Question

Consider a 4-digit number of the form abbb, i.e., the first digit is a (a > 0) and the last three digits are all b.
Which of the following conditions is both NECESSARY and SUFFICIENT to ensure that the 4- digit number is divisible by a?

• A. b is divisible by a
• B. b is equal to 0
• C. 21b is divisible by a
• D. 9b is divisible by a
• E. 3b is divisible by a

Answer 1

Answer: (E)

To determine the condition that is both necessary and sufficient for a 4-digit number of the form \(abbb\) to be divisible by \(a\), we start by expressing the number \(abbb\) in its numerical form:

\(abbb = 1000a + 111b\)

Here, \(a\) is the first digit and \(b\) is repeated three times at the end.

We need to find such a condition under which this number is divisible by \(a\). This implies:

\(1000a + 111b \equiv 0 \pmod{a}\)

This simplifies to:

\(111b \equiv 0 \pmod{a}\)

This means that \(111b\) must be divisible by \(a\). We can further simplify this by examining the factors of \(111\):

\(111 = 3 \times 37\)

Thus, for divisibility by \(a\), it is necessary and sufficient that:

\(3b \equiv 0 \pmod{a}\)

This indicates that the condition \(3b\) being divisible by \(a\) is both necessary and sufficient to ensure the number \(abbb\) is divisible by \(a\).

By checking the options, the correct choice is:

3b is divisible by a

 

Answer 2

To solve the problem, we have a 4-digit number of the form abbb, which means the number is represented as: \(1000a + 111b\).

We are tasked to find the condition that ensures this number is divisible by a. Here’s the step-by-step solution:

The number can be expressed as: \(1000a + 111b\).

For a number to be divisible by a, \(1000a + 111b\) must be divisible by a. Simplifying, we have:

1. \(1000a + 111b \equiv 0 \pmod{a}\).

Since \(1000a\) is already divisible by a (as it’s a multiple of a), it suffices to check the divisibility of \(111b\) by a. Thus, we require:

1. \(111b \equiv 0 \pmod{a}\)

The expression \(111b\) means a needs to divide \(111b\), or equivalently:

1. \(b\) multiplied by 111 should be divisible by a.

We factorize 111:

\(111 = 3 \times 37\)

This implies that b multiplied by 111 is divisible by a if 3b is divisible by a (the least factor that needs to be considered since if 3b is divisible by a, then \(111b\) will also be divisible by a).

Thus, the condition "3b is divisible by a" is both necessary and sufficient for the number \(abbb\) to be divisible by a.

The correct choice is: 3b is divisible by a.