Math, asked by vikas5717, 1 year ago

pqr is a three digit natural number so that pqr=p!+q!+r!.what is the value of(q+r)p

Answers

Answered by raisulrahat208pdm2pu
3
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Answered by SerenaBochenek
3

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


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