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Multiply Up to 20X20 In Your Head In just FIVE minutes you should learn to quickly multiply up to 20x20 in your head With this trick, you will be able to multiply any two numbers from 11 to 19 in your head quickly, without the use of a calculator I will assume that you know your multiplication table reasonably well up to 10x10 Try this: • Take 15 x 13 for an example • Always place the larger number of the two on top in your mind • Then draw the shape of Africa mentally so it covers the 15 and the from the 13 below Those covered numbers are all you need • First add 15 + = 18 • Add a zero behind it (multiply by 10) to get 180 • Multiply the covered lower x the single digit above it the "5" (3x5= 15) • Add 180 + 15 = 195 That is It! Wasn't that easy? Practice it on paper first! The 11 Rule You likely all know the 10 rule (to multiply by 10, just add a behind the number) but you know the 11 rule? It is as easy! You should be able to this one in you head for any two digit number Practice it on paper first! To multiply any two digit number by 11: • For this example we will use 54 • Separate the two digits in you mind (5 4) • Notice the hole between them! • Add the and the together (5+4=9) • Put the resulting in the hole 594 That's it! 11 x 54=594 The only thing tricky to remember is that if the result of the addition is greater than 9, you only put the "ones" digit in the hole and carry the "tens" digit from the addition For example 11 x 57 5+7=12 put the in the hole and add the from the 12 to the in to get for a result of 627 11 x 57 = 627 Practice it on paper first! Finger Math: 9X Rule To multiply by 9,try this: (1) Spread your two hands out and place them on a desk or table in front of you (2) To multiply by 3, fold down the 3rd finger from the left To multiply by 4, it would be the 4th finger and so on (3) the answer is 27 READ it from the two fingers on the left of the folded down finger and the fingers on the right of it This works for anything up to 9x10! Square a Digit Number Ending in For this example we will use 25 • Take the "tens" part of the number (the and add 1)=3 • Multiply the original "tens" part of the number by the new number (2x3) • Take the result (2x3=6) and put 25 behind it Result the answer 625 Try a few more 75 squared = 7x8=56 put 25 behind it is 5625 55 squared = 5x6=30 put 25 behind it is 3025 Another easy one! Practice it on paper first! Square Digit Number: UP-DOWN Method Square a Digit Number, for this example 37: • Look for the nearest 10 boundary • In this case up from 37 to 40 • Since you went UP to 40 go DOWN from 37 to 34 • Now mentally multiply 34x40 • The way I it is 34x10=340; • Double it mentally to 680 • Double it again mentally to 1360 • This 1360 is the FIRST interim answer • 37 is "3" away from the 10 boundary 40 • Square this "3" distance from 10 boundary • 3x3=9 which is the SECOND interim answer • Add the two interim answers to get the final answer • Answer: 1360 + = 1369 Multiply By To quickly multiply by four, double the number and then double it again Often this can be done in your head Multiply By To quickly multiply by 5, divide the number in two and then multiply it by 10 Often this can be done quickly in your head The 11 Rule Expanded You can directly write down the answer to any number multiplied by 11 • Take for example the number 51236 X 11 • First, write down the number with a zero in front of it 051236 The zero is necessary so that the rules are simpler • Draw a line under the number • Bear with me on this one It is simple if you work through it slowly To this, all you have to this is "Add the neighbor" Look at the in the "units" position of the number Since there is no number to the right of it, you can't add to its "neighbor" so just write down below the in the units col • For the "tens" place, add the to the its "neighbor" (the 6) Write the answer: below the • For the "hundreds" place, add the to the its "neighbor" (the 3) Write the answer: below the • For the "thousands" place, add the to the its "neighbor" (the 2) Write the answer: below the • For the "ten-thousands" place, add the to the its "neighbor" (the 1) Write the answer: below the • For the "hundred-thousands" place, add the to the its "neighbor" (the 5) Write the answer: below the That's it 11 X 051236 = 563596 Practice it on paper first! Divisibility Rules Dividing by Add up the digits: if the sum is divisible by three, then the number is as well Examples: 111111: the digits add to so the whole number is divisible by three 87687687 The digits add up to 57, and plus seven is 12, so the original number is divisible by three Why does the 'divisibility by 3' rule work? From: Dr Math To: Kevin