A 3 digit number is such that it's unit digit is equal to
the product of the other two digits which are prime. Also,
the difference between it's reverse and itself is 396.
What is the sum of the three digits?
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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.
| Is This Answer Correct ? | 14 Yes | 0 No |
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
| Is This Answer Correct ? | 6 Yes | 1 No |
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!!
| Is This Answer Correct ? | 4 Yes | 2 No |
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.
| Is This Answer Correct ? | 0 Yes | 2 No |
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