Number System
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Q.

pqr is a three digit natural number so that $pqr = p!+q!+r!$.

What is the value of $(q+r)^p$?

 A.

1296

 B.

3125

 C.

19683

 D.

9

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Solution:
Option(D) is correct

$100p+10q+r = p!+q!+r!$

Now, $6!=720$ and $7! =5040$.

If 7 is one of digits, then the sum of the factorials becomes four digits number or more.
Hence the numbers $7, 8, 9$ can be neglected.

Consider $6! =720$.
But 7 cannot be there in hundred's place.
Hence, we can neglect 6 also.

Now,$ 5!=120, 4!=24, 3!=6, 2!=2$ and $1!=1.$
To get a three digit number, 5 has to be present in the number.
But 5 cannot be in hundreds place as then the number greater than 500 which cannot be obtained as the sum of factorial.

Also maximum possible number is $5!+4!+3! = 150$
Also '$p$' cannot be zero as it is a three digit number.
Hence $p=1$.

Then different possible cases are $154, 153, 152, 125, 135, 145$
From this only 145 satisfies the condition
Thus, $(4+5) =$ 9.


(3) Comment(s)


Abhinav
 ()

can you explain more

not able to understand



PRATYUSH ANAND
 ()

Really Good and better explained by admin , Thanks.



Apoorv
 ()

i lyk d question...good question... Cool