three digit number
\n100*a[1]+10*b[1]+b[6]
", "definition": "100*a[1]+10*b[1]+b[6]", "templateType": "anything", "name": "g"}, "c": {"group": "Ungrouped variables", "description": "Two digit number generated from a and b
", "definition": "10*a[4] +b[4]", "templateType": "anything", "name": "c"}, "f": {"group": "Ungrouped variables", "description": "two digit number
\n10*a[2]+b[2]
", "definition": "10*a[2]+b[2]", "templateType": "anything", "name": "f"}, "b": {"group": "Ungrouped variables", "description": "", "definition": "shuffle(0..9)[0..10]", "templateType": "anything", "name": "b"}, "d": {"group": "Ungrouped variables", "description": "A three digit number generated from
\n100a[3] +10b[3]+b[5]
", "definition": "100*a[3]+10*b[3]+b[5]", "templateType": "anything", "name": "d"}}, "statement": "Complete the following without using a calculator
", "parts": [{"unitTests": [], "extendBaseMarkingAlgorithm": true, "variableReplacements": [], "showFeedbackIcon": true, "customMarkingAlgorithm": "", "type": "gapfill", "variableReplacementStrategy": "originalfirst", "sortAnswers": false, "prompt": "There are many different ways of carrying out multiplication by hand. For each method the calculation is set out in a different way. All have one thing in common, they involve finding the solution using only single digit multiplication, times tables facts, and place value knowledge. Use is made of the distributive property of multiplication, enabling the multiplication to be broken up and tackled in smaller chunks that then summed equal the same as the original multiplication.
\nFor example
\n\\[\\begin{align}26\\times15&= (20+6)\\times(10+5)\\\\&=20\\times(10+5)+6\\times(10+5)\\\\&=20\\times10+20\\times5+6\\times10+6\\times5\\\\&=200+100+60+30\\\\\\end{align}\\]
\n\n\n
Line up the numbers one on top of the other, usually with smaller number at the bottom, keeping the same place value in each column.
\nMultiplication is then carried out starting with the bottom right digit. This is multiplied with the top right digit and the answer placed in the bottom, carrying across any 10's, similarly to addition. This means that no matter what size the numbers being multiplied are only multof single idgits are needed so times e answer
\ni) Setting out the calculation
\n\\[\\begin{align}\\var{c}&\\\\\\underline{\\times\\hspace{2cm}\\var{a[1]}}&\\\\\\end{align}\\]
\nFirst step: Multiply the bottom right digit by the top right digit.
\n\\[\\begin{align}\\var{c}&\\\\\\underline{\\times\\hspace{2cm}\\var{a[1]}}&\\\\\\ _\\var{(a[1]*b[4]-mod(a[1]*b[4],10))/10} \\var{mod(a[1]*b[4],10)}\\\\\\end{align}\\]
\nThen this bottom digit multiplys the next digit along at the top, and the answer has the amount carried across added to this. So $\\var{a[1]} \\times \\var{a[4]}= \\var{a[1]*a[4]}$, adding to this the amount that has been carried over $\\var{a[1]*a[4]}+\\var{(a[1]*b[4]-mod(a[1]*b[4],10))/10}=\\var{a[1]*a[4]+(a[1]*b[4]-mod(a[1]*b[4],10))/10}$
\n\\[\\begin{align}\\var{c}&\\\\\\underline{\\times\\hspace{2cm}\\var{a[1]}}&\\\\\\var{a[1]*a[4]+(a[1]*b[4]-mod(a[1]*b[4],10))/10} \\var{mod(a[1]*b[4],10)}\\\\\\end{align}\\]
\nii) Similarly for slightly longer numbers with more place values the calculation is set out as
\n\\[\\begin{align}\\var{d}&\\\\\\underline{\\times\\hspace{2cm}\\var{a[2]}}&\\\\\\end{align}\\]
\nAgain the first step is to multiply the bottom right digit by the top right digit
\n\\[\\begin{align}\\var{d}&\\\\\\underline{\\times\\hspace{2cm}\\var{a[2]}}&\\\\\\ _\\var{(a[2]*b[5]-mod(a[2]*b[5],10))/10} \\var{mod(a[2]*b[5],10)}\\\\\\end{align}\\]
\nThen this bottom digit multiplys the next digit along at the top, and the answer has the amount carried across added to this. So $\\var{a[2]} \\times \\var{b[3]}= \\var{a[2]*b[3]}$, adding to this the amount that has been carried over $\\var{a[2]*b[3]}+\\var{(a[2]*b[5]-mod(a[2]*b[5],10))/10}=\\var{a[2]*b[3]+(a[2]*b[5]-mod(a[2]*b[5],10))/10}$ This is then placed at the bottom, keeping in mind the place value and making sure the location is correct for each digit.