Gallagher Subject: Re: Divisibility of a number by As Kevin Gallagher wrote to Dr Math On 5/11/96 at 21:35:40 (Eastern Time), >I'm looking for a SIMPLE way to explain to several very bright 2nd >graders why the divisibility by rule works, i.e add up all the >digits; if the sum is evenly divisible by 3, then the number is as well >Thanks! >Kevin Gallagher The only way that I can think of to explain this would be as follows: Look at a digit number: 10a+b=9a+(a+b) We know that 9a is divisible by 3, so 10a+b will be divisible by if and only if a+b is Similarly, 100a+10b+c=99a+9b+(a+b+c), and 99a+9b is divisible by 3, so the total will be iff a+b+c is This explanation also works to prove the divisibility by test It clearly originates from modular arithmetic ideas, and I'm not sure if it's simple enough, but it's the only explanation I can think of Doctor Darren, The Math Forum Check out our web site - http://mathforum.org/dr.math/ Dividing by Look at the last two digits If the number formed by its last two digits is divisible by 4, the original number is as well Examples: 100 is divisible by 4 1732782989264864826421834612 is divisible by four also, because 12 is divisible by four Dividing by If the last digit is a five or a zero, then the number is divisible by Dividing by Check and If the number is divisible by both and 2, it is divisible by as well Robert Rusher writes in: Another easy way to tell if a [multi-digit] number is divisible by six is to look at its [ones digit]: if it is even, and the sum of the [digits] is a multiple of 3, then the number is divisible by Dividing by To find out if a number is divisible by seven, take the last digit, double it, and subtract it from the rest of the number Example: If you had 203, you would double the last digit to get six, and subtract that from 20 to get 14 If you get an answer divisible by (including zero), then the original number is divisible by seven If you don't know the new number's divisibility, you can apply the rule again Matthew Correnti describes this method: If you not know if a two-digit number, call it ab, is divisible by 7, calculate 2a + 3b This will yield a smaller number, and if you the process enough times you will eventually if the number ab is divisible by end up with You can use a similar method if you have a three-digit number abc: take the digit a and multiply it by 2, then add it to the number bc, giving you 2a + bc; repeat and reduce until you recognize the result's divisibility by seven With a four-digit number abcd, take the digit a and multiply by 6, then add 6a to bcd giving This usually gives you a three-digit number; call it xyz Take that x and, as described previously, multiply x by two and add to yz (i.e., 2x + yz) Again, repeat and reduce until you recognize the result's divisibility by seven Ahmed Al Harthy writes in: To know if a number is a multiple of seven or not, we can use also coefficients (1 , , 3) We multiply the first number starting from the ones place by 1, then the second from the right by 3, the third by 2, the fourth by -1, the fifth by -3, the sixth by -2, and the seventh by 1, and so forth Example: 348967129356876 6 + 21 + 16 - - 15 - + + + - - 18 - 18 + + 12 + = 16 means the number is not multiple of seven If the number was 348967129356874, then the number is a multiple of seven because instead of 16, we would find 14 as a result, which is a multiple of So the pattern is as follows: for a number onmlkjihgfedcba, calculate a + 3b + 2c - d - 3e - 2f + g + 3h + 2i - j - 3k - 2l + m + 3n + 2o Example: 348967129356874 Below each digit let me write its respective figure -2 -3 -1 2 -2 -3 -1 (3×2) + (4×3) + (8×1) + (9×-2) + (6×-3) + (7×-1) + (1×2) + (2×3) + (9×1) + (3×-2) + (5×-3) + (6×-1) + (8×2) + (7×3) + (6×1) = 16 not a multiple of seven Another visitor observes: Here is one formula for seven 3X + L L = last digit X = everything in front of last digit All numbers that are divisible by seven have this in common There are no exceptions For example, 42: 3(4) + = 14 Seven divides into 14, so it divides into 42 Next example, 105: 3(10) + = 35 Seven divides into 35, so it divides into 105 Here is another formula for seven: 4X - L When using this formula, if you get zero, seven or a multiple of seven, the number will be divisible by seven For example, 56: 4(5) - = 14 Seven divides into 14, so it divides into 56 Next example, 168: 16(4) - = 56 Seven divides into 56, so it divides into 168 Similarly: The formula for is 2X + L The formula for is 4X + L The formula for is 6X + L The formula for is 5X + L The formula for is 2X + L and 4X + L in other words, the formulas for and must work before the number is divisible by The formula for is X + L The formula for 11 is X - L The formula