Education & Study

How to Teach Mental Math

Understand the value of 0., Understand the commutative property., Add by counting on., Make a ten when adding three or more numbers., Memorize doubles., Recognize doubles plus one., Use skip counting., Think of plus 9 as plus 10 minus 1.To do this...

30 Steps 5 min read Advanced

Step-by-Step Guide

  1. Step 1: Understand the value of 0.

    Adding zero to a number does not change its value.

    For example, if I have 6 apples, and you have 0 apples, together we have 6 apples:6+0=6{\displaystyle 6+0=6}

  2. Step 2: Understand the commutative property.

    The commutative property states that numbers can be added in any order.

    For example, 7 apples plus 4 apples is the same as 4 apples plus 7 apples.

    They both equal 11 apples:7+4=4+7{\displaystyle 7+4=4+7}11=11{\displaystyle 11=11} , Use the commutative property and start with the larger number, then count up the value of the smaller number.

    This strategy works best when one of the addends is less than five.

    Students can use their fingers or manipulatives to keep track of how many they count on.

    For example, to calculate 7+3{\displaystyle 7+3}, begin with 7 and count on three: “Seven, eight, nine, ten.” , Use the commutative property to make a ten, then add the remaining number.

    For example, to calculate 3+6+7{\displaystyle 3+6+7}, first make a ten by adding 7 and 3, then add 6:3+6+7{\displaystyle 3+6+7}(7+3)+6{\displaystyle (7+3)+6}(10)+6=16{\displaystyle (10)+6=16} , A double is an addition sentence that adds a number to itself.

    Adding a number to itself yields a number twice as big as the original number, so if students know how to multiply by two, they can use multiplication to help them add.

    For example, students can memorize doubles up to 10:1+1=2{\displaystyle 1+1=2}2+2=4{\displaystyle 2+2=4}3+3=6{\displaystyle 3+3=6}4+4=8{\displaystyle 4+4=8}5+5=10{\displaystyle 5+5=10}6+6=12{\displaystyle 6+6=12}7+7=14{\displaystyle 7+7=14}8+8=16{\displaystyle 8+8=16}9+9=18{\displaystyle 9+9=18}10+10=100{\displaystyle 10+10=100} , A doubles plus one is an addition sentence that would be a double, except one number is one larger than the other.

    Once students have their doubles memorized, they can simply add 1 to the doubles sum.

    For example, if a student knows that 6+6=12{\displaystyle 6+6=12}, they can recognize that 6+7=13{\displaystyle 6+7=13}, because 6+7=6+6+1{\displaystyle 6+7=6+6+1}. , Students can use skip counting when adding by twos, fives, or tens.

    Students should recognize that any even number plus two will equal an even number, and any odd number plus two will equal an odd number.For example, 5+5+5{\displaystyle 5+5+5} is the same as skip counting by fives three times: “Five, ten, fifteen.” , For example, to calculate 29+9{\displaystyle 29+9}, calculate:29+10=30{\displaystyle 29+10=30}30−1=29{\displaystyle 30-1=29} , For example, to calculate 58+32{\displaystyle 58+32}, you can break up 58 into 50+8{\displaystyle 50+8}, and you can break up 32 into 30+2{\displaystyle 30+2}.

    Then you can use the commutative property to add compatible numbers first:50+8+30+2{\displaystyle 50+8+30+2}(50+30)+(8+2){\displaystyle (50+30)+(8+2)}80+10=90{\displaystyle 80+10=90} , For example, to find 58+32{\displaystyle 58+32}, you could subtract 2 from 30, then add 2 to
    58.58+32{\displaystyle 58+32}(58+2)+(32−2){\displaystyle (58+2)+(32-2)}60+30=90{\displaystyle 60+30=90} , The result will be the answer, or difference.

    Students can use their fingers or manipulatives to count on.

    For example, to calculate 8−6{\displaystyle 8-6}, start with 6 and see how many you have to count on to get to 8: “Six, seven, eight.” You counted on 2, so 8−6=2{\displaystyle 8-6=2}. , When subtracting with pencil and paper you usually start from the ones place.

    When using the front end strategy, you work beginning from the other direction.

    This strategy only works when you do not have to borrow from other place values.

    You will know that the problem requires no borrowing if, when you line up the place values of each number, all the digits you are subtracting are smaller than the digits you are subtracting from.

    For example, to calculate 795−463{\displaystyle 795-463}, you would first subtract the hundreds place, then the tens place, then the ones place:7−4=3{\displaystyle 7-4=3}9−6=3{\displaystyle 9-6=3}5−3=2{\displaystyle 5-3=2}So 795−463=332{\displaystyle 795-463=332}. , You can also use this strategy to break up numbers into hundreds and tens, or larger place values, for easier subtraction.

