Difference between revisions of "1961 AHSME Problems/Problem 28"
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and we can see by Fermat's Theorem that this cycle repeats with the cyclicity of <math>4</math>. | and we can see by Fermat's Theorem that this cycle repeats with the cyclicity of <math>4</math>. | ||
− | Now <math>753</math> = <math>4k</math> + <math>1</math> \Rightarrow <math>U(</math>7< | + | Now <math>753</math> = <math>4k</math> + <math>1</math> \Rightarrow <math>U(</math>7<math>^753</math>)<math> = </math>7<math>. |
− | <math> ~GEOMETRY-WIZARD | + | </math> ~GEOMETRY-WIZARD $ |
==See Also== | ==See Also== |
Revision as of 09:02, 31 December 2023
Contents
Problem 28
If is multiplied out, the units' digit in the final product is:
Solution
has a unit digit of
.
has a unit digit of
.
has a unit digit of
.
has a unit digit of
.
has a unit digit of
.
Notice that the unit digit eventually cycles to itself when the exponent is increased by . It also does not matter what the other digits are in the base because the units digit is found by multiplying by only the units digit. Since
leaves a remainder of
after being divided by
, the units digit of
is
, which is answer choice
.
SOLUTION 2
(
): If
is a prime and
is an integer prime to
then we have
.
- Let's define
(
) as units digit funtion of
.
We can clearly observe that,
(
)=
. . . . . .
(
1$$ (Error compiling LaTeX. Unknown error_msg)
and we can see by Fermat's Theorem that this cycle repeats with the cyclicity of .
Now =
+
\Rightarrow
7
)
7
~GEOMETRY-WIZARD $
See Also
1961 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |
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