for 12 is 2X - L The formula for 13 is 3X - L The formula for 14 is 4X - L and 2X + L in other words, the formulas for and must work before the number is divisible by 14 The formula for 17 is 7X - L The formula for 21 is X - 2L The formula for 23 is 3X - 2L The formula for 31 is X - 3L Sara Heikali explains this way to test a number with three or more digits for divisibility by seven: Write down just the digits in the tens and ones places Take the other numbers to the left of those last two digits, and multiply them by two Add the answer from step two to the number from step one If the sum from step three is divisible by seven, then the original number is divisible by seven, as well If the sum is not divisible by seven, then the original number is not divisible by seven For example, if the number we are testing is 112, then Write down just the digits in the tens and ones places: 12 Take the other numbers to the left of those last two digits, and multiply them by two: × = Add the answer from step two to the number from step one: 12 + = 14 Fourteen is divisible be seven Therefore, our original number, one hundred twelve, is also divisible by seven See also Explaining the Divisibility Rule for in the Dr Math archives Dividing by Check the last three digits Since 1000 is divisible by 8, if the last three digits of a number are divisible by 8, then so is the whole number Example: 33333888 is divisible by 8; 33333886 isn't How can you tell whether the last three digits are divisible by 8? Phillip McReynolds answers: If the first digit is even, the number is divisible by if the last two digits are If the first digit is odd, subtract from the last two digits; the number will be divisible by if the resulting last two digits are So, to continue the last example, 33333888 is divisible by because the digit in the hundreds place is an even number, and the last two digits are 88, which is divisible by 33333886 is not divisible by because the digit in the hundreds place is an even number, but the last two digits are 86, which is not divisible by Sara Heikali explains this test of divisibility by eight for numbers with three or more digits: Write down the units digit of the original number Take the other numbers to the left of the last digit, and multiply them by two Add the answer from step two to the number from step one If the sum from step three is divisible by eight, then the original number is divisible by eight, as well If the sum is not divisible by eight, then the original number is not divisible by eight For example, if the number we are testing is 104, then Write down just the digits in ones place: Take the other numbers to the left of that last digit, and multiply them by two: 10 × = 20 Add the answer from step two to the number from step one: + 20 = 24 Twenty-four is divisible be eight Therefore, our original number, one hundred and four, is also divisible by eight Dividing by Add the digits If that sum is divisible by nine, then the original number is as well Dividing by 10 If the number ends in 0, it is divisible by 10 Dividing by 11 Let's look at 352, which is divisible by 11; the answer is 32 3+2 is 5; another way to say this is that 35 -2 is 33 Now look at 3531, which is also divisible by 11 It is not a coincidence that 353-1 is 352 and 11 × 321 is 3531 Here is a generalization of this system Let's look at the number 94186565 First we want to find whether it is divisible by 11, but on the way we are going to save the numbers that we use: in every step we will subtract the last digit from the other digits, then save the subtracted amount in order Start with Then Then Then Then Then 9418656 941865 94186 9418 941 93 - = = = = = = 9418651 941864 94182 9416 935 88 SAVE SAVE SAVE SAVE SAVE SAVE Then - = SAVE Now write the numbers we saved in reverse order, and we have 8562415, which multiplied by 11 is 94186565 Here's an even easier method, contributed by Chis Foren: Take any number, such as 365167484 Add the first, third, fifth, seventh, , digits + + + + = 22 Add the second, fourth, sixth, eighth, , digits + + + = 22 If the difference, including 0, is divisible by 11, then so is the number 22 - 22 = so 365167484 is evenly divisible by 11 See also Divisibility by 11 in the Dr Math archives Dividing by 12 Check for divisibility by and Dividing by 13 Here's a straightforward method supplied by Scott Fellows: Delete the last digit from the given number Then subtract nine times the deleted digit from the remaining number If what is left is divisible by 13, then so is the original number Rafael Ando contributes: Instead of deleting the last digit and subtracting it ninefold from the remaining number (which works), you could also add the deleted digit fourfold Both