mod(mod(2*a,10)+bu,10)}&\\\\\\end{align}\\]

Similarly for the hundreds column, $\\var{(mod(2*a,1000)-mod(2*a,100))/100}+\\var{bh}+\\var{((mod(2*a,100)-mod(2*a,10))/10+bt-mod((mod(2*a,100)-mod(2*a,10))/10+bt,10))/10}=\\var{(mod(2*a,1000)-mod(2*a,100))/100+bh+((mod(2*a,100)-mod(2*a,10))/10+bt-mod((mod(2*a,100)-mod(2*a,10))/10+bt,10))/10}$ This is then placed in the answer in the appropriate position, in the hundreds column, if the answer has two digits then the first of these is carried across and placed in the thousands position.

\\[\\begin{align} \\var{2*a}&\\\\\\underline{+\\hspace{0.5cm}\\var{b}}&\\\\=\\hspace{0.5cm}\\var{2*a+b}\\end{align}\\]

Addition of decimal numbers

Part c)

The process for addition of decimals is the same as that for whole numbers, lining up place values and starting addition with the right hand column. The important thing to remember when adding decimals is to ensure that all the numbers are aligned by the decimal point, this ensures each column contains the right place value.

Whole numbers are lined up so that they are directly to the left of the point. For ease you can add 0's in spaces where there are no numbers, making each number have the same number of decimal places.

\\[\\begin{align} \\var{a3}&\\\\\\var{c3}&\\\\\\underline{+\\hspace{0.5cm}\\var{d}}&\\\\=\\hspace{2cm}\\end{align}\\]

Then carrying out addition down each column, starting with right hand column $0+0+\\var{sigformat(mod((1000*d),10),1)}=\\var{sigformat(mod((1000*d),10),1)}$.

\\[\\begin{align} \\var{a3}&\\\\\\var{c3}&\\\\\\underline{+\\hspace{0.5cm}\\var{d}}&\\\\=\\hspace{2cm}\\var{sigformat(mod((1000*d),10),1)}&\\end{align}\\]

Continuing in this way $0+\\var{mod((100*c),10)}+\\var{(mod((1000*d)-mod(1000*d,10),100))/10}=\\var{mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10}$

In this case the answer has $\\var{(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10-(mod(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10,10)))/10}$ to carry across and $\\var{mod(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10,10)}$ which can go straight in for the answer.

\\[\\begin{align} \\var{a3}&\\\\\\var{c3}&\\\\\\underline{+\\hspace{0.5cm}\\var{d}}&\\\\=\\hspace{2cm}\\ _\\var{sigformat((mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10-(mod(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10,10)))/10,1)} \\var{sigformat(mod(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10,10),1)}\\var{sigformat(mod(1000*d,10),1)}&\\\\\\end{align}\\]

The next column is $0+\\var{(mod(100*c,100)-mod(100*c,10))/10}+\\var{(mod(1000*d,1000)-mod(1000*d,100))/100}=\\var{(mod(100*c,100)-mod(100*c,10))/10+(mod(1000*d,1000)-mod(1000*d,100))/100}$

Adding anything carried over $\\var{(mod(100*c,100)-mod(100*c,10))/10+(mod(1000*d,1000)-mod(1000*d,100))/100}+\\var{(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10-(mod(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10,10)))/10}=\\var{(mod(100*c,100)-mod(100*c,10))/10+(mod(1000*d,1000)-mod(1000*d,100))/100+(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10-(mod(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10,10)))/10}$

\\[\\begin{align} \\var{a3}&\\\\\\var{c3}&\\\\\\underline{+\\hspace{0.5cm}\\var{d}}&\\\\=\\hspace{2cm}\\ _\\var{((mod(100*c,100)-mod(100*c,10))/10+(mod(1000*d,1000)-mod(1000*d,100))/100+(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10-(mod(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10,10)))/10-mod((mod(100*c,100)-mod(100*c,10))/10+(mod(1000*d,1000)-mod(1000*d,100))/100+(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10-(mod(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10,10)))/10,10))/10} .\\var{mod((mod(100*c,100)-mod(100*c,10))/10+(mod(1000*d,1000)-mod(1000*d,100))/100+(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10-(mod(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10,10)))/10,10)}\\var{mod(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10,10)}\\var{sigformat(mod(1000*d,10),1)}&\\\\\\end{align}\\]

