2021 AMC 10B Spring Problem 8

Below is the video solution and professionally curated solution for Problem 8 of the 2021 AMC 10B Spring, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 10B Spring solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:pattern recognitionperfect square

Difficulty rating: 1420

8.

Mr. Zhou places all the integers from 11 to 225225 into a 1515 by 1515 grid. He places 11 in the middle square (eighth row and eighth column) and places other numbers one by one clockwise, as shown in part in the diagram below. What is the sum of the greatest number and the least number that appear in the second row from the top?

367 367

368 368

369 369

379 379

380 380

Video solution:
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Written solution:

In the outer 15×1515\times15 ring, the top row contains 211,212,,225211,212,\ldots,225, so the number just below 211211 in the second row is 210210. This is the greatest number in the second row.

The inner 13×1313\times13 spiral has 132=16913^2=169 in its upper-right corner. In the second row of the full grid, the inner-ring entries run from 157157 to 169169. Thus the least entry in that row is 157157.

The required sum is 210+157=367210+157=367.

Thus, the answer is A .

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