Three natural numbers p, q and r add up to 1000. The number r is a perfect square between 200 and 600, and p is at least 25% of q but at most 75% of q. Find the sum of the maximum and minimum possible values of p.
Answer and solution
Correct answer: 417
Solution
Answer: 417
1. Allowed squares are 225,256,289,324,361,400,441,484,529,576. Set S=p+q=1000−r.
2. From p≥q/4 and q=S−p, obtain p≥S/5. From p≤3q/4 obtain p≤3S/7.
3. To maximize p choose largest S=775, at r=225. Then p≤2325/7=332.142…, so max p=332. q=443 satisfies both ratio bounds.
4. To minimize p choose smallest S=424, at r=576. Then p≥424/5=84.8, so min p=85. q=339 satisfies both bounds.
5. Both extremes are attainable natural numbers. Required sum=332+85=417.
Why this method works:
The condition on p is relative to q, not to the combined sum S. Substituting q=S−p changes p≥q/4 into 5p≥S, and p≤3q/4 into 7p≤3S. This distinction explains the denominators 5 and 7.
Independent check and common trap:
For the minimum, ceil(424/5)=85 givesq=339 and 85≥84.75. For the maximum, floor(3×775/7)=332 givesq=443 and 332≤332.25. Check the opposite inequality too; both are satisfied, so both endpoint constructions are feasible.
Apply: Write S=1000−r. The ratio bounds give S/5≤p≤3S/7. Use largest square 576 for min p and smallest square 225 for max p; round inward.
Check: For the minimum, ceil(424/5)=85 givesq=339 and 85≥84.75. For the maximum, floor(3×775/7)=332 givesq=443 and 332≤332.25. Check the opposite inequality too; both are satisfied, so both endpoint constructions are feasible.
1. Allowed squares are 225,256,289,324,361,400,441,484,529,576. Set S=p+q=1000−r.
2. From p≥q/4 and q=S−p, obtain p≥S/5. From p≤3q/4 obtain p≤3S/7.
3. To maximize p choose largest S=775, at r=225. Then p≤2325/7=332.142…, so max p=332. q=443 satisfies both ratio bounds.
4. To minimize p choose smallest S=424, at r=576. Then p≥424/5=84.8, so min p=85. q=339 satisfies both bounds.
5. Both extremes are attainable natural numbers. Required sum=332+85=417.
Why this method works:
The condition on p is relative to q, not to the combined sum S. Substituting q=S−p changes p≥q/4 into 5p≥S, and p≤3q/4 into 7p≤3S. This distinction explains the denominators 5 and 7.
Independent check and common trap:
For the minimum, ceil(424/5)=85 givesq=339 and 85≥84.75. For the maximum, floor(3×775/7)=332 givesq=443 and 332≤332.25. Check the opposite inequality too; both are satisfied, so both endpoint constructions are feasible.
G Strategy shortcut
Spot: Number Properties.Apply: Write S=1000−r. The ratio bounds give S/5≤p≤3S/7. Use largest square 576 for min p and smallest square 225 for max p; round inward.
Check: For the minimum, ceil(424/5)=85 givesq=339 and 85≥84.75. For the maximum, floor(3×775/7)=332 givesq=443 and 332≤332.25. Check the opposite inequality too; both are satisfied, so both endpoint constructions are feasible.
G Strategy: Spot: Number Properties. Apply: Write S=1000−r. The ratio bounds give S/5≤p≤3S/7. Use largest square 576 for min p and smallest square 225 for max p; round inward. Check: For the minimum, ceil(424/5)=85 givesq=339 and 85≥84.75. For the maximum, floor(3×775/7)=332 givesq=443 and 332≤332.25. Check the opposite inequality too; both are satisfied, so both endpoint constructions are feasible.