Answer / guest

The required number is 236 and the sum is 11.

It is given that the first two digits of the required number
are prime numbers i.e. 2, 3, 5 or 7. Note that 1 is neither
prime nor composite. Also, the third digit is the
multiplication of the first two digits. Thus, first two
digits must be either 2 or 3 i.e. 22, 23, 32 or 33 which
means that there are four possible numbers - 224, 236, 326
and 339.

Now, it is also given that - the difference between it's
reverse and itself is 396. Only 236 satisfies this
condition. Hence, the sum of the three digits is 11.

Answer / vignesh1988i

the number is : x*1+y*10+z*100 (236)

second eq: (z*1+y*10+x*100)-(x*1+y*10+z*100)=396 ,
solving this we get x-z=4

given that z&y are prime so it can only take 2,3,5,7 oly.
putting this trial we can solve and we will end up in
236....

thank u

Answer / lalit

Let the number be 100x + 10y + xy.

ofcourse xy > x. therefore 10+x-xy=6, (unit digit of the diff).

Hence x(y-1)=4

x=2, y=3, QED!!

Answer / shailendra singh

let the no. is xyz,
z=xy
suppose that no. is 236,
6=2*3,
reverse no is 632
diff=632-236-396
hence 236 is the no.

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