Difference between revisions of "Mock AIME 1 2007-2008 Problems/Problem 4"
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Since <math>x</math> is odd, let <math>x = 2n-1</math>. The expression becomes | Since <math>x</math> is odd, let <math>x = 2n-1</math>. The expression becomes | ||
<cmath>4(10n-4)(10n-2)(10n)=32(5n-2)(5n-1)(5n)</cmath> | <cmath>4(10n-4)(10n-2)(10n)=32(5n-2)(5n-1)(5n)</cmath> | ||
| − | Consider just the product of the last three terms, <math>5n-2,5n-1,5n</math>, which are consecutive. At least one term must be divisible by <math>2</math> and one term must be divisible by <math>3</math> then. Also, since there is the <math>5n</math> term, the expression must be divisible by <math>5</math>. Therefore, the minimum integer that always divides the expression must be <math>32 \cdot 2 \cdot 3 \cdot 5 = \boxed{960}</math>. | + | Consider just the product of the last three terms, <math>5n-2,5n-1,5n</math>, which are consecutive. At least one term must be divisible by <math>2</math> and one term must be divisible by <math>3</math> then. Also, since there is the <math>5n</math> term, the expression must be divisible by <math>5</math>. Therefore, the minimum integer that always divides the expression must be <math>32 \cdot 2 \cdot 3 \cdot 5 = \Rightarrow{\boxed{960}}</math>. |
| − | To prove that the number is the largest integer to work, consider when <math>x=1</math> and <math>x = 5</math>. These respectively evaluate to be <math>1920,\ 87360</math>; their [[greatest common factor]] is indeed <math>960</math>. | + | To prove that the number is the largest integer to work, consider when <math>x=1</math> and <math>x = 5</math>. These respectively evaluate to be <math>1920,\ 87360</math>; their [[greatest common factor]] is indeed <math>960</math>. |
== See also == | == See also == | ||
Latest revision as of 00:04, 10 August 2019
Problem
If 
is an odd number, then find the largest integer that always divides the expression
![\[(10x+2)(10x+6)(5x+5)\]](http://latex.artofproblemsolving.com/3/d/8/3d87c25db3731139b72d00b8d2232e15f92cb468.png)
Solution
Rewrite the expression as
![\[4(5x + 1)(5x + 3)(5x+5)\]](http://latex.artofproblemsolving.com/a/1/4/a14042fc131e4f712264a8a8916e4ce0d98e444e.png)
Since is odd, let 
. The expression becomes
![\[4(10n-4)(10n-2)(10n)=32(5n-2)(5n-1)(5n)\]](http://latex.artofproblemsolving.com/5/f/d/5fd22487faa25d82b05aaa1f0fdd6cde5ab18f99.png)
Consider just the product of the last three terms, 
, which are consecutive. At least one term must be divisible by 
and one term must be divisible by 
then. Also, since there is the 
term, the expression must be divisible by 
. Therefore, the minimum integer that always divides the expression must be 
.
To prove that the number is the largest integer to work, consider when 
and 
. These respectively evaluate to be 
; their greatest common factor is indeed 
.
See also
| Mock AIME 1 2007-2008 (Problems, Source) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
