Difference between revisions of "1993 AJHSME Problems/Problem 22"
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<math>\text{(A)}\ 22 \qquad \text{(B)}\ 99 \qquad \text{(C)}\ 112 \qquad \text{(D)}\ 119 \qquad \text{(E)}\ 199</math> | <math>\text{(A)}\ 22 \qquad \text{(B)}\ 99 \qquad \text{(C)}\ 112 \qquad \text{(D)}\ 119 \qquad \text{(E)}\ 199</math> | ||
| + | |||
| + | ==Solution== | ||
| + | There is <math>1</math> two in the one-digit numbers. | ||
| + | |||
| + | The number of two-digit numbers with a two in the tens place is <math>10</math> and the number with a two in the ones place is <math>9</math>. Thus the digit two is used <math>10+9=19</math> times for the two digit numbers. | ||
| + | |||
| + | Now, Pat Peano only has <math>22-1-19=2</math> remaining twos. You must subtract 1 because 22 is counted twice. The last numbers with a two that he can write are <math>102</math> and <math>112</math>. He can continue numbering the last couple pages without a two until <math>120</math>, with the last number he writes being <math>\boxed{\text{(D)}\ 119}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AJHSME box|year=1993|num-b=21|num-a=23}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 20:49, 9 December 2020
Problem
Pat Peano has plenty of 0's, 1's, 3's, 4's, 5's, 6's, 7's, 8's and 9's, but he has only twenty-two 2's. How far can he number the pages of his scrapbook with these digits?
Solution
There is 
two in the one-digit numbers.
The number of two-digit numbers with a two in the tens place is 
and the number with a two in the ones place is 
. Thus the digit two is used 
times for the two digit numbers.
Now, Pat Peano only has 
remaining twos. You must subtract 1 because 22 is counted twice. The last numbers with a two that he can write are 
and 
. He can continue numbering the last couple pages without a two until 
, with the last number he writes being 
.
See Also
| 1993 AJHSME (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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