\n\\[\\begin{align}\\var{d}&\\\\\\underline{\\times\\hspace{2cm}\\var{a[2]}}&\\\\\\ _\\var{((10*b[3]+b[5])*a[2]- mod((10*b[3]+b[5])*a[2],100))/100} \\var{mod(a[2]*b[3]+(a[2]*b[5]-mod(a[2]*b[5],10))/10,10)}\\var{mod((10*b[3]+b[5])*a[2],10)}\\\\\\end{align}\\]
\nOn to the next digit and the process is repeated
\nSo $\\var{a[2]} \\times \\var{a[3]}= \\var{a[2]*a[3]}$, adding to this the amount that has been carried over $\\var{((10*b[3]+b[5])*a[2]- mod((10*b[3]+b[5])*a[2],100))/100} $ gives $\\var{a[2]*a[3]+((10*b[3]+b[5])*a[2]- mod((10*b[3]+b[5])*a[2],100))/100}$. Placing this answer at the bottom gives the final answer.
\n\\[\\begin{align}\\var{d}&\\\\\\underline{\\times\\hspace{2cm}\\var{a[2]}}&\\\\\\var{d*a[2]}\\\\\\end{align}\\]
\n\nThis method inlvolves breaking the numbers being multiplied into place value parts, multiplying these parts and then adding the answers to obtain the final result.
\nBy doing this any size multiplication can be carried out using single digit multiplication facts and then counting the number of zeros and placing these in each answer to ensure the correct place value is obtained for each answer.
\ni) Set the grid out and then start completing. The grid can be filled in any order, multiplying the number at the top of the column with the number to the left of the row.
\nSo to complete the grid the following calculations are needed
\n$\\var{a[4]}\\times\\var{a[1]}=\\var{a[4]*a[1]}$ now to take into consideration the place value. There is one zero in the sum so there needs to be one added to the answer to give $\\var{a[4]*10}\\times\\var{a[1]}=\\var{a[4]*10*a[1]}$
\n$\\var{b[4]}\\times\\var{a[1]}=\\var{b[4]*a[1]}$ Placing these in the relevant parts of the grid
\n| \n | $\\var{a[4]*10}$ | \n$\\var{b[4]}$ | \n
| $\\var{a[1]}$ | \n$\\var{a[4]*10*a[1]}$ | \n$\\var{b[4]*a[1]}$ | \n
Then add these numbers together
\n\\[\\begin{align}\\var{a[4]*10*a[1]}&\\\\\\underline{+{\\hspace0.5cm}\\var{b[4]*a[1]}}&\\\\\\var{a[1]*c}&\\\\\\end{align}\\]
\nSo this gives the solution by $\\var{a[4]*10*a[1]}+\\var{b[4]*a[1]}=\\var{a[1]*c}$
\nii) Again the numbers are broken down into place value parts
\n| \n | $\\var{a[3]*100}$ | \n$\\var{b[3]*10}$ | \n$\\var{b[5]}$ | \n
| $\\var{a[2]}$ | \n\n | \n | \n |
Carring out the multiplication
\n$\\var{a[3]}\\times\\var{a[2]}=\\var{a[3]*a[2]}$ now to take into consideration the place value. There are two zeros in the sum so there needs to be two added to the answer to give $\\var{a[3]*100}\\times\\var{a[2]}=\\var{a[3]*100*a[2]}$
\n$\\var{b[3]}\\times\\var{a[2]}=\\var{b[3]*a[2]}$ now to take into consideration the place value. There is one zero in the sum so there needs to be one added to the answer to give $\\var{b[3]*10}\\times\\var{a[2]}=\\var{b[3]*10*a[2]}$
