Difference between revisions of "1997 PMWC Problems/Problem I2"
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==Problem== | ==Problem== | ||
| − | In the multiplication in the image, each letter and each box represent a single digit. Different letters represent different digits but a box can represent any digit. What does the five-digit number HAPPY stand for? | + | In the multiplication in the image, each letter and each box represent a single digit. Different letters represent different digits but a box can represent any digit. What does the five-digit number <math>\mathrm{HAPPY}</math> stand for? |
| − | + | <cmath> \begin{array}{c c c c c}& &\Box & 1 &\Box\\ &\times & & 9 &\Box\\ \hline &\Box &\Box & 9 &\Box\\ \Box &\Box &\Box & 7 &\\ \hline H & A & P & P & Y\end{array} </cmath> | |
==Solution== | ==Solution== | ||
| Line 30: | Line 30: | ||
20661 | 20661 | ||
| − | ==See | + | ==See Also== |
| + | {{PMWC box|year=1997|num-b=I1|num-a=I3}} | ||
[[Category:Logic Problems]] | [[Category:Logic Problems]] | ||
Latest revision as of 19:32, 10 March 2015
Problem
In the multiplication in the image, each letter and each box represent a single digit. Different letters represent different digits but a box can represent any digit. What does the five-digit number 
stand for?
![\[\begin{array}{c c c c c}& &\Box & 1 &\Box\\ &\times & & 9 &\Box\\ \hline &\Box &\Box & 9 &\Box\\ \Box &\Box &\Box & 7 &\\ \hline H & A & P & P & Y\end{array}\]](http://latex.artofproblemsolving.com/2/c/7/2c799be34e34a45250f0654f364be808f2c0e2b3.png)
Solution
Following the rules of multiplication, we see that 9 times the units digit of the three digit number ends in 7, which means that the digit must be a 3. Carrying out the multiplication, we see that the last two digits of the second product are 17, which means that the hundreds digit in the first product must be a 4. We now have
-13 *9- ___ -49- --17 _____ HA66Y
The only digit that would work as the units digit of 9- is 7. Therefore we have
-13 *97 ___ -491 --17 _____ HA661
The only multiple of 7 that is two digits and is 7 times a digit is 14. Therefore we have
213 *97 ___ 1491 1917 _____ 20661
See Also
| 1997 PMWC (Problems) | ||
| Preceded by Problem I1 |
Followed by Problem I3 | |
| I: 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 T: 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
