Difference between revisions of "2016 AMC 10B Problems/Problem 6"
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Adding the numbers <math>257</math> and <math>143</math>, we get <math>400</math> which means our answer is <math>\boxed{\textbf{(B)}\ 4}</math>. | Adding the numbers <math>257</math> and <math>143</math>, we get <math>400</math> which means our answer is <math>\boxed{\textbf{(B)}\ 4}</math>. | ||
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+ | ==Video Solution (CREATIVE THINKING)== | ||
+ | https://youtu.be/6cwlGbTpp2Y | ||
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+ | ~Education, the Study of Everything | ||
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==Video Solution== | ==Video Solution== | ||
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~savannahsolver | ~savannahsolver | ||
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+ | == Video Solution == | ||
+ | https://youtu.be/HISL2-N5NVg?t=1046 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=B|num-b=5|num-a=7}} | {{AMC10 box|year=2016|ab=B|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 12:47, 2 July 2023
Contents
Problem
Laura added two three-digit positive integers. All six digits in these numbers are different. Laura's sum is a three-digit number . What is the smallest possible value for the sum of the digits of
?
Solution 1
Let the two three-digit numbers she added be and
with
and
. The hundreds digits of these numbers must be at least
and
, so
and
.
Say and
; then we just need
with
and
having different digits which aren't
or
.There are many solutions, but
and
give
which proves that
is attainable.
Solution 2
For this problem, to find the -digit integer with the smallest sum of digits, one should make the units and tens digit add to
. To do that, we need to make sure the digits are all distinct. For the units digit, we can have a variety of digits that work.
works best for the top number which makes the bottom digit
. The tens digits need to add to
because of the
that needs to be carried from the addition of the units digits. We see that
and
work the best as we can't use
and
. Finally, we use
and
for our hundreds place digits.
Adding the numbers and
, we get
which means our answer is
.
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
~savannahsolver
Video Solution
https://youtu.be/HISL2-N5NVg?t=1046
~ pi_is_3.14
See Also
2016 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.