\n$\\var{b[5]}\\times\\var{a[2]}=\\var{b[5]*a[2]}$ Placing these in the relevant parts of the grid
\n\n| \n | $\\var{a[3]*100}$ | \n$\\var{b[3]*10}$ | \n$\\var{b[5]}$ | \n
| $\\var{a[2]}$ | \n$\\var{a[3]*100*a[2]}$ | \n$\\var{b[3]*10*a[2]}$ | \n$\\var{b[5]*a[2]}$ | \n
Adding these gives
\n\\[\\begin{align}\\var{((a[3])*(a[2]))*100}&\\\\\\var{a[2]*b[3]*10}&\\\\\\underline{+{\\hspace 0.5cm}\\var{b[5]*a[2]}}&\\\\\\var{a[2]*d}&\\\\\\end{align}\\]
\n\n\nThis method is a combination of the traditional and the grid method. It is particuarly useful if you need to calculate large numbers by hand. It can take some time to get familiar with setting out the grid and placing the numbers. The calculations are all up to 10 x 10 and so are no harder than than the other methods. The main difference is that in placing your answers you separate the tens from the unist using a diagonal divider within each box. The final addition is then carried out diagonally. Use the link below to open an App that has step by step demonstration of how the Lattice method is used. Place your numbers into the boxes above and to the right of the grid and then click on step to see the calculation carried out.
\nLink to App for Lattice Method - Geogebra
\n\ni) Setting out the calculation
\n\\[\\begin{align}\\var{c}&\\\\\\underline{\\times\\hspace{2cm}\\var{f}}&\\\\\\end{align}\\]
\nFirst step: Multiply the bottom right digit by the top right digit.
\n\\[\\begin{align}\\var{c}&\\\\\\underline{\\times\\hspace{2cm}\\var{f}}&\\\\\\ _\\var{(b[4]*b[2]-mod(b[4]*b[2],10))/10} \\var{mod(b[4]*b[2],10)}\\\\\\end{align}\\]
\nThen this bottom digit multiplys the next digit along at the top, and the answer has the amount carried across added to this. So $\\var{b[2]} \\times \\var{a[4]}= \\var{b[2]*a[4]}$, adding to this the amount that has been carried over $\\var{b[2]*a[4]}+\\var{(b[2]*b[4]-mod(b[2]*b[4],10))/10}=\\var{b[2]*a[4]+(b[2]*b[4]-mod(b[2]*b[4],10))/10}$
\n\\[\\begin{align}\\var{c}&\\\\\\underline{\\times\\hspace{2cm}\\var{f}}&\\\\\\var{b[2]*(10*a[4]+b[4])}\\\\\\end{align}\\]
\nSo far $\\var{c}\\times\\var{b[2]}$ has been calculated, this is $\\var{b[2]*(10*a[4]+b[4])}$
\nThe next number on the bottom row is now used to multiply each digit in the top row. As this next number $\\var{a[2]}$ is in the tens column it has to be remembered when carrying out the calculation that the numbers will need to be multiplied by 10, we do this by placing a 0 in the units column below the existing answer and positioning the answer found for this next digit to the left of this 0.
\n\\[\\begin{align}\\var{c}&\\\\\\underline{\\times\\hspace{2cm}\\var{f}}&\\\\\\var{b[2]*(10*a[4]+b[4])}&\\\\0&\\\\\\end{align}\\]
\nSo the answer to the calculation $\\var{a[2]}\\times\\var{b[4]}$ is placed below the existing answer to the left of the 0, carrying across in the usual way.
