Difference between revisions of "1985 AJHSME Problem 7"

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<math>\text{(A)}\ 34 \qquad \text{(B)}\ 35 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 37 \qquad \text{(E)}\ 38</math>
 
<math>\text{(A)}\ 34 \qquad \text{(B)}\ 35 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 37 \qquad \text{(E)}\ 38</math>
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== In-depth Solution by Boundless Brain!==
  
 
== Solution==
 
== Solution==
 
Notice that in the <math>n</math>th row, there are <math>n</math> white squares and <math>n-1</math> black squares. So, the <math>37th</math> row will have <math>\boxed{\text{(C) 36}}</math> black squares.
 
Notice that in the <math>n</math>th row, there are <math>n</math> white squares and <math>n-1</math> black squares. So, the <math>37th</math> row will have <math>\boxed{\text{(C) 36}}</math> black squares.

Revision as of 23:48, 29 June 2023

Problem

A "stair-step" figure is made of alternating black and white squares in each row. Rows $1$ through $4$ are shown. All rows begin and end with a white square. The number of black squares in the $37\text{th}$ row is

[asy] draw((0,0)--(7,0)--(7,1)--(0,1)--cycle); draw((1,0)--(6,0)--(6,2)--(1,2)--cycle); draw((2,0)--(5,0)--(5,3)--(2,3)--cycle); draw((3,0)--(4,0)--(4,4)--(3,4)--cycle); fill((1,0)--(2,0)--(2,1)--(1,1)--cycle,black); fill((3,0)--(4,0)--(4,1)--(3,1)--cycle,black); fill((5,0)--(6,0)--(6,1)--(5,1)--cycle,black); fill((2,1)--(3,1)--(3,2)--(2,2)--cycle,black); fill((4,1)--(5,1)--(5,2)--(4,2)--cycle,black); fill((3,2)--(4,2)--(4,3)--(3,3)--cycle,black); [/asy]

$\text{(A)}\ 34 \qquad \text{(B)}\ 35 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 37 \qquad \text{(E)}\ 38$

In-depth Solution by Boundless Brain!

Solution

Notice that in the $n$th row, there are $n$ white squares and $n-1$ black squares. So, the $37th$ row will have $\boxed{\text{(C) 36}}$ black squares.