Question

A mystery number, its successor, and the sum of the number and its successor are all add up to give a three-digit number that is a multiple of the lowest odd prime. What can you say about the predecessor of the mystery number?

• A. Such a mystery number cannot exist.
• B. The predecessor is not a multiple of 3.
• C. The predecessor is a multiple of 3.
• D. The predecessor is a prime.

Answer 1

The correct option is C The predecessor is a multiple of 3.

Let the mystery number be "k".

$⟹Successorof{}^{\prime }{k}^{\prime }=k+1⟹Sumofthenumberanditssuccessor=k+(k+1)=2k+1$

Now, the lowest odd prime number is 3.

Hence, the sum of the three numbers is a multiple of 3.

So, for an integer "N", we have:

$k+(k+1)+(2k+1)=3\times N⟹4k+2=3N$

Now, we know that "4k + 2" is a three-digit number.

So, we take:

$Hundredsdigitof(4k+2)=aTensdigitof(4k+2)=bOnesdigitof(4k+2)=c$

$⟹4k+2=(100\times a)+(10\times b)+c=99a+9b+a+b+c$

Now, as "4k + 2" is a multiple of 3, the sum of its digits will also be a multiple of 3.

So, for an integer M, we have:

$⟹a+b+c=3\times M$

$⟹4k+2=99a+9b+3\times M⟹3k+(k+2)=3(33a+3b+M)⟹k+2=3(33a+3b+M–k)⟹k+2isamultipleof3.$

Now, the predecessor of "k" is "k – 1".

$k–1=(k+2)–3=3(33a+3b+M–k)+3=3(33a+3b+M–k+1)⟹(k–1)isamultipleof3$

Hence, the predecessor of the unknown number is a multiple of 3.