pqr is a three digit natural number so that pqr=p!+q!+r!.what is the value of(q+r)p
Answers
Answer:
The value of (q+r)p=(4+5)1=9
Step-by-step explanation:
Given 3 digit natural number pqr such that pqr=p!+q!+r!
We have to find out the value of (q+r)p
we have to choose the value of p, q and r . let us choose 1 digit with the greatest number to form 3 digit natural number. Let it be 9. we can't choose this number as the factorial of itself is of 6 digits but we have to make 3 digit natural number.
Due to same reason as above mentioned 8 and 7 can't be considered. Therefore, discarded.
Now, if we take 6 its factorial equals to 720 but this number includes 7 which we can't consider mentioned above.
So we get the greatest number to start with as 5.
Lets take one digit 5, to satisfying the condition pqr=p!+q!+r! we can't take any of the number less than 5. Now as we know that 5!=120 and we also know that all the digits in the number must be different, we cannot exceed the value of 150 (150=5!+4!+3!). So, one digit will be 1
we know 2 digits and last one digit can be find out by hit and trial method
Lets take 4, which gives the number 1!+4!+5!=145
Hence, the number formed is 145
⇒p=1, q=4, r=5
Therefore, (q+r)p=(4+5)1=9