Mental Arithmetic Tricks - Learn number tricks for fast multiplication and roots

Aug 25, 2026

Uploaded by Aria

A structured set of mental-arithmetic techniques for fast calculations, including multiplication tricks for specific numbers (such as 37 and 25), mirror-based products, and quick use of algebraic identities. It then extends to power and root tricks: learning cubes and cube roots from selected digits, determining square and fifth roots via modular patterns and leading-digit tables, and handling large-number cube roots with memorized remainder tables. The material emphasizes short explanations suitable for classroom use and demonstrates worked examples to verify results.

LanguageEnglish

CategoryExam

FormatPDF

Pages5

Reading Time13 min

AccessFree to download

Mental Arithmetic Tricks

Multiplication tricks (37, 25)

Mirror tricks (at 5 and 10)

Squares up to 20

Third powers and cube roots

Fifth roots and square roots

Cube roots of large numbers

Mental Arithmetic Tricks Andreas Klein The normal tuxmath questions a simple. With the options TRICK_ _ALLOWED you can enable some tricky problems which need explanation. The ﬁrst ﬁve tricks are suitable for disciples at 5th or 6th grade. I add a short explanation for each trick. Teachers should work out longer explanations with more examples before presenting them to the class. 1. The trick with the 37 (TRICK_37_ALLOWED) 37 × 3 = 111 . This give us 37 × (3n) = nnn. Example: 37 × 12 = 444 12 ÷ 3 = 4 2. The trick with the 25 (TRICK_37_ALLOWED) 25 × 4 = 100 . This makes it easy to multiply 25 with any other two digit number. Example: 25 × 35 = 8 75 35 ÷ 4 = 8 remainder 3, 3 × 25 = 75 3. The mirror at 5 (TRICK_MIRROR_5_ALLOWED) For the next trick we need two numbers with the property that the ﬁrst digits are the same and the sum of the last digits is 10 . Then it easy to calculate the product (ab)10 × (ac)10 = (xy zw)10 a × (a + 1) = (xy)10 b × c = (zw)10 Example: 62 × 68 = 42 16 6 × 7 = 42 2 × 8 = 16 (Important: if the last two digits are 1 and 9, you must write 1 × 9 = 09 don’t forget the leading 0!) The proof of the trick is just a little exercise in calculus. 4. The mirror at 10 (TRICK_MIRROR_ 10_ALLOWED) The thrid binomial theorem (x + y)(x − y) = x2 − y 2 give us a nice trick for mental arithmetic, if we specialise x to a multiple of 10 and y to a number less than 10 . Example: 76 × 64 = 48 64 7 × 7 − 1 = 48 6 × 6 = 36, 100 − 36 = 64 5. Squares (TRICK_SQUARE20 _ALLOWED) This is not really a trick, but it is useful to know the squares up to 20 . The next tricks are designed to make a show out of your mental arithmetic skills. 1. Exercising third powers (TRICK_CUBES_ALLOWED) You must learn the third power for the numbers from 1 to 10 . x 1 2 3 4 5 6 7 8 9 10 x3 1 8 27 64 125 216 343 512 729 1000 2. Cube roots (TRICK_CUBEROOT_ALLOWED) Ask one in the audience to choose a two digit number n and compute n3 . When you hear n3 you can tell immediately n. The trick is quite easy x 7→ x3 mod 10 is a bijection on {0 , . . . , 9} (see table above) . You know all thrid powers, so you get directly the last digit. The ﬁrst digit you get from the size of the number. Example You hear 658503 . Remove the last three digits. You get 658, since 83 ≤ 658 < 93 the result must be between 80 and 89 . Since last digit is 3 and 73 = 343 the last digit of the result must be 7 . So you have the answer 87 . 3. 5th roots (TRICK_FIFTH_ROOT_ALLOWED) Taking the 5th root is even easier than the cuberoot. You have x5 ≡ x mod 10, so the last digit is simple. For the ﬁrst digit you learn the following table: x Size of x5 1 100 thousand 2 3 million 3 24 million 4 100 million 5 300 million 6 777 million 7 1.6 milliards 8 3 milliards 9 6 milliards 10 10 milliards (Our American friends must replace the word milliards by billions.) You hear 8587340257 . As soon as you hear 8 milliards you know by the table that the ﬁrst digit must be 9 . You ignore the rest and wait for the last digit which is 7, so the result is 97 . 4. Square roots (TRICK_SQUAREROOT_ALLOWED) You can also take fast square roots, if you know that the solution must bean integer. Consider the problem √7569. Since 82 ≤ 75 < 92 you know that the ﬁst digit must be an 8 . x2 ≡ 9 mod 10 has two solutions x = 3 and x = 7, so the last digit is either 3 or 7 . To decide which one is correct you calculate 852 = 7225 (use the trick mirror at 5) . 7569 > 7225, so the last digit must be 7 . Solution √7569 = 87 . 5. Cube roots of large numbers (TRICK_LARGE_CUBEROOT_ALLOWED) Now it gets really hard. Ask one of the audience to compute the cube of a three digit number. You will tell the cube root. I explain two ways to solve the problem First method: Learn the following table. x 0 1 2 3 4 5 6 7 8 9 10 x3 mod 11 0 1 8 5 9 4 7 2 6

FAQs

What are the key multiplication tricks taught in the document?

The document presents specific fast methods for cases like 37×3 leading to 111, and 25×4 leading to 100, along with mirror-based multiplication strategies that rely on digit properties.

How does the audience determine a two-digit cube root after hearing n³?

It uses the fact that the last digit of x³ modulo 10 is a bijection for {0..9}, so the last digit of n can be inferred from n³, while the first digit comes from the magnitude bounds of n³.

How are cube roots of large numbers handled in the advanced section?

The approach combines memorized remainder/power tables to infer first and last digits, then determines middle digits by computing the value modulo 11 or modulo 100 using digit alternating sums and solving a congruence.

Aug 25, 2026

Uploaded by Aria