Difference between revisions of "2007 AMC 8 Problems/Problem 18"

Latest revision as of 04:43, 31 December 2022

Problem

The product of the two $99$ -digit numbers

$303,030,303,...,030,303$ and $505,050,505,...,050,505$

has thousands digit $A$ and units digit $B$ . What is the sum of $A$ and $B$ ?

$\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 5 \qquad \mathrm{(C)}\ 6 \qquad \mathrm{(D)}\ 8 \qquad \mathrm{(E)}\ 10$

Solution

We can first make a small example to find out $A$ and $B$ . So,

$303\times505=153015$

The ones digit plus thousands digit is $5+3=8$ .

Note that the ones and thousands digits are, added together, $8$ . (and so on...) So the answer is $\boxed{\textbf{(D)}\ 8}$ This is a direct multiplication way.

Video Solution by OmegaLearn

https://youtu.be/7an5wU9Q5hk?t=2085

~ pi_is_3.14

Video Solution by WhyMath

https://youtu.be/_goaFuScO6M

~savannahsolver

2007 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 1: / Line 1: @@
-When you multiply <math> 303 </math> by <math> 505 </math>, it's <math> 153015 </math>. The ones digit plus thousands digit is <math> 5 </math> <math> + </math> <math> 3 </math> <math> = </math> <math> 8 </math>. <math> 30303 </math> <math> \times </math> <math> 50505 </math> <math> = </math> <math> 1530453015 </math>. Note that the ones and thousands digits are, added together, <math> 8 </math>. (and so on...) So the answer is <math> \boxed{D} </math> <math> (8) </math>.
+==Problem==
+The product of the two <math>99</math>-digit numbers
+<math>303,030,303,...,030,303</math> and <math>505,050,505,...,050,505</math>
+has thousands digit <math>A</math> and units digit <math>B</math>. What is the sum of <math>A</math> and <math>B</math>?
+<math>\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 5 \qquad \mathrm{(C)}\ 6 \qquad \mathrm{(D)}\ 8 \qquad \mathrm{(E)}\ 10</math>
+==Solution==
+We can first make a small example to find out <math>A</math> and <math>B</math>. So,
+<math>303\times505=153015 </math>
+The ones digit plus thousands digit is <math>5+3=8</math>.
+Note that the ones and thousands digits are, added together, <math>8</math>. (and so on...) So the answer is <math>\boxed{\textbf{(D)}\ 8}</math>
+This is a direct multiplication way.
+==Video Solution by OmegaLearn==
+https://youtu.be/7an5wU9Q5hk?t=2085
+~ pi_is_3.14
+==Video Solution by WhyMath==
+https://youtu.be/_goaFuScO6M
+~savannahsolver
+==See Also==
+{{AMC8 box|year=2007|num-b=17|num-a=19}}
+{{MAA Notice}}

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