\n\\[\\begin{align}\\var{c}&\\\\\\underline{\\times\\hspace{2cm}\\var{f}}&\\\\\\var{b[2]*(10*a[4]+b[4])}&\\\\\\ _\\var{(a[2]*b[4]-mod(a[2]*b[4],10))/10}\\var{mod(a[2]*b[4],10)}0&\\\\\\end{align}\\]
\nNow moving along again to the next number at the top with the same bottom number $\\var{a[2]}\\times\\var{a[4]}$ which is $\\var{a[2]*a[4]}$. adding to this anything that was carried across makes it $\\var{a[2]*a[4]}+\\var{(a[2]*b[4]-mod(a[2]*b[4],10))/10}=\\var{a[2]*a[4]+(a[2]*b[4]-mod(a[2]*b[4],10))/10}$. Placing this in carefully gives
\n\\[\\begin{align}\\var{c}&\\\\\\underline{\\times\\hspace{2cm}\\var{f}}&\\\\\\var{b[2]*(10*a[4]+b[4])}&\\\\\\var{a[2]*(10*a[4]+b[4])}0&\\\\\\end{align}\\]
\nSo the calculation has been broken down into two parts $\\var{b[2]}\\times\\var{a[4]*10+b[4]}=\\var{b[2]*(a[4]*10+b[4])}$ and $\\var{a[2]*10}\\times\\var{(a[4]*10+b[4])}=\\var{a[2]*10*(a[4]*10+b[4])}$. These two parts together make the final answer as so adding them together gives
\n\\[\\begin{align}\\var{c}&\\\\\\underline{\\times\\hspace{2cm}\\var{f}}&\\\\\\var{b[2]*(10*a[4]+b[4])}&\\\\\\underline{+{\\hspace {0.5cm}}\\var{a[2]*(10*a[4]+b[4])}0}&\\\\\\var{c*f}&\\\\\\end{align}\\]
\n\nii) Similarly for more digits the method continues in the same way
\n\n\\[\\begin{align}\\var{d}&\\\\\\underline{\\times\\hspace{2cm}\\var{c}}&\\\\\\end{align}\\]
\nAgain the first step is to multiply the bottom right digit by the top right digit
\n\\[\\begin{align}\\var{d}&\\\\\\underline{\\times\\hspace{2cm}\\var{c}}&\\\\\\ _\\var{(b[4]*b[5]-mod(b[4]*b[5],10))/10} \\var{mod(b[4]*b[5],10)}\\\\\\end{align}\\]
\n\nThen this bottom digit multiplys the next digit along at the top, and the answer has the amount carried across added to this. So $\\var{b[4]} \\times \\var{b[3]}= \\var{b[4]*b[3]}$, adding to this the amount that has been carried over $\\var{b[4]*b[3]}+\\var{(b[4]*b[5]-mod(b[4]*b[5],10))/10}=\\var{b[4]*b[3]+(b[4]*b[5]-mod(b[4]*b[5],10))/10}$ This is then placed at the bottom, keeping in mind the place value and making sure the location is correct for each digit.
\n\\[\\begin{align}\\var{d}&\\\\\\underline{\\times\\hspace{2cm}\\var{c}}&\\\\\\ _\\var{((10*b[3]+b[5])*b[4]- mod((10*b[3]+b[5])*b[4],100))/100} \\var{mod(b[4]*b[3]+(b[4]*b[5]-mod(b[4]*b[5],10))/10,10)}\\var{mod((10*b[3]+b[5])*b[4],10)}\\\\\\end{align}\\]
\nOn to the next digit and the process is repeated
\nSo $\\var{b[4]} \\times \\var{a[3]}= \\var{b[4]*a[3]}$, adding to this the amount that has been carried over $\\var{((10*b[3]+b[5])*b[4]- mod((10*b[3]+b[5])*b[4],100))/100} $ gives $\\var{b[4]*a[3]+((10*b[3]+b[5])*b[4]- mod((10*b[3]+b[5])*b[4],100))/100}$. Placing this answer at the bottom gives the final answer.