So for the above the carry across is $\\var{((mod(100*c,100)-mod(100*c,10))/10+(mod(1000*d,1000)-mod(1000*d,100))/100+(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10-(mod(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10,10)))/10-mod((mod(100*c,100)-mod(100*c,10))/10+(mod(1000*d,1000)-mod(1000*d,100))/100+(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10-(mod(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10,10)))/10,10))/10}$ and the addition continues in the same way as for the whole numbers to produce

\\[\\begin{align} \\var{a3}&\\\\\\var{c3}&\\\\\\underline{+\\hspace{0.5cm}\\var{d}}&\\\\=\\hspace{2cm}\\var{d+a+c}&\\end{align}\\]

Part d)

Again the first thing to do is to line the numbers up by thier decimal points. You do not need to put the numbers into size order but some people find this helpful.

\\[\\begin{align} \\var{j}&\\\\\\var{k}&\\\\\\underline{+\\hspace{0.5cm}\\var{l}}&\\\\=\\hspace{2cm}\\end{align}\\]

Then carrying out addition down each column, starting with right hand column $0+0+\\var{sigformat(mod((1000*l),10),1)}=\\var{sigformat(mod((1000*l),10),1)}$.

\\[\\begin{align} \\var{j}&\\\\\\var{k}&\\\\\\underline{+\\hspace{0.5cm}\\var{l}}&\\\\=\\hspace{2cm}\\var{sigformat(mod((1000*l),10),1)}&\\end{align}\\]

Continuing in this way $0+\\var{mod((100*k1),10)}+\\var{(mod((1000*l)-mod(1000*l,10),100))/10}=\\var{mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10}$

In this case the answer has $\\var{(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10-(mod(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10,10)))/10}$ to carry across and $\\var{mod(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10,10)}$ which can go straight in for the answer.

\\[\\begin{align} \\var{j}&\\\\\\var{k}&\\\\\\underline{+\\hspace{0.5cm}\\var{l}}&\\\\=\\hspace{2cm}\\ _\\var{(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10-(mod(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10,10)))/10} \\var{mod(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10,10)}\\var{sigformat(mod(1000*l,10),1)}&\\\\\\end{align}\\]

The next column is $0+\\var{(mod(100*k1,100)-mod(100*k1,10))/10}+\\var{(mod(1000*l,1000)-mod(1000*l,100))/100}=\\var{(mod(100*k1,100)-mod(100*k1,10))/10+(mod(1000*l,1000)-mod(1000*l,100))/100}$

Adding anything carried over $\\var{(mod(100*k1,100)-mod(100*k1,10))/10+(mod(1000*l,1000)-mod(1000*l,100))/100}+\\var{(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10-(mod(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10,10)))/10}=\\var{(mod(100*k1,100)-mod(100*k1,10))/10+(mod(1000*l,1000)-mod(1000*l,100))/100+(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10-(mod(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10,10)))/10}$

\\[\\begin{align} \\var{j}&\\\\\\var{k}&\\\\\\underline{+\\hspace{0.5cm}\\var{l}}&\\\\=\\hspace{2cm}\\ _\\var{((mod(100*k1,100)-mod(100*k1,10))/10+(mod(1000*l,1000)-mod(1000*l,100))/100+(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10-(mod(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10,10)))/10-mod((mod(100*k1,100)-mod(100*k1,10))/10+(mod(1000*l,1000)-mod(1000*l,100))/100+(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10-(mod(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10,10)))/10,10))/10} .\\var{mod((mod(100*k1,100)-mod(100*k1,10))/10+(mod(1000*l,1000)-mod(1000*l,100))/100+(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10-(mod(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10,10)))/10,10)}\\var{mod(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10,10)}\\var{sigformat(mod(1000*l,10),1)}&\\\\\\end{align}\\]

And carrying on in the same way until all columns have been completed to give

\\[\\begin{align} \\var{j}&\\\\\\var{k}&\\\\\\underline{+\\hspace{0.5cm}\\var{l}}&\\\\=\\hspace{2cm}\\var{j1+k1+l}&\\end{align}\\]

Part e)

Again lining the numbers up around the decimal point and adding each column in turn gives

\\[\\begin{align} \\var{b3}&\\\\\\var{l}&\\\\\\underline{+\\hspace{0.5cm}\\var{c3}}&\\\\=\\hspace{2cm}\\var{b+l+c}&\\end{align}\\]

Additional resources

For further examples and questions

Multi Digit Addition - Khan Academy

Adding Decimals - Khan Academy

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