Math Problem Statement

I have a set of four numbers P, Q, R, S in ascending order. the average of P, Q, and S is 22, and the average of Q, R, and S is 24. the largest number is 27. then, the smallest number in this set: (a) is prime, (b) is a multiple of 9, (c) is a multiple of 11, (d) is a multiple of 7, and (e) cannot be determined with this information.

Solution

We are given that P,Q,R,SP, Q, R, S are in ascending order, and the following information:

  1. The average of P,Q,SP, Q, S is 22: P+Q+S3=22    P+Q+S=66\frac{P + Q + S}{3} = 22 \implies P + Q + S = 66

  2. The average of Q,R,SQ, R, S is 24: Q+R+S3=24    Q+R+S=72\frac{Q + R + S}{3} = 24 \implies Q + R + S = 72

  3. The largest number is S=27S = 27.

Using these, let's find the values of P,Q,R,SP, Q, R, S.

Step 1: Substitute S=27S = 27 into both equations.

From P+Q+S=66P + Q + S = 66, we get: P+Q+27=66    P+Q=39P + Q + 27 = 66 \implies P + Q = 39

From Q+R+S=72Q + R + S = 72, we get: Q+R+27=72    Q+R=45Q + R + 27 = 72 \implies Q + R = 45

Step 2: Solve for P,Q,RP, Q, R.

We now have two equations:

  1. P+Q=39P + Q = 39
  2. Q+R=45Q + R = 45

Solve for PP and RR in terms of QQ:

  • P=39QP = 39 - Q
  • R=45QR = 45 - Q

Since P,Q,R,SP, Q, R, S are in ascending order, PQRSP \leq Q \leq R \leq S.

Step 3: Determine possible values of QQ.

We know that PQRP \leq Q \leq R, and using P=39QP = 39 - Q and R=45QR = 45 - Q, let's impose the condition PQRP \leq Q \leq R:

  • 39QQ39 - Q \leq Q (from PQP \leq Q) 392Q    Q19.539 \leq 2Q \implies Q \geq 19.5 Therefore, Q20Q \geq 20.

  • Q45QQ \leq 45 - Q (from QRQ \leq R) 2Q45    Q22.52Q \leq 45 \implies Q \leq 22.5 Therefore, Q22Q \leq 22.

Thus, QQ must be either 20, 21, or 22.

Step 4: Find corresponding values of P,RP, R and verify the conditions.

  • If Q=20Q = 20: P=3920=19,R=4520=25P = 39 - 20 = 19, \quad R = 45 - 20 = 25 The set P,Q,R,S=19,20,25,27P, Q, R, S = 19, 20, 25, 27 satisfies the ascending order.

  • If Q=21Q = 21: P=3921=18,R=4521=24P = 39 - 21 = 18, \quad R = 45 - 21 = 24 The set P,Q,R,S=18,21,24,27P, Q, R, S = 18, 21, 24, 27 also satisfies the ascending order.

  • If Q=22Q = 22: P=3922=17,R=4522=23P = 39 - 22 = 17, \quad R = 45 - 22 = 23 The set P,Q,R,S=17,22,23,27P, Q, R, S = 17, 22, 23, 27 satisfies the ascending order as well.

Step 5: Analyze the smallest numbers.

  • In the first case, P=19P = 19, which is prime.
  • In the second case, P=18P = 18, which is a multiple of 9.
  • In the third case, P=17P = 17, which is prime.

Conclusion:

The smallest number PP can be either 17, 18, or 19, depending on the value of QQ. Since the smallest number could be prime (e.g., 17 or 19), the answer is:

(a) is prime.

Would you like further details or have any questions?

Here are 5 related questions:

  1. How would the result change if the average of P,Q,RP, Q, R were given instead?
  2. What is the significance of ascending order in solving this problem?
  3. Can there be multiple solutions in similar problems, and how do we choose between them?
  4. What are the properties of prime numbers that make them significant here?
  5. How could inequalities help in solving problems with unknowns?

Tip: Prime numbers are natural numbers greater than 1 with no divisors other than 1 and themselves. Keep an eye out for these when solving problems involving number properties.

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Math Problem Analysis

Mathematical Concepts

Algebra
Averages
Number Properties

Formulas

Average Formula: (Sum of values) / (Number of values) = Average

Theorems

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Suitable Grade Level

Grades 6-8