\n\\[\\begin{align}\\var{d}&\\\\\\underline{\\times\\hspace{2cm}\\var{c}}&\\\\\\var{d*b[4]}\\\\\\end{align}\\]
\nNow along to the next digit on the bottom to repeat the process, remembering to place a 0 in first as the digit is in the 10's column.
\n\\[\\begin{align}\\var{d}&\\\\\\underline{\\times\\hspace{2cm}\\var{c}}&\\\\\\var{d*b[4]}&\\\\0&\\\\\\end{align}\\]
\n$\\var{a[4]}\\times\\var{b[5]}=\\var{a[4]*b[5]}$
\n\\[\\begin{align}\\var{d}&\\\\\\underline{\\times\\hspace{2cm}\\var{c}}&\\\\\\var{d*b[4]}&\\\\\\ _\\var{(a[4]*b[5]-mod(a[4]*b[5],10))/10}\\var{mod(a[4]*b[5],10)}0&\\\\\\end{align}\\]
\n$\\var{a[4]}\\times\\var{b[3]}=\\var{a[4]*b[3]}$ adding anything carried across, $\\var{a[4]*b[3]}+\\var{(a[4]*b[5]-mod(a[4]*b[5],10))/10}=\\var{a[4]*b[3]+(a[4]*b[5]-mod(a[4]*b[5],10))/10}$
\n\\[\\begin{align}\\var{d}&\\\\\\underline{\\times\\hspace{2cm}\\var{c}}&\\\\\\var{d*b[4]}&\\\\\\ _\\var{(a[4]*(10*b[3]+b[5])-mod(a[4]*(10*b[3]+b[5]),100))/100}\\var{mod(a[4]*b[3]+(a[4]*b[5]-mod(a[4]*b[5],10))/10,10)}\\var{mod(a[4]*b[5],10)}0&\\\\\\end{align}\\]
\n$\\var{a[4]}\\times\\var{a[3]}=\\var{a[4]*a[3]}$ adding on the carry across $\\var{a[4]*a[3]}+\\var{(a[4]*(10*b[3]+b[5])-mod(a[4]*(10*b[3]+b[5]),100))/100}=\\var{a[4]*a[3]+(a[4]*(10*b[3]+b[5])-mod(a[4]*(10*b[3]+b[5]),100))/100}$
\n\\[\\begin{align}\\var{d}&\\\\\\underline{\\times\\hspace{2cm}\\var{c}}&\\\\\\var{d*b[4]}&\\\\\\var{a[4]*d}0&\\\\\\end{align}\\]
\nNow to add these together
\n\\[\\begin{align}\\var{d}&\\\\\\underline{\\times\\hspace{2cm}\\var{c}}&\\\\\\var{d*b[4]}&\\\\\\underline{+\\hspace{0.5cm}\\var{a[4]*d}0}&\\\\\\var{d*c}&\\\\\\end{align}\\]
\niii) Following the same process but needing to add an additional line with two zeros at the end as the third number along has a place value of 100's
\n\\[\\begin{align}\\var{g}&\\\\\\underline{\\times\\hspace{2cm}\\var{d}}&\\\\\\var{b[5]*g}&\\\\\\var{b[3]*g}0&\\\\\\underline{+\\hspace{1cm}00}&\\\\\\end{align}\\]
\nNow using the same method as above the third sum is calculated $\\var{a[3]}\\times\\var{g}=\\var{a[3]*g}$ and placed in the correct position, the last digit in the 100's column.
\n\\[\\begin{align}\\var{g}&\\\\\\underline{\\times\\hspace{2cm}\\var{d}}&\\\\\\var{b[5]*g}&\\\\\\var{b[3]*g}0&\\\\\\underline{+\\hspace{1cm}\\var{a[3]*g}00}&\\\\\\end{align}\\]
\nAdding these gives the final answer as $\\var{g*d}$
\n\\[\\begin{align}\\var{g}&\\\\\\underline{\\times\\hspace{2cm}\\var{d}}&\\\\\\var{b[5]*g}&\\\\\\var{b[3]*g}0&\\\\\\underline{+\\hspace{1cm}\\var{a[3]*g}00}&\\\\\\var{g*d}\\\\\\end{align}\\]
\n\ni)
\nSet the grid out, with an additional row, then start completing. The grid can be filled in any order, multiplying the number at the top of the column with the number to the left of the row.
