Difference between revisions of "2007 AMC 12A Problems/Problem 11"
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<math>\mathrm{(A)}\ 3\qquad \mathrm{(B)}\ 7\qquad \mathrm{(C)}\ 13\qquad \mathrm{(D)}\ 37\qquad \mathrm{(E)}\ 43</math> | <math>\mathrm{(A)}\ 3\qquad \mathrm{(B)}\ 7\qquad \mathrm{(C)}\ 13\qquad \mathrm{(D)}\ 37\qquad \mathrm{(E)}\ 43</math> | ||
| − | == | + | ==Solutions== |
A given digit appears as the hundreds digit, the tens digit, and the units digit of a term the same number of times. Let <math>k</math> be the sum of the units digits in all the terms. Then <math>S=111k=3 \cdot 37k</math>, so <math>S</math> must be divisible by <math>37\ \mathrm{(D)}</math>. To see that it need not be divisible by any larger prime, the sequence <math>123, 231, 312</math> gives <math>S=666=2 \cdot 3^2 \cdot 37\Rightarrow \mathrm{\boxed{(D)}}</math>. | A given digit appears as the hundreds digit, the tens digit, and the units digit of a term the same number of times. Let <math>k</math> be the sum of the units digits in all the terms. Then <math>S=111k=3 \cdot 37k</math>, so <math>S</math> must be divisible by <math>37\ \mathrm{(D)}</math>. To see that it need not be divisible by any larger prime, the sequence <math>123, 231, 312</math> gives <math>S=666=2 \cdot 3^2 \cdot 37\Rightarrow \mathrm{\boxed{(D)}}</math>. | ||
Revision as of 03:42, 12 January 2021
- The following problem is from both the 2007 AMC 12A #11 and 2007 AMC 10A #22, so both problems redirect to this page.
Problem
A finite sequence of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might begin with the terms 247, 475, and 756 and end with the term 824. Let 
be the sum of all the terms in the sequence. What is the largest prime factor that always divides ?
Solutions
A given digit appears as the hundreds digit, the tens digit, and the units digit of a term the same number of times. Let 
be the sum of the units digits in all the terms. Then 
, so must be divisible by 
. To see that it need not be divisible by any larger prime, the sequence 
gives 
.
See also
| 2007 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 10 |
Followed by Problem 12 |
| 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 | |
| 2007 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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. 
