Question

PQR is a three-digit number such that ((100P+10Q+R/(P+Q+R)) is least. What is the product of the tens and the unit digits of PQR?​

Answer 1

Answer:

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 (Ans)

Mark as Brainlist please bro

Answer 2

Answer:

Answer: 49

Step-by-step explanation:

Minimizing Three-digit Fractions.

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PQR is a three-digit number such that ((100P+10Q+R/(P+Q+R)) is least. What is the product of the tens and the unit digits of PQR?

We want to minimize the value of the expression:

(100P + 10Q + R)/(P + Q + R)

To do this, we can try different values of P, Q, and R. However, notice that the expression is symmetric in P, Q, and R, so we can assume without loss of generality that P ≤ Q ≤ R.

If P = 1, then we have:

(100 + 10Q + R)/(1 + Q + R)

Since Q ≤ R, we have Q + R ≥ 2Q, so:

(100 + 10Q + R)/(1 + Q + R) ≤ (100 + 10Q + R)/(2Q + 1)

We want to minimize this expression. Since Q ≤ R, we have 10Q + R ≤ 10R + R = 11R. Therefore:

(100 + 10Q + R)/(2Q + 1) ≤ (100 + 11R)/(2R + 1)

We want to minimize this expression as well. Notice that:

(100 + 11R)/(2R + 1) = (99 + 10R + R)/(2R + 1) = 49 + (10R + R - 49)/(2R + 1)

Since R is a digit, we have 0 ≤ R ≤ 9. We can check that this expression is minimized when R = 0, which gives us:

(100 + 10Q + R)/(2Q + 1) ≤ 49

Therefore, the smallest possible value of the expression is 49, which occurs when P = 1, Q = 0, and R = 0. Therefore, the three-digit number is 100, and the product of the tens and units digits is 0