Finding Numbers Divisible by 7 and 11

To find a number that is divisible by both 7 and 11, we need to check each option using the divisibility rules for these numbers. A number divisible by both 7 and 11 must also be divisible by their least common multiple (LCM). Since 7 and 11 are prime numbers, their LCM is simply their product, which is $7 \times 11 = 77$. Thus, we are looking for a number divisible by 77.

Divisibility Rule for 7

One way to check for divisibility by 7 is to repeatedly subtract twice the last digit from the number formed by the remaining digits until you get a small number. If the result is 0 or a multiple of 7, the original number is divisible by 7.

Divisibility Rule for 11

To check for divisibility by 11, find the alternating sum of the digits, starting from the rightmost digit and alternating between adding and subtracting. If the result is 0 or a multiple of 11, the original number is divisible by 11.

Checking Each Option for Divisibility

Option 1: 16,425

• Divisibility by 7: Let's apply the rule:
  • Take the number 16425. The last digit is 5. Twice the last digit is $2 \times 5 = 10$. Subtract this from the remaining digits (1642): $1642 - 10 = 1632$.
  • Now consider 1632. The last digit is 2. Twice the last digit is $2 \times 2 = 4$. Subtract this from the remaining digits (163): $163 - 4 = 159$.
  • Now consider 159. The last digit is 9. Twice the last digit is $2 \times 9 = 18$. Subtract this from the remaining digits (15): $15 - 18 = -3$.
  Since -3 is not a multiple of 7, the number 16,425 is not divisible by 7. We can stop here, as it must be divisible by both 7 and 11.

Option 2: 12,235

• Divisibility by 7: Let's apply the rule:
  • Take the number 12235. The last digit is 5. Twice the last digit is $2 \times 5 = 10$. Subtract this from the remaining digits (1223): $1223 - 10 = 1213$.
  • Now consider 1213. The last digit is 3. Twice the last digit is $2 \times 3 = 6$. Subtract this from the remaining digits (121): $121 - 6 = 115$.
  • Now consider 115. The last digit is 5. Twice the last digit is $2 \times 5 = 10$. Subtract this from the remaining digits (11): $11 - 10 = 1$.
  Since 1 is not a multiple of 7, the number 12,235 is not divisible by 7. We can stop here.

Option 3: 16,257

• Divisibility by 7: Let's apply the rule:
  • Take the number 16257. The last digit is 7. Twice the last digit is $2 \times 7 = 14$. Subtract this from the remaining digits (1625): $1625 - 14 = 1611$.
  • Now consider 1611. The last digit is 1. Twice the last digit is $2 \times 1 = 2$. Subtract this from the remaining digits (161): $161 - 2 = 159$.
  • Now consider 159. The last digit is 9. Twice the last digit is $2 \times 9 = 18$. Subtract this from the remaining digits (15): $15 - 18 = -3$.
  Since -3 is not a multiple of 7, the number 16,257 is not divisible by 7. We can stop here.

Option 4: 16,324

• Divisibility by 7: Let's apply the rule:
  • Take the number 16324. The last digit is 4. Twice the last digit is $2 \times 4 = 8$. Subtract this from the remaining digits (1632): $1632 - 8 = 1624$.
  • Now consider 1624. The last digit is 4. Twice the last digit is $2 \times 4 = 8$. Subtract this from the remaining digits (162): $162 - 8 = 154$.
  • Now consider 154. The last digit is 4. Twice the last digit is $2 \times 4 = 8$. Subtract this from the remaining digits (15): $15 - 8 = 7$.
  Since 7 is a multiple of 7 (as $7 \div 7 = 1$), the number 16,324 is divisible by 7.
• Divisibility by 11: Let's apply the rule by finding the alternating sum of digits: Start from the rightmost digit (4) and alternate signs: $4 - 2 + 3 - 6 + 1$. Calculate the sum: $(4 + 3 + 1) - (2 + 6) = 8 - 8 = 0$. Since the alternating sum is 0, which is a multiple of 11 ($0 \div 11 = 0$), the number 16,324 is divisible by 11.

Since the number 16,324 is divisible by both 7 and 11, it is the correct option.

Revision Table: Checking Divisibility

Additional Information: Understanding Divisibility

Divisibility rules are shortcuts that help us determine if a number can be evenly divided by another number without performing long division. Knowing these rules can save time, especially in tests.

When a number is divisible by two different prime numbers, like 7 and 11, it means the number is also divisible by the product of those primes. In this case, the product is 77. So, any number divisible by both 7 and 11 must also be divisible by 77. This is because 7 and 11 share no common factors other than 1; they are coprime. If the two numbers were not coprime (e.g., 4 and 6), a number divisible by both would be divisible by their Least Common Multiple (LCM), not necessarily their product. For 4 and 6, the LCM is 12, and a number divisible by both 4 and 6 is divisible by 12. Since 7 and 11 are prime, their LCM is simply $7 \times 11 = 77$.