Question

The possible number that can be formed by using unit's digit and ten's digit in $(274\times 243\times 131)$ lies between:

• A. $23$ and $45$
• B. $23$ and $44$
• C. $26$ and $46$
• D. $25$ and $50$

Answer 1

Answer: (A), (B)

The correct options are
A $23$ and $45$
B $23$ and $44$

We can observe that the unit digit will be formed by the product of $4\ast 3\ast 1=12$ So the unit digit will be 2.
Now for the tens place $(200+70+4)\ast (200+40+3)\ast (100+30+1)$
We are concerned with only tens digit so we can omit the hundred place part.
$(70+4)\ast (40+3)\ast (30+1)$
$=\left[\right(70+4\left)\right(40+3\left)\right](30+1)$
$(2800+210+160+12)(30+1)$
Omitting the hundreds place after multiplying the first two terms in the expression, we get
$(00+10+60+12)(30+1)$
$82\ast (30+1)$
$2460+82$
$(60+82)$ (again omitting all places to the left of ten's place)
$\left(142\right)$
So the last two digits are 4 and 2 so the numbers can be 42 and 24.
Option A and B