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 | + | Simplifiying the ins384)^{10}<math> |
− | Again, we can disregard the tens and hundreds digit of <math>384< | + | 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< | + | 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>4^1 = 4<math>. |
− | <math>4^2 = 4\cdot 4 = 1\underline{6}< | + | </math>4^2 = 4\cdot 4 = 1\underline{6}<math>, discard the </math>1<math>. |
− | <math>4^3 = 1 \cdot 4 = \underline{4}< | + | </math>4^3 = 1 \cdot 4 = \underline{4}<math> |
− | <math>4^4 = 4 \cdot 4 = 1\underline{6}< | + | </math>4^4 = 4 \cdot 4 = 1\underline{6}<math>, discard the </math>1<math>. |
− | <math>4^5 = 1 \cdot 4 = \underline{4}< | + | </math>4^5 = 1 \cdot 4 = \underline{4}<math> |
− | Those equalities are, in reality, congruences <math>\mod {10}< | + | 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\}< | + | 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 17:57, 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?
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
There will be groups of numbers. The number now can be rewritten as
Simplifiying the ins384)^{10}3844^{10}$.
Now, we try to find a pattern to the units digit of$ (Error compiling LaTeX. Unknown error_msg)4^n4^1 = 44^2 = 4\cdot 4 = 1\underline{6}14^3 = 1 \cdot 4 = \underline{4}$$ (Error compiling LaTeX. Unknown error_msg)4^4 = 4 \cdot 4 = 1\underline{6}14^5 = 1 \cdot 4 = \underline{4}\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\}nn+26\boxed{D}$
See Also
1997 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Question | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.