Question

If $AB\times 5=CAB$, then how many possible values of CAB exist?

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is C 3

Here,
If we take $B$ = 0, 1 or 5, then
$0\times 5=0$
$1\times 5=5$
$5\times 5=25$
The values of B that satisfies the condition $AB\times 5=CAB$ are 0 and 5 as 1 x 5 $\ne$ 1.

(i) If $B=0$, then
$A0\times 5=CA0$
If we take $A=5$, then
$50\times 5=250$
$C=2,A=5andB=0$
$\therefore CAB=250$

(ii) If B = 5, then
A5 x 5 = CA5
If A = 2
25 x 5 = 125
$C=1,A=2andB=5$
$\therefore CAB=125$

If A = 7
75 x 5 = 375
$C=3,A=7andB=5$
$\therefore CAB=375$

Therefore, there are three possible combinations of CAB which are 125, 250 and 375.