methods work because 91 and 39 are each multiples of 13 For any prime p (except and 5), a rule of divisibility could be "created" using this method: Find m, such that m is the (preferably) smallest multiple of p that ends in either or Delete the last digit and add (if multiple ends in 9) or subtract (if it ends in 1) the deleted digit times the integer nearest to m/10 For example, if m = 91, the integer closest to 91/10 = 9.1 is 9; and for 3.9, it's Verify if the result is a multiple of p Use this process until it's obvious Example 1: Let's see if 14281581 is a multiple of 17 In this case, m = 51 (which is 17×3), so we'll be deleting the last number and subtracting it fivefold 1428158 - 5×1 = 1428153 142815 - 5×3 = 142800 14280 - 5×0 = 14280 1428 - 5×0 = 1428 142 - 5×8 = 102 10 - 5×2 = 0, which is a multiple of 17, so 14281581 is multiple of 17 Example 2: Let's see if 7183186 is a multiple of 46 First, note that 46 is not a prime number, and its factorization is 2×23 So, 7183186 needs to be divisible by both and 23 Since it's an even number, it's obviously divisible by So let's verify that it is a multiple of 23: m = 3×23 = 69, which means we'll be adding the deleted digit sevenfold 718318 + 7×6 = 718360 71836 + 7×0 = 71836 7183 + 7×6 = 7225 722 + 7×5 = 757 75 + 7×7 = 124 12 + 7×4 = 40 + 7×0 = (not divisible by 23), so 7183186 is not divisible by 46 Note that you could've stopped calculating whenever you find the result to be obvious (i.e., you don't need to it until the end) For example, in example if you recognize 102 as divisible by 17, you don't need to continue (likewise, if you recognized 40 as not divisible by 23) The idea behind this method it that you're either subtracting m×(last digit) and then dividing by 10, or adding m×(last digit) and then dividing by 10 Jeremy Lane adds: It may be noted that while applying these rules, it is possible to loop among numbers as results Example: Is 1313 divisible by 13? Using the procedure given we take 13×3 and obtain 39 This multiple ends in so we add four-fold the last digit 131 + 4×3 = 143 14 + 4×3 = 26 + 4×6 = 26 Example: Is 1326 divisible by 13? Using the procedure given we take 13×7 = 91 This is not the smallest multiple, but it does show looping The smaller multiple does loop at 39 as well There are some examples where we would still need to recognize certain multiples So we subtract nine-fold the last digit 132 - 9×6 = 78 - 9×8 = -65 (factor out -1) - 9×5 = -39 (again factor out -1) - 9×9 = -78 (factor out -1) This only occurs though if the number does happen to be divisible by the prime divisor Otherwise, eventually you will have a number that is less than the prime divisor And here's a more complex method that can be extended to other formulas: = (mod 13) 10 = -3 (mod 13) (i.e., 10 - -3 is divisible by 13) 100 = -4 (mod 13) (i.e., 100 - -4 is divisible by 13) 1000 = -1 (mod 13) (i.e., 1000 - -1 is divisible by 13) 10000 = (mod 13) 100000 = (mod 13) 1000000 = (mod 13) Call the ones digit a, the tens digit b, the hundreds digit c, and you get: a - 3×b - 4×c - d + 3×e + 4×f + g - If this number is divisible by 13, then so is the original number You can keep using this technique to get other formulas for divisibility for prime numbers For composite numbers just check for divisibility by divisors Dividing by 14 Sara Heikali builds on her divisibility test for seven: How can you know if a number with three or more digits is divisible by the number fourteen? Check if the last digit of the original number is odd or even If the number is odd, then the number is not divisible by fourteen If the number is even, then apply the divisibility rule for seven (Keep in mind, the odd and even test is to see if the number is divisible by two.) If the original even number is divisible by seven, then it is also divisible by fourteen If the original even number is not divisible by seven, it is not divisible by fourteen [...]...1 32 - 9×6 = 78 7 - 9×8 = -65 (factor out -1) 6 - 9×5 = -39 (again factor out -1) 3 - 9×9 = -78 (factor out -1) This only occurs though if the number does happen to be divisible by the prime divisor Otherwise, ... 3b + 2c - d - 3e - 2f + g + 3h + 2i - j - 3k - 2l + m + 3n + 2o Example: 348967 129 356874 Below each digit let me write its respective figure -2 -3 -1 2 -2 -3 -1 (3 2) + (4×3) + (8×1) + (9× -2) +... seventh, , digits + + + + = 22 Add the second, fourth, sixth, eighth, , digits + + + = 22 If the difference, including 0, is divisible by 11, then so is the number 22 - 22 = so 365167484 is evenly... 1 428 15 - 5×3 = 1 428 00 1 428 0 - 5×0 = 1 428 0 1 428 - 5×0 = 1 428 1 42 - 5×8 = 1 02 10 - 5 2 = 0, which is a multiple of 17, so 1 428 1581 is multiple of 17 Example 2: Let's see if 7183186 is a multiple