\nThe completed grid looks like
\n| \n | $\\var{10*a[4]}$ | \n$\\var{b[4]}$ | \n
| $\\var{10*a[2]}$ | \n$\\var{100*a[2]*a[4]}$ | \n$\\var{10*a[2]*b[4]}$ | \n
| $\\var{b[2]}$ | \n$\\var{10*b[2]*a[4]}$ | \n$\\var{b[2]*b[4]}$ | \n
Then add these numbers together
\n\\[\\begin{align}\\var{((a[4])*(a[2]))*100}&\\\\\\var{a[2]*b[4]*10}&\\\\\\var{b[2]*a[4]*10}&\\\\\\underline{+{\\hspace 0.5cm}\\var{b[2]*b[4]}}&\\\\\\var{c*f}&\\\\\\end{align}\\]
\nii)
\nAgain with larger numbers the process is the same - set the grid out - fill it in - add them up.
\nThe completed grid is
\n| \n | $\\var{100*a[3]}$ | \n$\\var{10*b[3]}$ | \n$\\var{b[5]}$ | \n
| $\\var{10*a[4]}$ | \n$\\var{1000*a[3]*a[4]}$ | \n$\\var{100*a[4]*b[3]}$ | \n$\\var{10*a[4]*b[5]}$ | \n
| $\\var{b[4]}$ | \n$\\var{100*a[3]*b[4]}$ | \n$\\var{10*b[3]*b[4]}$ | \n$\\var{b[4]*b[5]}$ | \n
Adding up the numbers in the grid gives
\n\\[\\begin{align}\\var{((a[4])*(a[3]))*1000}&\\\\\\var{a[4]*b[3]*100}&\\\\\\var{b[5]*a[4]*10}&\\\\\\var{100*a[3]*b[4]}&\\\\\\var{10*b[3]*b[4]}&\\\\\\underline{+{\\hspace 0.5cm}\\var{b[5]*b[4]}}&\\\\\\var{d*c}&\\\\\\end{align}\\]
\niii)
\n| \n | $\\var{100*a[1]}$ | \n$\\var{10*b[2]}$ | \n$\\var{b[6]}$ | \n
| $\\var{100*a[3]}$ | \n$\\var{10000*a[1]*a[3]}$ | \n$\\var{1000*a[3]*b[2]}$ | \n$\\var{100*a[3]*b[6]}$ | \n
| $\\var{10*b[3]}$ | \n$\\var{1000*b[3]*a[1]}$ | \n$\\var{100*b[3]*b[2]}$ | \n$\\var{10*b[3]*b[6]}$ | \n
| $\\var{b[6]}$ | \n$\\var{100*b[6]*a[1]}$ | \n$\\var{10*b[6]*b[2]}$ | \n$\\var{b[6]*b[6]}$ | \n
You may wish to add up each row or column first and then add these together. Check you get the same total as below.
\n\\[\\begin{align}\\var{((a[1])*(a[3]))*10000}&\\\\\\var{a[3]*b[2]*1000}&\\\\\\var{b[6]*a[3]*100}&\\\\\\var{1000*a[1]*b[3]}&\\\\\\var{100*b[3]*b[2]}&\\\\\\var{10*b[3]*b[6]}&\\\\\\var{100*b[6]*a[1]}&\\\\\\var{10*b[6]*b[2]}&\\\\\\underline{+{\\hspace 0.5cm}\\var{b[6]*b[6]}}&\\\\\\var{d*g}&\\\\\\end{align}\\]
\n\nFor further examples and questions
\nMulti Digit Multiplication- Khan Academy
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