Difference between revisions of "1998 AJHSME Problems/Problem 7"

Latest revision as of 23:28, 4 July 2013

$100\times 19.98\times 1.998\times 1000=$

$\text{(A)}\ (1.998)^2 \qquad \text{(B)}\ (19.98)^2 \qquad \text{(C)}\ (199.8)^2 \qquad \text{(D)}\ (1998)^2 \qquad \text{(E)}\ (19980)^2$

$19.98\times100=1998$

$1.998\times1000=1998$

$1998\times1998=(1998)^2=\boxed{D}$

1998 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

@@ Line 1: / Line 1: @@
-==Problem 7==
+==Problem==
 <math>100\times 19.98\times 1.998\times 1000=</math>
@@ Line 8: / Line 8: @@
 <math>19.98\times100=1998</math>
 <math>1.998\times1000=1998</math>
 <math>1998\times1998=(1998)^2=\boxed{D}</math>
 == See also ==
-{{AJHSME box|year=1998|before=[[1997 AJHSME Problems|1997 AJHSME]]|after=[[1999 AMC 8 Problems|1999 AMC 8]]}}
+{{AJHSME box|year=1998|num-b=6|num-a=8}}
 * [[AJHSME]]
 * [[AJHSME Problems and Solutions]]
 * [[Mathematics competition resources]]
+{{MAA Notice}}