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Divisibility Rule for 7 and 13

Posted by EduEvoke on May 29, 2025
Math Maths quantitative aptitude
Divisibility Rules for 7 and 13

Divisibility Rules for 7 and 13

Divisibility rules are shortcuts to determine if a number is divisible by another number without performing long division. Here are the rules for 7 and 13, including some alternative methods:

Divisibility Rule for 7

Method 1: Subtracting Twice the Last Digit

  • Take the last digit of the number.
  • Double it.
  • Subtract this doubled value from the remaining part of the number (the number without its last digit).
  • If the result is $0$ or a multiple of $7$, then the original number is divisible by $7$.
  • If the result is a large number, you can repeat the process until you get a number you can easily check for divisibility by $7$.
Example 1: Is $343$ divisible by $7$?
  1. Last digit is $3$.
  2. Double $3$: $3 \times 2 = 6$.
  3. Remaining part of the number is $34$. Subtract $6$ from $34$: $34 - 6 = 28$.
  4. Since $28$ is a multiple of $7$ ($28 = 7 \times 4$), $343$ is divisible by $7$.
Example 2: Is $458409$ divisible by $7$?
  1. Last digit is $9$. Double it: $9 \times 2 = 18$. Remaining: $45840$. $45840 - 18 = 45822$.
  2. For $45822$: Last digit is $2$. Double it: $2 \times 2 = 4$. Remaining: $4582$. $4582 - 4 = 4578$.
  3. For $4578$: Last digit is $8$. Double it: $8 \times 2 = 16$. Remaining: $457$. $457 - 16 = 441$.
  4. For $441$: Last digit is $1$. Double it: $1 \times 2 = 2$. Remaining: $44$. $44 - 2 = 42$.
  5. Since $42$ is a multiple of $7$ ($42 = 7 \times 6$), $458409$ is divisible by $7$.

Method 2: Oscillating Series (Right to Left with Weights $1, 3, 2, -1, -3, -2...$)

  • Multiply each digit of the number, starting from the right (units place), by the repeating sequence of multipliers: $1, 3, 2, -1, -3, -2$.
  • Sum these products.
  • If the sum is divisible by $7$, then the original number is divisible by $7$.
Example: Is $2016$ divisible by $7$?
  1. $6 \times 1 = 6$
  2. $1 \times 3 = 3$
  3. $0 \times 2 = 0$
  4. $2 \times (-1) = -2$
  5. Sum: $6 + 3 + 0 - 2 = 7$.
  6. Since $7$ is divisible by $7$, $2016$ is divisible by $7$.

Method 3: Alternating Sum of Three-Digit Blocks (The $1001$ Rule)

This rule works for $7$, $11$, and $13$ because $7 \times 11 \times 13 = 1001$.

  • Divide the number into blocks of three digits, starting from the right. Add leading zeros if needed to complete a block.
  • Take the alternating sum of these blocks.
  • If the resulting sum is divisible by $7$ (or $11$ or $13$), then the original number is divisible by $7$ (or $11$ or $13$).
Example: Is $13587$ divisible by $7$?
  1. Divide into blocks: $013$ and $587$.
  2. Alternating sum: $587 - 13 = 574$.
  3. Now, check $574$ using Method 1 for $7$: $57 - (4 \times 2) = 57 - 8 = 49$.
  4. Since $49$ is divisible by $7$, $13587$ is divisible by $7$.

Method 4: Adding Five Times the Last Digit

  • Take the last digit, multiply it by $5$, and add it to the rest of the number.
  • Repeat if necessary.
Example: Is $133$ divisible by $7$?
  1. Last digit is $3$. Multiply by $5$: $3 \times 5 = 15$.
  2. Remaining part is $13$. Add: $13 + 15 = 28$.
  3. Since $28$ is divisible by $7$, $133$ is divisible by $7$.

Divisibility Rule for 13

Method 1: Adding Four Times the Last Digit

  • Take the last digit of the number.
  • Multiply it by $4$.
  • Add this product to the remaining part of the number (the number without its last digit).
  • If the result is $0$ or a multiple of $13$, then the original number is divisible by $13$.
  • If the result is a large number, you can repeat the process until you get a number you can easily check for divisibility by $13$.
Example 1: Is $429$ divisible by $13$?
  1. Last digit is $9$.
  2. Multiply $9$ by $4$: $9 \times 4 = 36$.
  3. Remaining part of the number is $42$. Add $36$ to $42$: $42 + 36 = 78$.
  4. Since $78$ is a multiple of $13$ ($78 = 13 \times 6$), $429$ is divisible by $13$.
Example 2: Is $1352$ divisible by $13$?
  1. Last digit is $2$. Multiply by $4$: $2 \times 4 = 8$. Remaining: $135$. $135 + 8 = 143$.
  2. For $143$: Last digit is $3$. Multiply by $4$: $3 \times 4 = 12$. Remaining: $14$. $14 + 12 = 26$.
  3. Since $26$ is a multiple of $13$ ($26 = 13 \times 2$), $1352$ is divisible by $13$.

Method 2: Subtracting Nine Times the Last Digit

  • Take the last digit of the number.
  • Multiply it by $9$.
  • Subtract this product from the remaining part of the number.
  • If the result is $0$ or a multiple of $13$, then the original number is divisible by $13$.
  • Repeat if necessary.
Example: Is $858$ divisible by $13$?
  1. Last digit is $8$. Multiply by $9$: $8 \times 9 = 72$.
  2. Remaining part is $85$. Subtract: $85 - 72 = 13$.
  3. Since $13$ is a multiple of $13$, $858$ is divisible by $13$.

Method 3: Alternating Sum of Three-Digit Blocks (The $1001$ Rule)

As mentioned for $7$, this rule also applies to $13$.

  • Divide the number into blocks of three digits, starting from the right.
  • Take the alternating sum of these blocks.
  • If the resulting sum is divisible by $13$, then the original number is divisible by $13$.
Example: Is $1,139,502$ divisible by $13$?
  1. Blocks: $001$, $139$, $502$.
  2. Alternating sum: $502 - 139 + 1 = 363 + 1 = 364$.
  3. Now, check $364$ using Method 1 for $13$: $36 + (4 \times 4) = 36 + 16 = 52$.
  4. Since $52$ is a multiple of $13$ ($52 = 13 \times 4$), $1,139,502$ is divisible by $13$.
While these alternative rules exist, the ones involving doubling/quadrupling the last digit and adding/subtracting are generally the most straightforward for manual calculation. The alternating sum of 3-digit blocks is very efficient for large numbers, as it simultaneously tests for divisibility by $7$, $11$, and $13$.

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