    For example, to calculate 42−24{\displaystyle 42-24}, break up 24 into 20 and 4:42−24{\displaystyle 42-24}(42−20)−4{\displaystyle (42-20)-4}(22)−4=18{\displaystyle (22)-4=18} , A number multiplied by 0 will always equal
    0.

    For example, 5 apples zero times is zero: 5×0=0{\displaystyle 5\times 0=0}. , A number multiplied by 1 will always equal the number.

    For example, 5 apples 1 time is 5: 5×1=5{\displaystyle 5\times 1=5}. , The shortcut is that, when multiplying any number by a multiple of ten, simply add the number of zeroes in the multiple to the other number.

    For example:27×10=270{\displaystyle 27\times 10=270}27×100=2,700{\displaystyle 27\times 100=2,700}27×1000=27,000{\displaystyle 27\times 1000=27,000} , The associative property states that you can change the order of groupings you multiply first.

    For example, to calculate 27×5×2{\displaystyle 27\times 5\times 2}, multiplying the 5 and 2 first will make a ten, which makes the problem easier:27×5×2{\displaystyle 27\times 5\times 2}27×(5×2){\displaystyle 27\times (5\times 2)}27×(10)=270{\displaystyle 27\times (10)=270} , To do this, whenever you are multiplying a number by 5, multiply by 10 instead, and then half the product.

    For example, to calculate 27×5{\displaystyle 27\times 5}, change the problem to 27×10{\displaystyle 27\times 10}, then divide the answer in half:27×5=12(27×10){\displaystyle 27\times 5={\frac {1}{2}}(27\times 10)}27×5=12(270){\displaystyle 27\times 5={\frac {1}{2}}(270)}27×5=135{\displaystyle 27\times 5=135} , For example, to calculate 125×8{\displaystyle 125\times 8}, you can factor 125 as 25×5{\displaystyle 25\times 5} and 8 as 4×2{\displaystyle 4\times 2}.

    You can then use the commutative and associative property to multiply the factors in any order or combination.

    Thus:125×8{\displaystyle 125\times 8}(25×5)×(4×2){\displaystyle (25\times 5)\times (4\times 2)}(25×4)×(5×2){\displaystyle (25\times 4)\times (5\times 2)}100×10=1000{\displaystyle 100\times 10=1000} , For example, to calculate 8×45{\displaystyle 8\times 45}, you could half the 8 and double the 45:8×45=4×90{\displaystyle 8\times 45=4\times 90}8×45=360{\displaystyle 8\times 45=360} , To do this, break up the number you are dividing into smaller numbers that can easily be divided by the divisor.

    Then, sum the quotients.

    For example, to calculate 104÷8{\displaystyle 104\div 8}, break up the 104 into 64 and 40:104÷8{\displaystyle 104\div 8}(64+40)÷8{\displaystyle (64+40)\div 8}(64÷8)+(40÷8){\displaystyle (64\div 8)+(40\div 8)}(8)+(5)=13{\displaystyle (8)+(5)=13} , The shortcut is that, when dividing any number by a multiple of ten, simply subtract the number of zeroes in the multiple from the other number.

    For example:27,000÷10=2,700{\displaystyle 27,000\div 10=2,700}27,000÷100=270{\displaystyle 27,000\div 100=270}27,000÷1,000=27{\displaystyle 27,000\div 1,000=27} , Whenever you are dividing a number by five, you can instead divide the number by ten, then multiply the quotient by
    2.

    For example, to calculate 1230÷5{\displaystyle 1230\div 5}, instead divide 1230 by ten, then multiply the answer by 2:1230÷5=2(1230÷10){\displaystyle 1230\div 5=2(1230\div 10)}1230÷5=2(123){\displaystyle 1230\div 5=2(123)}1230÷5=246{\displaystyle 1230\div 5=246}

  3. Step 3: Add by counting on.

  4. Step 4: Make a ten when adding three or more numbers.

  5. Step 5: Memorize doubles.

  6. Step 6: Recognize doubles plus one.

  7. Step 7: Use skip counting.

  8. Step 8: Think of plus 9 as plus 10 minus 1.To do this

  9. Step 9: whenever you add by 9

  10. Step 10: add by 10 instead

  11. Step 11: and then subtract 1 from the sum.

  12. Step 12: Break up larger numbers to make compatible numbers.Compatible numbers are numbers that are easier to add together.

  13. Step 13: Balance numbers before adding.To balance numbers

  14. Step 14: you can subtract from one number and add the same amount to the other.

  15. Step 15: Count on from the number you are subtracting (the subtrahend) to the number you are subtracting from (the minuend).

  16. Step 16: Use the front end strategy for problems that require no borrowing.To do this

  17. Step 17: subtract the digits beginning with the largest place value

  18. Step 18: ending with the lowest place value.

  19. Step 19: Break up the subtrahend into tens and ones.Then subtract your group of tens

  20. Step 20: then subtract the group of ones.

  21. Step 21: Understand the value of 0.

  22. Step 22: Understand the value of 1.

  23. Step 23: Use the shortcut for multiples of ten.

  24. Step 24: Use the associative property.

  25. Step 25: Use the factor of 5 as half the factor of 10.

  26. Step 26: Break up numbers into compatible factors.Compatible numbers are numbers that are easier to multiply.

  27. Step 27: Double one number and half the other.This is another way of finding compatible numbers that are easier to multiply.

  28. Step 28: Use the distributive property.

  29. Step 29: Use the shortcut for multiples of ten.

  30. Step 30: Use the divisor 5 as half the divisor of 10.

Detailed Guide

Adding zero to a number does not change its value.

For example, if I have 6 apples, and you have 0 apples, together we have 6 apples:6+0=6{\displaystyle 6+0=6}

The commutative property states that numbers can be added in any order.

For example, 7 apples plus 4 apples is the same as 4 apples plus 7 apples.

They both equal 11 apples:7+4=4+7{\displaystyle 7+4=4+7}11=11{\displaystyle 11=11} , Use the commutative property and start with the larger number, then count up the value of the smaller number.

This strategy works best when one of the addends is less than five.

Students can use their fingers or manipulatives to keep track of how many they count on.

For example, to calculate 7+3{\displaystyle 7+3}, begin with 7 and count on three: “Seven, eight, nine, ten.” , Use the commutative property to make a ten, then add the remaining number.

For example, to calculate 3+6+7{\displaystyle 3+6+7}, first make a ten by adding 7 and 3, then add 6:3+6+7{\displaystyle 3+6+7}(7+3)+6{\displaystyle (7+3)+6}(10)+6=16{\displaystyle (10)+6=16} , A double is an addition sentence that adds a number to itself.

Adding a number to itself yields a number twice as big as the original number, so if students know how to multiply by two, they can use multiplication to help them add.

For example, students can memorize doubles up to 10:1+1=2{\displaystyle 1+1=2}2+2=4{\displaystyle 2+2=4}3+3=6{\displaystyle 3+3=6}4+4=8{\displaystyle 4+4=8}5+5=10{\displaystyle 5+5=10}6+6=12{\displaystyle 6+6=12}7+7=14{\displaystyle 7+7=14}8+8=16{\displaystyle 8+8=16}9+9=18{\displaystyle 9+9=18}10+10=100{\displaystyle 10+10=100} , A doubles plus one is an addition sentence that would be a double, except one number is one larger than the other.

Once students have their doubles memorized, they can simply add 1 to the doubles sum.

For example, if a student knows that 6+6=12{\displaystyle 6+6=12}, they can recognize that 6+7=13{\displaystyle 6+7=13}, because 6+7=6+6+1{\displaystyle 6+7=6+6+1}. , Students can use skip counting when adding by twos, fives, or tens.

Students should recognize that any even number plus two will equal an even number, and any odd number plus two will equal an odd number.For example, 5+5+5{\displaystyle 5+5+5} is the same as skip counting by fives three times: “Five, ten, fifteen.” , For example, to calculate 29+9{\displaystyle 29+9}, calculate:29+10=30{\displaystyle 29+10=30}30−1=29{\displaystyle 30-1=29} , For example, to calculate 58+32{\displaystyle 58+32}, you can break up 58 into 50+8{\displaystyle 50+8}, and you can break up 32 into 30+2{\displaystyle 30+2}.

Then you can use the commutative property to add compatible numbers first:50+8+30+2{\displaystyle 50+8+30+2}(50+30)+(8+2){\displaystyle (50+30)+(8+2)}80+10=90{\displaystyle 80+10=90} , For example, to find 58+32{\displaystyle 58+32}, you could subtract 2 from 30, then add 2 to
58.58+32{\displaystyle 58+32}(58+2)+(32−2){\displaystyle (58+2)+(32-2)}60+30=90{\displaystyle 60+30=90} , The result will be the answer, or difference.

Students can use their fingers or manipulatives to count on.

