Difference between revisions of "2005 AMC 12B Problems/Problem 15"
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<math>221</math> can be written as the sum of eight two-digit numbers, let's say <math>\overline{ae}</math>, <math>\overline{bf}</math>, <math>\overline{cg}</math>, and <math>\overline{dh}</math>. Then <math>221= 10(a+b+c+d)+(e+f+g+h)</math>. The last digit of <math>221</math> is <math>1</math>, and <math>10(a+b+c+d)</math> won't affect the units digits, so <math>(e+f+g+h)</math> must end with <math>1</math>. The smallest value <math>(e+f+g+h)</math> can have is <math>(1+2+3+4)=10</math>, and the greatest value is <math>(6+7+8+9)=30</math>. Therefore, <math>(e+f+g+h)</math> must equal <math>11</math> or <math>21</math>. | <math>221</math> can be written as the sum of eight two-digit numbers, let's say <math>\overline{ae}</math>, <math>\overline{bf}</math>, <math>\overline{cg}</math>, and <math>\overline{dh}</math>. Then <math>221= 10(a+b+c+d)+(e+f+g+h)</math>. The last digit of <math>221</math> is <math>1</math>, and <math>10(a+b+c+d)</math> won't affect the units digits, so <math>(e+f+g+h)</math> must end with <math>1</math>. The smallest value <math>(e+f+g+h)</math> can have is <math>(1+2+3+4)=10</math>, and the greatest value is <math>(6+7+8+9)=30</math>. Therefore, <math>(e+f+g+h)</math> must equal <math>11</math> or <math>21</math>. | ||
Revision as of 16:53, 31 January 2013
Contents
Problem
The sum of four two-digit numbers is . None of the eight digits is
and no two of them are the same. Which of the following is not included among the eight digits?
Solution 1
can be written as the sum of eight two-digit numbers, let's say
,
,
, and
. Then
. The last digit of
is
, and
won't affect the units digits, so
must end with
. The smallest value
can have is
, and the greatest value is
. Therefore,
must equal
or
.
Case 1:
The only distinct positive integers that can add up to is
. So,
,
,
, and
must include four of the five numbers
. We have
, or
. We can add all of
, and try subtracting one number to get to
, but to no avail. Therefore,
cannot add up to
.
Case 2:
Checking all the values for ,
,
,and
each individually may be time-consuming, instead of only having
solution like Case 1. We can try a different approach by looking at
first. If
,
, or
. That means
. We know
, so the missing digit is
Solution 2
Alternatively, we know that a number is congruent to the sum of its digits mod 9, so , where
is some digit. Clearly,
.
See also
2005 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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 12 Problems and Solutions |