Finding the Least Number to Add for Divisibility by 3, 4, 5, and 7

The problem asks for the smallest number that, when added to 35460, makes the resulting sum exactly divisible by 3, 4, 5, and 7. For a number to be divisible by multiple numbers simultaneously, it must be divisible by their Least Common Multiple (LCM).

First, we need to find the LCM of 3, 4, 5, and 7.

Step 1: Calculate the LCM of 3, 4, 5, and 7

To find the LCM, we can use prime factorization:

• Prime factors of 3: 3
• Prime factors of 4: \(2^2\)
• Prime factors of 5: 5
• Prime factors of 7: 7

Since 3, 5, and 7 are prime numbers and 4 is \(2^2\), and there are no common factors other than 1 among these numbers (they are pairwise coprime), the LCM is the product of these numbers:

\[ \text{LCM}(3, 4, 5, 7) = 3 \times 4 \times 5 \times 7 \]

\[ \text{LCM}(3, 4, 5, 7) = 12 \times 35 \]

\[ \text{LCM}(3, 4, 5, 7) = 420 \]

So, the least common multiple is 420. This means the number we are looking for (35460 plus the added number) must be a multiple of 420.

Step 2: Find the Remainder When 35460 is Divided by the LCM (420)

We need to determine how far 35460 is from the next multiple of 420. We do this by dividing 35460 by 420 and finding the remainder.

Let's perform the division: \(35460 \div 420\).

We can simplify the division by removing a zero from both numbers: \(3546 \div 42\).

We can use long division:

To verify the remainder for \(3546 \div 42\): \(42 \times 84 + 18 = 3528 + 18 = 3546\). The remainder is 18.

Now, let's relate this back to \(35460 \div 420\). Since \(35460 = 3546 \times 10\) and \(420 = 42 \times 10\), the division is equivalent to \( (3546 \times 10) \div (42 \times 10) \). The quotient will be the same (84), but the remainder will be 10 times the remainder of \(3546 \div 42\).

Remainder for \(35460 \div 420\) is \(18 \times 10 = 180\).

Alternatively, using standard division:

\[ 35460 = 420 \times Q + R \]

\[ 35460 = 420 \times 84 + 180 \]

The quotient is 84 and the remainder is 180.

Step 3: Determine the Least Number to Add

The number 35460 is not a perfect multiple of 420; it is 180 less than the next multiple of 420. The next multiple of 420 after \(420 \times 84\) is \(420 \times 85\).

\[ 420 \times 85 = 420 \times (84 + 1) = (420 \times 84) + 420 = 35280 + 420 = 35700 \]

To get from 35460 to the next multiple of 420 (which is 35700), we need to add the difference:

Number to add \( = \text{Next multiple of LCM} - \text{Current number} \)

Number to add \( = 35700 - 35460 \)

Number to add \( = 240 \)

Alternatively, the number to add to make a number divisible by the divisor is \( \text{Divisor} - \text{Remainder} \).

Number to add \( = \text{LCM} - \text{Remainder} \)

Number to add \( = 420 - 180 \)

Number to add \( = 240 \)

Adding 240 to 35460 gives \(35460 + 240 = 35700\). We already checked that 35700 is divisible by 3, 4, 5, and 7 because it is a multiple of 420 (specifically, \(35700 = 420 \times 85\)).

Thus, the least number that should be added to 35460 so that the sum is exactly divisible by 3, 4, 5 and 7 is 240.

This corresponds to option 3.

Revision Table: Divisibility Problem

Additional Information: LCM and Divisibility Rules

Understanding LCM and basic divisibility rules is crucial for solving problems like this. Here's a brief recap:

Least Common Multiple (LCM)

The LCM is useful in problems involving cycles, simultaneous events, or, as in this case, finding a number divisible by several others. Methods to find LCM include prime factorization or listing multiples.

Divisibility Rules

• Divisibility by 3: A number is divisible by 3 if the sum of its digits is divisible by 3.
• Divisibility by 4: A number is divisible by 4 if the number formed by its last two digits is divisible by 4.
• Divisibility by 5: A number is divisible by 5 if its last digit is 0 or 5.
• Divisibility by 7: There's a rule, but it's often easier to just divide the number by 7 or check if it's a multiple of 7 if the numbers are large.

In our calculation, we found that adding 240 resulted in 35700. Let's quickly check the divisibility rules for 35700:

• Sum of digits: \(3+5+7+0+0 = 15\). 15 is divisible by 3, so 35700 is divisible by 3.
• Last two digits: 00. 00 is divisible by 4, so 35700 is divisible by 4.
• Last digit: 0. 35700 is divisible by 5.
• For 7: \(35700 = 357 \times 100\). \(357 = 7 \times 51\). Since 357 is divisible by 7, 35700 is also divisible by 7.

All divisibility rules confirm that 35700 is divisible by 3, 4, 5, and 7.