For example, to calculate 8−6{\displaystyle 8-6}, start with 6 and see how many you have to count on to get to 8: “Six, seven, eight.” You counted on 2, so 8−6=2{\displaystyle 8-6=2}. , When subtracting with pencil and paper you usually start from the ones place.

When using the front end strategy, you work beginning from the other direction.

This strategy only works when you do not have to borrow from other place values.

You will know that the problem requires no borrowing if, when you line up the place values of each number, all the digits you are subtracting are smaller than the digits you are subtracting from.

For example, to calculate 795−463{\displaystyle 795-463}, you would first subtract the hundreds place, then the tens place, then the ones place:7−4=3{\displaystyle 7-4=3}9−6=3{\displaystyle 9-6=3}5−3=2{\displaystyle 5-3=2}So 795−463=332{\displaystyle 795-463=332}. , You can also use this strategy to break up numbers into hundreds and tens, or larger place values, for easier subtraction.

For example, to calculate 42−24{\displaystyle 42-24}, break up 24 into 20 and 4:42−24{\displaystyle 42-24}(42−20)−4{\displaystyle (42-20)-4}(22)−4=18{\displaystyle (22)-4=18} , A number multiplied by 0 will always equal
0.

For example, 5 apples zero times is zero: 5×0=0{\displaystyle 5\times 0=0}. , A number multiplied by 1 will always equal the number.

For example, 5 apples 1 time is 5: 5×1=5{\displaystyle 5\times 1=5}. , The shortcut is that, when multiplying any number by a multiple of ten, simply add the number of zeroes in the multiple to the other number.

For example:27×10=270{\displaystyle 27\times 10=270}27×100=2,700{\displaystyle 27\times 100=2,700}27×1000=27,000{\displaystyle 27\times 1000=27,000} , The associative property states that you can change the order of groupings you multiply first.

For example, to calculate 27×5×2{\displaystyle 27\times 5\times 2}, multiplying the 5 and 2 first will make a ten, which makes the problem easier:27×5×2{\displaystyle 27\times 5\times 2}27×(5×2){\displaystyle 27\times (5\times 2)}27×(10)=270{\displaystyle 27\times (10)=270} , To do this, whenever you are multiplying a number by 5, multiply by 10 instead, and then half the product.

For example, to calculate 27×5{\displaystyle 27\times 5}, change the problem to 27×10{\displaystyle 27\times 10}, then divide the answer in half:27×5=12(27×10){\displaystyle 27\times 5={\frac {1}{2}}(27\times 10)}27×5=12(270){\displaystyle 27\times 5={\frac {1}{2}}(270)}27×5=135{\displaystyle 27\times 5=135} , For example, to calculate 125×8{\displaystyle 125\times 8}, you can factor 125 as 25×5{\displaystyle 25\times 5} and 8 as 4×2{\displaystyle 4\times 2}.

You can then use the commutative and associative property to multiply the factors in any order or combination.

Thus:125×8{\displaystyle 125\times 8}(25×5)×(4×2){\displaystyle (25\times 5)\times (4\times 2)}(25×4)×(5×2){\displaystyle (25\times 4)\times (5\times 2)}100×10=1000{\displaystyle 100\times 10=1000} , For example, to calculate 8×45{\displaystyle 8\times 45}, you could half the 8 and double the 45:8×45=4×90{\displaystyle 8\times 45=4\times 90}8×45=360{\displaystyle 8\times 45=360} , To do this, break up the number you are dividing into smaller numbers that can easily be divided by the divisor.

Then, sum the quotients.

For example, to calculate 104÷8{\displaystyle 104\div 8}, break up the 104 into 64 and 40:104÷8{\displaystyle 104\div 8}(64+40)÷8{\displaystyle (64+40)\div 8}(64÷8)+(40÷8){\displaystyle (64\div 8)+(40\div 8)}(8)+(5)=13{\displaystyle (8)+(5)=13} , The shortcut is that, when dividing any number by a multiple of ten, simply subtract the number of zeroes in the multiple from the other number.

For example:27,000÷10=2,700{\displaystyle 27,000\div 10=2,700}27,000÷100=270{\displaystyle 27,000\div 100=270}27,000÷1,000=27{\displaystyle 27,000\div 1,000=27} , Whenever you are dividing a number by five, you can instead divide the number by ten, then multiply the quotient by
2.

For example, to calculate 1230÷5{\displaystyle 1230\div 5}, instead divide 1230 by ten, then multiply the answer by 2:1230÷5=2(1230÷10){\displaystyle 1230\div 5=2(1230\div 10)}1230÷5=2(123){\displaystyle 1230\div 5=2(123)}1230÷5=246{\displaystyle 1230\div 5=246}

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