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Difference between revisions of "1997 AJHSME Problems/Problem 25"
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There will be <math>10</math> groups of <math>4</math> numbers.  The number now can be rewritten as <math>(2\cdot 4 \cdot 6 \cdot 8)^{10}</math>
 
There will be <math>10</math> groups of <math>4</math> numbers.  The number now can be rewritten as <math>(2\cdot 4 \cdot 6 \cdot 8)^{10}</math>
  
−
Simplifiying the inside, we get <math>(384)^{10}</math>
+
Simplifiying the ins384)^{10}<math>
  
−
Again, we can disregard the tens and hundreds digit of <math>384</math>, since we only want the units digit of the number, leaving <math>4^{10}</math>.
+
Again, we can disregard the tens and hundreds digit of </math>384<math>, since we only want the units digit of the number, leaving </math>4^{10}<math>.
  
−
Now, we try to find a pattern to the units digit of <math>4^n</math>.  To compute this quickly, we once again discard all tens digits and higher.
+
Now, we try to find a pattern to the units digit of </math>4^n<math>.  To compute this quickly, we once again discard all tens digits and higher.
  
−
<math>4^1 = 4</math>.
+
</math>4^1 = 4<math>.
  
−
<math>4^2 = 4\cdot 4 = 1\underline{6}</math>, discard the <math>1</math>.
+
</math>4^2 = 4\cdot 4 = 1\underline{6}<math>, discard the </math>1<math>.
  
−
<math>4^3 = 1 \cdot 4 = \underline{4}</math>
+
</math>4^3 = 1 \cdot 4 = \underline{4}<math>
  
−
<math>4^4 = 4 \cdot 4 = 1\underline{6}</math>, discard the <math>1</math>.
+
</math>4^4 = 4 \cdot 4 = 1\underline{6}<math>, discard the </math>1<math>.
  
−
<math>4^5 = 1 \cdot 4 = \underline{4}</math>
+
</math>4^5 = 1 \cdot 4 = \underline{4}<math>
  
−
Those equalities are, in reality, congruences <math>\mod {10}</math>.
+
Those equalities are, in reality, congruences </math>\mod {10}<math>.
  
−
Thus, the pattern of the units digits is <math>\{4, 6, 4, 6, 4, 6, 4, 6, 4, 6\}</math>.  The cycle repeats so that term <math>n</math> is the same as term <math>n+2</math>.  The tenth number in the cycle is <math>6</math>, giving an answer of <math>\boxed{D}</math>
+
Thus, the pattern of the units digits is </math>\{4, 6, 4, 6, 4, 6, 4, 6, 4, 6\}<math>.  The cycle repeats so that term </math>n<math> is the same as term </math>n+2<math>.  The tenth number in the cycle is </math>6<math>, giving an answer of </math>\boxed{D}$
  
 
==See Also==
 
==See Also==

Revision as of 18:57, 31 March 2023

Problem

All of the even numbers from 2 to 98 inclusive, excluding those ending in 0, are multiplied together. What is the rightmost digit (the units digit) of the product?

$\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$


Solution

All the tens digits of the product will be irrelevant to finding the units digit. Thus, we are searching for the units digit of $(2\cdot 4\cdot 6 \cdot 8) \cdot (2 \cdot 4 \cdot 6 \cdot 8) \cdot (2\cdot 4\cdot 6 \cdot 8) \cdot ...$

There will be $10$ groups of $4$ numbers. The number now can be rewritten as $(2\cdot 4 \cdot 6 \cdot 8)^{10}$

Simplifiying the ins384)^{10}$Again, we can disregard the tens and hundreds digit of$384$, since we only want the units digit of the number, leaving$4^{10}$.

Now, we try to find a pattern to the units digit of$ (Error compiling LaTeX. Unknown error_msg)4^n$. To compute this quickly, we once again discard all tens digits and higher.$4^1 = 4$.$4^2 = 4\cdot 4 = 1\underline{6}$, discard the$1$.$4^3 = 1 \cdot 4 = \underline{4}$$ (Error compiling LaTeX. Unknown error_msg)4^4 = 4 \cdot 4 = 1\underline{6}$, discard the$1$.$4^5 = 1 \cdot 4 = \underline{4}$Those equalities are, in reality, congruences$\mod {10}$.

Thus, the pattern of the units digits is$ (Error compiling LaTeX. Unknown error_msg)\{4, 6, 4, 6, 4, 6, 4, 6, 4, 6\}$. The cycle repeats so that term$n$is the same as term$n+2$. The tenth number in the cycle is$6$, giving an answer of$\boxed{D}$

See Also

1997 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Question
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All AJHSME/AMC 8 Problems and Solutions

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