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| title | chunk | source | category | tags | date_saved | instance |
|---|---|---|---|---|---|---|
| Divisibility rule | 2/8 | https://en.wikipedia.org/wiki/Divisibility_rule | reference | science, encyclopedia | 2026-05-05T08:13:56.367542+00:00 | kb-cron |
=== Divisibility by 5 === Divisibility by 5 is easily determined by checking the last digit in the number (475), and seeing if it is either 0 or 5. If the last number is either 0 or 5, the entire number is divisible by 5. If the last digit in the number is 0, then the result will be the remaining digits multiplied by 2. For example, the number 40 ends in a zero, so take the remaining digits (4) and multiply that by two (4 × 2 = 8). The result is the same as the result of 40 divided by 5(40/5 = 8). If the last digit in the number is 5, then the result will be the remaining digits multiplied by two, plus one. For example, the number 125 ends in a 5, so take the remaining digits (12), multiply them by two (12 × 2 = 24), then add one (24 + 1 = 25). The result is the same as the result of 125 divided by 5 (125/5=25). Example. If the last digit is 0
110 (The original number) 11 0 (Take the last digit of the number, and check if it is 0 or 5) 11 0 (If it is 0, take the remaining digits, discarding the last) 11 × 2 = 22 (Multiply the result by 2) 110 ÷ 5 = 22 (The result is the same as the original number divided by 5) If the last digit is 5
85 (The original number) 8 5 (Take the last digit of the number, and check if it is 0 or 5) 8 5 (If it is 5, take the remaining digits, discarding the last) 8 × 2 = 16 (Multiply the result by 2) 16 + 1 = 17 (Add 1 to the result) 85 ÷ 5 = 17 (The result is the same as the original number divided by 5)
=== Divisibility by 6 === Divisibility by 6 is determined by checking the original number to see if it is both an even number (divisible by 2) and divisible by 3. If the final digit is even the number is divisible by two, and thus may be divisible by 6. If it is divisible by 2, continue by adding the digits of the original number and checking if that sum is a multiple of 3. Any number which is both a multiple of 2 and of 3 is a multiple of 6. Example.
324 (The original number) Final digit 4 is even, so 324 is divisible by 2, and may be divisible by 6. 3 + 2 + 4 = 9 which is a multiple of 3. Therefore, the original number is divisible by both 2 and 3 and is divisible by 6.
=== Divisibility by 7 === Divisibility by 7 can be tested by a recursive method. A number of the form 10x + y is divisible by 7 if and only if x − 2y is divisible by 7. In other words, subtract twice the last digit from the number formed by the remaining digits. Continue to do this until a number is obtained for which it is known whether it is divisible by 7. The original number is divisible by 7 if and only if the number obtained using this procedure is divisible by 7. For example, the number 371: 37 − (2×1) = 37 − 2 = 35; 3 − (2 × 5) = 3 − 10 = −7; thus, since −7 is divisible by 7, 371 is divisible by 7. Similarly, a number of the form 10x + y is divisible by 7 if and only if x + 5y is divisible by 7. So add five times the last digit to the number formed by the remaining digits, and continue to do this until a number is obtained for which it is known whether it is divisible by 7. Another method is multiplication by 3. A number of the form 10x + y has the same remainder when divided by 7 as 3x + y. One must multiply the leftmost digit of the original number by 3, add the next digit, take the remainder when divided by 7, and continue from the beginning: multiply by 3, add the next digit, etc. For example, the number 371: 3×3 + 7 = 16 remainder 2, and 2×3 + 1 = 7. This method can be used to find the remainder of division by 7. A more complicated algorithm for testing divisibility by 7 uses the fact that 100 ≡ 1, 101 ≡ 3, 102 ≡ 2, 103 ≡ 6, 104 ≡ 4, 105 ≡ 5, 106 ≡ 1, ... (mod 7). Take each digit of the number (371) in reverse order (173), multiplying them successively by the digits 1, 3, 2, 6, 4, 5, repeating with this sequence of multipliers as long as necessary (1, 3, 2, 6, 4, 5, 1, 3, 2, 6, 4, 5, ...), and adding the products (1×1 + 7×3 + 3×2 = 1 + 21 + 6 = 28). The original number is divisible by 7 if and only if the number obtained using this procedure is divisible by 7 (hence 371 is divisible by 7 since 28 is). This method can be simplified by removing the need to multiply. All it would take with this simplification is to memorize the sequence above (132645...), and to add and subtract, but always working with one-digit numbers. The simplification goes as follows: