BTGmoderatorDC wrote: ↑

Tue Jul 19, 2022 6:16 pm

If A, B and C represent different digits in the multiplication, AAB * B = CB5B, then A + B + C = ?

A. 9
B. 12
C. 14
D. 15
E. 17


OA E

Source: Magoosh

Step 1: Value of \(B\).
If we look at the units digit of the numbers being multiplied and the product, we get \(B*B=_B\).
Thus, B can only be \(0, 1, 5\) and \(6\).
If \(0\), then the product will be \(0\). Discard
If \(1\), then the product will be \(AA1\). Discard
If \(5\), then \(AA5*5=C555\). But the product of two multiples of \(5\) should end in \(00, 25, 50\) or \(75\). Discard
So, \(B=6\).

Step 2: Value of \(C\).
So, the product is \(\Rightarrow \, AA6\ast 6=C656\)
Now, \(C656\) should be a multiple of \(6\), so \(C\) can be \(1, 4\) or \(7\).

If \(C\) is \(1\), then \(\dfrac{1656}{6}=276 \neq AA6\)

If \(C\) is \(4\), then \(\dfrac{4656}{6}=776 = AA6\)

So, \(C=4\) and \(A=7\)

SUM \(= 7+4+6 = 17\)