Difference between revisions of "2020 AMC 10A Problems/Problem 22"
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== Solution == | == Solution == | ||
Let <math>a = \left\lfloor \frac{998}n \right\rfloor</math>. If the expression is not divisible by <math>3</math>, then the three terms in the expression must be <math>(a, a + 1, a + 1)</math>, which would imply that <math>n</math> is divisible by <math>999</math> but not <math>1000</math>, or <math>(a, a, a + 1)</math>, which would imply that <math>n</math> is divisible by <math>1000</math> but not <math>999</math>. <math>999 = 3^3 \cdot 37</math> has <math>4 \cdot 2 = 8</math> factors, and <math>1000 = 2^3 \cdot 5^3</math> has <math>4 \cdot 4 = 16</math> factors. However, <math>n = 1</math> does not work because <math>1</math> is divisible by both <math>999</math> and <math>1000</math>, and since <math>1</math> is counted twice, the answer is <math>16 + 8 - 2 = \boxed{\textbf{(A) }22}</math>. | Let <math>a = \left\lfloor \frac{998}n \right\rfloor</math>. If the expression is not divisible by <math>3</math>, then the three terms in the expression must be <math>(a, a + 1, a + 1)</math>, which would imply that <math>n</math> is divisible by <math>999</math> but not <math>1000</math>, or <math>(a, a, a + 1)</math>, which would imply that <math>n</math> is divisible by <math>1000</math> but not <math>999</math>. <math>999 = 3^3 \cdot 37</math> has <math>4 \cdot 2 = 8</math> factors, and <math>1000 = 2^3 \cdot 5^3</math> has <math>4 \cdot 4 = 16</math> factors. However, <math>n = 1</math> does not work because <math>1</math> is divisible by both <math>999</math> and <math>1000</math>, and since <math>1</math> is counted twice, the answer is <math>16 + 8 - 2 = \boxed{\textbf{(A) }22}</math>. | ||
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+ | ==Video Solution== | ||
+ | https://youtu.be/Ozp3k2464u4 | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== |
Revision as of 23:23, 31 January 2020
For how many positive integers is
not divisible by
? (Recall that
is the greatest integer less than or equal to
.)
Solution
Let . If the expression is not divisible by
, then the three terms in the expression must be
, which would imply that
is divisible by
but not
, or
, which would imply that
is divisible by
but not
.
has
factors, and
has
factors. However,
does not work because
is divisible by both
and
, and since
is counted twice, the answer is
.
Video Solution
~IceMatrix
See Also
2020 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 |
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