Aptitude Discussion

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 |

**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.**

**Abhinav**

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**PRATYUSH ANAND**

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Really Good and better explained by admin , Thanks.

**Apoorv**

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

can you explain more

not able to understand