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MathRevolution
[GMAT math practice question]

\(a, b\) and \(c\) are different positive integers between \(1\) and \(9\), inclusive. The \(5\)-digit integer \(ababc\) is a multiple of \(12\), and the \(2\)-digit number \(ab\) is equal to \(c^2\). What is the value of the \(3\)-digit number \(abc\)?

\(A. 164\)

\(B. 255\)

\(C. 366\)

\(D. 497\)

\(E. 648\)

a*b = c^2

Since a,b and c are different numbers hence c can't be 5/6.

Let's look at other possibilities of c: 4, 7, 8 and 9 only.

Put values of ababc to check whether it's divisible by both 3 and 2- only possibility: abc = 164
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First of all, a,b, and c are all different integers between \(1\)and \(9\).
It is given that the digits \(ab = c^2\)

So ab will have \(6\) possible values for \(ab\) - \(16, 25, 36, 49, 64, and 81.\)

\(abc\) is can be drawn from the above.
Suppose if \(c = 4\), then \(abc\) will be \(164\). If \(c = 5\), \(abc\) will be \(255\), and so on!

We need to find out \(ababc\), a five-digit integer.
A very important constraint! \(ababc\) must be divisible by \(12\), which further means it is divisible by \(3\)as well as \(4\), as we know from the divisibility test of \(12\).

Take the options, You'll arrive at the solutions in less than a minute!

A.\(164\)->\(16164\). We see that the number formed by the last two digits is divisible 4. Okay!
the sum of the digits is divisible by \(3\). Alright! Satisfies all the conditions. This is the answer.
B.\(255\)->\(25255\). Neither divisible by \(3\) nor \(4\)! OUT.
C.\(366\)->\(36366\). Divisible by \(3\). Not divisible by 4. OUT.
D.\(497\)->\(49497\). Not divisible by \(4\). OUT.
E.\(648\)->\(64648\). Not Divisible by \(3\). OUT.
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